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MEASUREMENTS
The foundation of physics rests upon physical quantities in terms of which the laws of physics are
expressed. Therefore, these quantities have to be measured accurately. Among these are mass, length,
time, velocity, force, density, temperature, electric current, and numerous others.
Physical quantities are often divided into two categories: base quantities and derived quantities.
Derived quantities are those whose definitions are based on other physical quantities. Velocity
acceleration and force etc. are usually viewed as derived quantities. Base quantities are not defined in
terms of other physical quantities. The base quantities are the minimum number of those physical
quantities in terms of which other physical quantities can be defined.
Typical examples of base quantities are length, mass and time.
Length
Length is the distance between two points. S.I. unit of the length is meter (m).
Note: Length is a scalar quantity
Length-Measurement
Length can be measured using micrometers, vernier calipers, rulers and measuring tapes.
Length to be measured
Example
Measuring instrument
Long length (Several
meters)
Length of the class room
Measuring tape
Medium length
Length of table
Meter rule
Short length
External or internal diameter of the test
tube
Vernier calliper
Very short length
Diameter of the wire
Micrometer screwgauge
Using the vernier calipers and micrometer screw gauge
The precision of length measurements may be increased by using a device that uses a sliding
vernier scale. Two such instruments that are based on a vernier scale which you will use in the
laboratory to measure lengths of objects are the vernier callipers and the micrometer screw
gauge. These instruments have a main scale (in millimetres) and a sliding or rotating vernier
scale. In figure 1 below, the vernier scale (below) is divided into 10 equal divisions and thus the
least count of the instrument is 0.1 mm. Both the main scale and the vernier scale readings are
taken into account while making a measurement. The main scale reading is the first reading on
the main scale immediately to the left of the zero of the vernier scale (3 mm), while the vernier
scale reading is the mark on the vernier scale which exactly coincides with a mark on the main
scale (0.7 mm). The reading is therefore 3.7 mm.
Figure 1 : The reading here is 3.7 mm.
Figure 1 : The reading here is 15.8 mm.
The vernier calipers
The vernier calipers found in the laboratory incorporates a main scale and a sliding vernier scale
which allows readings to the nearest 0.02 mm. This instrument may be used to measure outer
dimensions of objects (using the main jaws), inside dimensions (using the smaller jaws at the
top), and depths (using the stem).
Figure 3: The vernier calipers
To measure outer dimensions of an object, the object is placed between the jaws, which are then
moved together until they secure the object. The screw clamp may then be tightened to ensure
that the reading does not change while the scale is being read.
The first significant figures are read immediately to the left of the zero of the vernier scale and
the remaining digits are taken as the vernier scale division that lines up with any main scale
division.
Some examples:
Note that the important region of the vernier scale is enlarged in the upper right hand corner of
each figure.
Figure 4: The reading is 37.46 mm.
In figure 4 above, the first significant figures are taken as the main scale reading to the left of the
vernier zero, i.e. 37 mm. The remaining two digits are taken from the vernier scale reading that
lines up with any main scale reading, i.e. 46 on the vernier scale. Thus the reading is 37.46 mm.
Figure 5: The reading is 34.60 mm.
In figure 5 above, the first significant figures are taken as the main scale reading to the left of the
vernier zero, i.e. 34 mm. The remaining two digits are taken from the vernier scale reading that
lines up with any main scale reading, i.e. 60 on the vernier scale. Note that the zero must be
included because the scale can differentiate between fiftieths of a millimetre. Therefore the
reading is 34.60 mm.
Figure 6: The reading is 40.00 mm.
In figure 6 the zero and the ten on the vernier scale both line up with main scale readings,
therefore the reading is 40.00 cm.
Try the following for yourself.
Figure 7
Figure 8
Figure 9
The micrometer screw gauge
The micrometer screw gauge is used to measure even smaller dimensions than the vernier
callipers. The micrometer screw gauge also uses an auxiliary scale (measuring hundredths of a
millimetre) which is marked on a rotary thimble. Basically it is a screw with an accurately
constant pitch (the amount by which the thimble moves forward or backward for one complete
revolution). The micrometers in our laboratory have a pitch of 0.50 mm (two full turns are
required to close the jaws by 1.00 mm). The rotating thimble is subdivided into 50 equal
divisions. The thimble passes through a frame that carries a millimetre scale graduated to 0.5
mm. The jaws can be adjusted by rotating the thimble using the small ratchet knob. This
includes a friction �clutch� which prevents too much tension being applied. The thimble must
be rotated through two revolutions to open the jaws by 1 mm.
Figure 10: The micrometer screw gauge
In order to measure an object, the object is placed between the jaws and the thimble is rotated
using the ratchet until the object is secured. Note that the ratchet knob must be used to secure the
object firmly between the jaws, otherwise the instrument could be damaged or give an
inconsistent reading. The manufacturer recommends 3 clicks of the ratchet before taking the
reading. The lock may be used to ensure that the thimble does not rotate while you take the
reading.
The first significant figure is taken from the last graduation showing on the sleeve directly to the
left of the revolving thimble. Note that an additional half scale division (0.5 mm) must be
included if the mark below the main scale is visible between the thimble and the main scale
division on the sleeve. The remaining two significant figures (hundredths of a millimetre) are
taken directly from the thimble opposite the main scale.
Figure 11: The reading is 7.38 mm.
In figure 11 the last graduation visible to the left of the thimble is 7 mm and the thimble lines up
with the main scale at 38 hundredths of a millimetre (0.38 mm); therefore the reading is 7.38
mm.
Figure 12: The reading is 7.72 mm.
In figure 12 the last graduation visible to the left of the thimble is 7.5 mm; therefore the reading
is 7.5 mm plus the thimble reading of 0.22 mm, giving 7.72 mm.
Figure 13: The reading is 3.46 mm.
In figure 13 the main scale reading is 3 mm while the reading on the drum is 0.46 mm; therefore,
the reading is 3.46 mm.
Figure 14: The reading is 3.56 mm.
In figure 14 the 0.5 mm division is visible below the main scale; therefore the reading is 3.5 mm
+ 0.06 mm = 3.56 mm.
Try the following for yourself.
TIME
The S.I. unit of time is second.
Initially a second was taken as 1/86400 of a mean solar day of the whole year of 1900 AD. In the General
Conference on Weight and Measure in 1967 the second was redfined as "the time interval taken by
9192631770 vibrations of cesium-133 atom under specified condition".
Multiples / submultiples of time
1 day
24 hours (h)
1 hour (h)
60 minute (min)
1 minute (min)
60 second (s)
1 second (s)
1000 millisecond (ms)
1 milli second
1000 microsecond (µs)
STOPWATCH
A stopwatch is a handheld timepiece designed to measure the amount of time elapsed from a particular
time when activated to when the piece is deactivated. A large digital version of a stopwatch designed for
viewing at a distance, as in a sports stadium, is called a stopclock.
It is a specially designed watch by which time of an event can be measured accurately in minutes and
seconds.It consists of two hands , a small minutes hand and a long second hand. Both the hands are set at
zero by pressing the button "Reset". When the second handle completes one rotation equal to sixty
seconds , the minute hand advances by one division.
When we want to start the watch we press the button "start/stop" and when we we want to stop it, we
press the button again "start/stop".This makes the position of the minutes and second hand stationary and
record the time interval lapsed by noting the positions of the hands.
PARALLAX ERROR
Whenever using a metre ruler or any other similar instrument for measurement a common experimental
error, called parallax error, should be avoided.
Parallax error occurs because the markings of the scale is not physically touching the object ( there exist
some distance between markings and object ) and the marking is not viewed directly from above.
This will bring about relative movement between the object and markings on the scale when an observer’s
eye is moved from side to side.
PENDULUM
A pendulum is a weight suspended from a pivot so that it can swing freely.[1] When a pendulum is
displaced sideways from its resting equilibrium position, it is subject to a restoring force due to gravity
that will accelerate it back toward the equilibrium position. When released, the restoring force
combined with the pendulum's mass causes it to oscillate about the equilibrium position, swinging back
and forth. The time for one complete cycle, a left swing and a right swing, is called the period. A
pendulum swings with a specific period which depends (mainly) on its length.
VOLUME MEASUREMENT
Introduction: Remember that volume is the amount of space an object takes up. Liquids have
volume just like solids. However, since liquids take the shape of the container they are in, we
can use this property and place liquids into calibrated (known size) containers to read the
volume. The base unit of measurement of liquids is the LITER. The most accurate container to
measure a liquid in is a graduated cylinder. You must ALWAYS check the side of the cylinder to
determine what gradations it is marked in – sometimes it will be in single mL but other times
might be in 2 mL or even 5 mL increments.
Another aspect of water in particular, is that it clings to the side of the container. This creates a
meniscus, or lower area in the middle. This meniscus is where you read the volume of the
liquid, not along the sides!
Check your understanding
Mass
The mass of a body is the amount of matter in the body. The S.I. unit of mass is kilogram (kg).Mass can
be measured with the help of beam balance.
Weight
Weight is the force of gravity on the object. It is measured in Newtons (N). The weight of object
depends upon its mass and the gravitational field strength (gravitational acceleration). Any mass near the
surface of earth has weight due to the earth's gravitational pull.
Weight can be calculated using the equation:
Weight = mass (m) X gravitational field strength (g)
w = mg
The value of the gravitational field strength on earth is 9.81 N/kg, through we round it up to 10 N/kg or
10 m / s2 to make the calculations easier. Gravitational field is the region around earth for any body, in
which object enters will get the force of attraction towards the center of earth. If you will stand on the
moon, you will feel the gravity of the moon pulling you downward towards the center of the earth.
The gravitational field strength on the moon is very less as compared to the earth. On the moon
gravitational field strength is 1.6 N / kg or 1.6 m / s2.
Spring Balance
A spring balance can be used to determine the mass of the objects.
The upper end of the spring is hung from a hook and the spring is stretched by the weight of the pan
attached to its lower end. The scale then can be adjusted so that the pointer is aligned with the zero
mark.The balance can be graduated by placing known mass in the pan. If the spring balance shows 30 N
then the mass would be W / g = 30 / 10 = 3 kg, Where g is the acceleration due to gravity.
If this balance is moved to the moon the weight would be less and spring would not stretch so far. In fact
the pointer would indicate a weight of 4.8 N (Because the value of g on the moon is 1.6 m / s2, which is
the 1 / 6th of the gravitational acceleration of earth)
So spring balance measures the weight of the object in Newtons.
Remember: Mass of the object remains same everywhere (either on the moon or the earth)
Gravitational Acceleration
The earth exerts a constant gravitational pull on a body at any point on its surface.
This gravitational pull will give an uniform acceleration to a free falling body (the body falling freely under the
action of gravity ignoring air resistance), increasing its velocity by approximately 9.81 m/s (10m/s) every second.
Hence the value of g is approximately 9.8m/s2 (10 m/s2).
Difference between mass and weight
1.
2.
3.
4.
5.
Mass is amount of matter in an object but weight is force of gravity on the object.
Mass is measured in kilograms but weight is measured in newtons as it is a force.
Mass is a scalar quantity but weight is the vector quantity.
Mass can be measured by using level(physical) balance while weight is measured by using spring balance.
Mass remains constant everywhere but weight varies from place to place.
For example if your weight on earth is 500 N so your mass is 50 kg.
If you stand on moon still your mass would be 50 kg, but your weight would be 80N only because g = 1.6 N/kg or
1.6 m / s2 on moon.
Density
Density of a substance is defined as "the mass of substance per unit volume".
Density = Mass / Volume
The S.I unit of density is kg/m3 or g/cm3. Mass of the substance can be found by using common level
balance. Volume of a substance can either be found by calculation from linear measurements or by using
measuring cylinder.
Density of water = 1 g / cm3 = 1000 kg / m3
KINEMATICS
Describing Motion with Words
MOTION ALONG A STRAIGHT LINE
Preview Questions
What is needed to describe a position?
What quantities are the basis for describing motion?
The term position refers to the location of an object. To designate the position of an object, a
reference point and a measurement scale are needed. For example, the entrance to campus is 1.6 km from
the intersection with the traffic light. The book is 15 cm from the corner of the table(corner of the table is
the reference point). The Cartesian coordinates of the point on a graph are (x, y)= (2.0 cm, 3.0 cm).Here
the reference point is the origin of the coordinate system.
If an object changes position, we say that motion has occurred. That is, an object is in motion when it is
undergoing a continuous change in position. Consider an automobile traveling on a straight highway. The
motion of the automobile may or may not be occurring at a constant rate. In either case, the motion is
described by using the fundamental units of length and time. That length and time describe motion is
evident in running. Combining length and time to give the time rate of change of position is the basis of
describing motion in terms of speed and velocity, as discussed in the following section.
Did You Learn?
To designate a position or location, both a reference point and a measurement scale are needed.
Motion involves a continuous change of position.
Introduction to the Language of Kinematics
A typical physics course concerns itself with a variety of broad topics. One such topic is mechanics - the
study of the motion of objects. The first six units of The Physics Classroom tutorial
will involve an investigation into the physics of motion. As we focus on the language,
principles, and laws that describe and explain the motion of objects, your efforts
should center on internalizing the meaning of the information. Avoid memorizing the
information; and avoid abstracting the information from the physical world that
it describes and explains. Rather, contemplate the information, thinking about its
meaning and its applications.
Kinematics is the science of describing the motion of objects using words, diagrams, numbers, graphs,
and equations. Kinematics is a branch of mechanics. The goal of any study of kinematics is to develop
sophisticated mental models that serve to describe (and ultimately, explain) the motion of real-world
objects.
In this lesson, we will investigate the words used to describe the motion of objects. That is, we will focus
on the language of kinematics. The hope is to gain a comfortable foundation with the language that is
used throughout the study of mechanics. We will study such terms as scalars, vectors, distance,
displacement, speed, velocity and acceleration. These words are used with regularity to describe the
motion of objects. Your goal should be to become very familiar with their meaning.
Scalars and Vectors
Physics is a mathematical science. The underlying concepts and principles have a mathematical basis.
Throughout the course of our study of physics, we will encounter a variety of concepts that have a
mathematical basis associated with them. While our emphasis will often be upon the conceptual nature of
physics, we will give considerable and persistent attention to its mathematical aspect.
The motion of objects can be described by words. Even a person without a background in physics has a
collection of words that can be used to describe moving objects. Words and phrases such as going fast,
stopped, slowing down, speeding up, and turning provide a sufficient vocabulary for describing the
motion of objects. In physics, we use these words and many more. We will be expanding upon this
vocabulary list with words such as distance, displacement, speed, velocity, and acceleration. As we will
soon see, these words are associated with mathematical quantities that have strict definitions. The
mathematical quantities that are used to describe the motion of objects can be divided into two categories.
The quantity is either a vector or a scalar. These two categories can be distinguished from one another by
their distinct definitions:


Scalars are quantities that are fully described by a magnitude (or numerical value) alone.
Vectors are quantities that are fully described by both a magnitude and a direction.
The remainder of this lesson will focus on several examples of vector and scalar quantities (distance,
displacement, speed, velocity, and acceleration). As you proceed through the lesson, give careful attention
to the vector and scalar nature of each quantity. As we proceed through other units at The Physics
Classroom Tutorial and become introduced to new mathematical quantities, the discussion will often
begin by identifying the new quantity as being either a vector or a scalar.
Check Your Understanding
1. To test your understanding of this distinction, consider the following quantities listed below.
Categorize each quantity as being either a vector or a scalar. Click the button to see the answer.
Quantity
Category
a. 5 m
b. 30 m/sec, East
c. 5 mi., North
d. 20 degrees Celsius
e. 256 bytes
f. 4000 Calories
Distance and Displacement
Distance and displacement are two quantities that may seem to mean the same thing yet have distinctly
different definitions and meanings.


Distance is a scalar quantity that refers to "how much ground an object has covered" during its
motion.
Displacement is a vector quantity that refers to "how far out of place an object is"; it is the
object's overall change in position.
To test your understanding of this distinction, consider the motion depicted in the diagram below. A
physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North.
Even though the physics teacher has walked a total distance of 12 meters, her displacement is 0 meters.
During the course of her motion, she has "covered 12 meters of ground" (distance = 12 m). Yet when she
is finished walking, she is not "out of place" - i.e., there is no displacement for her motion (displacement
= 0 m). Displacement, being a vector quantity, must give attention to direction. The 4 meters east cancels
the 4 meters west; and the 2 meters south cancels the 2 meters north. Vector quantities such as
displacement are direction aware. Scalar quantities such as distance are ignorant of direction. In
determining the overall distance traveled by the physics teachers, the various directions of motion can be
ignored.
Now consider another example. The diagram below shows the position of a cross-country skier at various
times. At each of the indicated times, the skier turns around and reverses the direction of travel. In other
words, the skier moves from A to B to C to D.
Use the diagram to determine the resulting displacement and the distance traveled by the skier
during these three minutes.
As a final example, consider a football coach pacing back and forth along the sidelines. The diagram
below shows several of coach's positions at various times. At each marked position, the coach makes a
"U-turn" and moves in the opposite direction. In other words, the coach moves from position A to B to C
to D.
What is the coach's resulting displacement and distance of travel?
To understand the distinction between distance and displacement, you must know the definitions. You
must also know that a vector quantity such as displacement is direction-aware and a scalar quantity such
as distance is ignorant of direction. When an object changes its direction of motion, displacement takes
this direction change into account; heading the opposite direction effectively begins to cancel whatever
displacement there once was.
Check Your Understanding
1. What is the displacement of the cross-country team if they begin at the school, run 10 miles and finish
back at the school?
2. What is the distance and the displacement of the race car drivers in the Indy 500?
Speed and Velocity
Just as distance and displacement have distinctly different meanings (despite their similarities), so do
speed and velocity. Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be
thought of as the rate at which an object covers distance. A fast-moving object has a high speed and
covers a relatively large distance in a short amount of time. Contrast this to a slow-moving object that has
a low speed; it covers a relatively small amount of distance in the same amount of time. An object with no
movement at all has a zero speed.
Velocity is a vector quantity that refers to "the rate at which an object changes its position." Imagine a
person moving rapidly - one step forward and one step back - always returning to the original starting
position. While this might result in a frenzy of activity, it would result in a zero velocity. Because the
person always returns to the original position, the motion would never result in a change in position.
Since velocity is defined as the rate at which the position changes, this motion results in zero velocity. If a
person in motion wishes to maximize their velocity, then that person must make every effort to maximize
the amount that they are displaced from their original position. Every step must go into moving that
person further from where he or she started. For certain, the person should never change directions and
begin to return to the starting position.
Velocity is a vector quantity. As such, velocity is direction aware. When evaluating the velocity of an
object, one must keep track of direction. It would not be enough to say that an object has a velocity of 55
mi/hr. One must include direction information in order to fully describe the velocity of the object. For
instance, you must describe an object's velocity as being 55 mi/hr, east. This is one of the essential
differences between speed and velocity. Speed is a scalar quantity and does not
keep track of direction; velocity is a vector quantity and is direction aware.
The task of describing the direction of the velocity vector is easy. The direction of
the velocity vector is simply the same as the direction that an object is moving. It
would not matter whether the object is speeding up or slowing down. If an object is moving rightwards,
then its velocity is described as being rightwards. If an object is moving downwards, then its velocity is
described as being downwards. So an airplane moving towards the west with a speed of 300 mi/hr has a
velocity of 300 mi/hr, west. Note that speed has no direction (it is a scalar) and the velocity at any instant
is simply the speed value with a direction.
As an object moves, it often undergoes changes in speed. For example, during an average trip to school,
there are many changes in speed. Rather than the speed-o-meter maintaining a steady reading, the needle
constantly moves up and down to reflect the stopping and starting and the accelerating and decelerating.
One instant, the car may be moving at 50 mi/hr and another instant, it might be stopped (i.e., 0 mi/hr). Yet
during the trip to school the person might average 32 mi/hr. The average speed during an entire motion
can be thought of as the average of all speedometer readings. If the speedometer readings could be
collected at 1-second intervals (or 0.1-second intervals or ... ) and then averaged together, the average
speed could be determined. Now that would be a lot of work. And fortunately, there is a shortcut. Read
on.
Calculating Average Speed and Average Velocity
The average speed during the course of a motion is often computed using the following formula:
In contrast, the average velocity is often computed using this formula
Let's begin implementing our understanding of these formulas with the following problem:
Q: While on vacation, Lisa Carr traveled a total distance of 440 miles. Her trip took 8 hours. What
was her average speed?
To compute her average speed, we simply divide the distance of travel by the time of travel.
That was easy! Lisa Carr averaged a speed of 55 miles per hour. She may not have been traveling at a
constant speed of 55 mi/hr. She undoubtedly, was stopped at some instant in time (perhaps for a bathroom
break or for lunch) and she probably was going 65 mi/hr at other instants in time. Yet, she averaged a
speed of 55 miles per hour. The above formula represents a shortcut method of determining the average
speed of an object.
Average Speed versus Instantaneous Speed
Since a moving object often changes its speed during its motion, it is common to
distinguish between the average speed and the instantaneous speed. The distinction is
as follows.


Instantaneous Speed - the speed at any given instant in time.
Average Speed - the average of all instantaneous speeds; found simply by a
distance/time ratio.
You might think of the instantaneous speed as the speed that the speedometer reads at any given instant in
time and the average speed as the average of all the speedometer readings during the course of the trip.
Since the task of averaging speedometer readings would be quite complicated (and maybe even
dangerous), the average speed is more commonly calculated as the distance/time ratio.
Moving objects don't always travel with erratic and changing speeds. Occasionally, an object will move at
a steady rate with a constant speed. That is, the object will cover the same distance every regular interval
of time. For instance, a cross-country runner might be running with a constant speed of 6 m/s in a straight
line for several minutes. If her speed is constant, then the distance traveled every second is the same. The
runner would cover a distance of 6 meters every second. If we could measure her position (distance from
an arbitrary starting point) each second, then we would note that the position would be changing by 6
meters each second. This would be in stark contrast to an object that is changing its speed. An object with
a changing speed would be moving a different distance each second. The data tables below depict objects
with constant and changing speed.
Now let's consider the motion of that physics teacher again. The physics teacher walks 4 meters East, 2
meters South, 4 meters West, and finally 2 meters North. The entire motion lasted for 24 seconds.
Determine the average speed and the average velocity.
The physics teacher walked a distance of 12 meters in 24 seconds; thus, her average speed was 0.50 m/s.
However, since her displacement is 0 meters, her average velocity is 0 m/s. Remember that the
displacement refers to the change in position and the velocity is based upon this position change. In this
case of the teacher's motion, there is a position change of 0 meters and thus an average velocity of 0 m/s.
Here is another example similar to what was seen before in the discussion of distance and displacement.
The diagram below shows the position of a cross-country skier at various times. At each of the indicated
times, the skier turns around and reverses the direction of travel. In other words, the skier moves from A
to B to C to D.
Use the diagram to determine the average speed and the average velocity of the skier during these
three minutes. When finished, click the button to view the answer.
As a last example, consider a football coach pacing back and forth along the sidelines. The diagram below
shows several of coach's positions at various times. At each marked position, the coach makes a "U-turn"
and moves in the opposite direction. In other words, the coach moves from position A to B to C to D.
What is the coach's average speed and average velocity? When finished, click the button to view
the answer.
In conclusion, speed and velocity are kinematic quantities that have distinctly different definitions. Speed,
being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance (a
scalar quantity) per time ratio. Speed is ignorant of direction. On the other hand, velocity is a vector
quantity; it is direction-aware. Velocity is the rate at which the position changes. The average velocity is
the displacement or position change (a vector quantity) per time ratio.
Acceleration
The final mathematical quantity discussed in Lesson 1 is acceleration. An often confused quantity,
acceleration has a meaning much different than the meaning associated with it by sports announcers and
other individuals. The definition of acceleration is:

Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity.
An object is accelerating if it is changing its velocity.
Sports announcers will occasionally say that a person is accelerating if he/she is
moving fast. Yet acceleration has nothing to do with going fast. A person can be
moving very fast and still not be accelerating. Acceleration has to do with
changing how fast an object is moving. If an object is not changing its velocity,
then the object is not accelerating. The data at the right are representative of a
northward-moving accelerating object. The velocity is changing over the course
of time. In fact, the velocity is changing by a constant amount - 10 m/s - in each
second of time. Anytime an object's velocity is changing, the object is said to be
accelerating; it has an acceleration.
The Meaning of Constant Acceleration
Sometimes an accelerating object will change its velocity by the same amount each second. As mentioned
in the previous paragraph, the data table above show an object changing its velocity by 10 m/s in each
consecutive second. This is referred to as a constant acceleration since the velocity is changing by a
constant amount each second. An object with a constant acceleration should not be confused with an
object with a constant velocity. Don't be fooled! If an object is changing its velocity -whether by a
constant amount or a varying amount - then it is an accelerating object. And an object with a constant
velocity is not accelerating. The data tables below depict motions of objects with a constant acceleration
and a changing acceleration. Note that each object has a changing velocity.
Since accelerating objects are constantly changing their velocity, one can say that the distance
traveled/time is not a constant value. A falling object for instance usually accelerates as it falls. If we were
to observe the motion of a free-falling object (free fall motion will be discussed in detail later), we would
observe that the object averages a velocity of approximately 5 m/s in the first second, approximately 15
m/s in the second second, approximately 25 m/s in the third second, approximately 35 m/s in the fourth
second, etc. Our free-falling object would be constantly accelerating. Given these average velocity values
during each consecutive 1-second time interval, we could say that the object would fall 5 meters in the
first second, 15 meters in the second second (for a total distance of 20 meters), 25 meters in the third
second (for a total distance of 45 meters), 35 meters in the fourth second (for a total distance of 80 meters
after four seconds). These numbers are summarized in the table below.
Time
Interval
0-1s
1 -2 s
2-3s
3-4s
Ave. Velocity During
Time Interval
~ 5 m/s
~ 15 m/s
~ 25 m/s
~ 35 m/s
Distance Traveled During
Time Interval
~5m
~ 15 m
~ 25 m
~ 35 m
Total Distance Traveled from 0s to
End of Time Interval
~5m
~ 20 m
~ 45 m
~ 80 m
Note: The ~ symbol as used here means approximately.
This discussion illustrates that a free-falling object that is accelerating at a constant rate will cover
different distances in each consecutive second. Further analysis of the first and last columns of the data
above reveal that there is a square relationship between the total distance traveled and the time of travel
for an object starting from rest and moving with a constant acceleration. The total distance traveled is
directly proportional to the square of the time. As such, if an object travels for twice the time, it will cover
four times (2^2) the distance; the total distance traveled after two seconds is four times the total distance
traveled after one second. If an object travels for three times the time, then it will cover nine times (3^2)
the distance; the distance traveled after three seconds is nine times the distance traveled after one second.
Finally, if an object travels for four times the time, then it will cover 16 times (4^2) the distance; the
distance traveled after four seconds is 16 times the distance traveled after one second. For objects with a
constant acceleration, the distance of travel is directly proportional to the square of the time of travel.
Calculating the Average Acceleration
The average acceleration (a) of any object over a given interval of time (t) can be calculated using the
equation
This equation can be used to calculate the acceleration of the object whose motion is depicted by the
velocity-time data table above. The velocity-time data in the table shows that the object has an
acceleration of 10 m/s/s. The calculation is shown below.
Acceleration values are expressed in units of velocity/time. Typical acceleration units include the
following:
m/s/s
mi/hr/s
km/hr/s
m/s2
These units may seem a little awkward to a beginning physics
student. Yet they are very reasonable units when you begin to
consider the definition and equation for acceleration. The reason for the units becomes obvious upon
examination of the acceleration equation.
Since acceleration is a velocity change over a time, the units on acceleration are velocity units divided by
time units - thus (m/s)/s or (mi/hr)/s. The (m/s)/s unit can be mathematically simplified to m/s2.
The Direction of the Acceleration Vector
Since acceleration is a vector quantity, it has a direction associated with it. The direction of the
acceleration vector depends on two things:


whether the object is speeding up or slowing down
whether the object is moving in the + or - direction
The general RULE OF THUMB is:
If an object is slowing down, then its acceleration is in the opposite direction of
its motion.
This RULE OF THUMB can be applied to determine whether the sign of the
acceleration of an object is positive or negative, right or left, up or down, etc.
Consider the two data tables below. In each case, the acceleration of the object
is in the positive direction. In Example A, the object is moving in the positive direction (i.e., has a positive
velocity) and is speeding up. When an object is speeding up, the acceleration is in the same direction as
the velocity. Thus, this object has a positive acceleration. In Example B, the object is moving in the
negative direction (i.e., has a negative velocity) and is slowing down. According to our RULE OF
THUMB, when an object is slowing down, the acceleration is in the opposite direction as the velocity.
Thus, this object also has a positive acceleration.
This same RULE OF THUMB can be applied to the motion of the objects represented in the two data
tables below. In each case, the acceleration of the object is in the negative direction. In Example C, the
object is moving in the positive direction (i.e., has a positive velocity) and is slowing down. According to
our RULE OF THUMB, when an object is slowing down, the acceleration is in the apposite direction as
the velocity. Thus, this object has a negative acceleration. In Example D, the object is moving in the
negative direction (i.e., has a negative velocity) and is speeding up. When an object is speeding up, the
acceleration is in the same direction as the velocity. Thus, this object also has a negative acceleration.
Observe the use of positive and negative as used in the discussion above (Examples A - D). In physics,
the use of positive and negative always has a physical meaning. It is more than a mere mathematical
symbol. As used here to describe the velocity and the acceleration of a moving object, positive and
negative describe a direction. Both velocity and acceleration are vector quantities and a full description of
the quantity demands the use of a directional adjective. North, south, east, west, right, left, up and down
are all directional adjectives. Physics often borrows from mathematics and uses the + and - symbols as
directional adjectives. Consistent with the mathematical convention used on number lines and graphs,
positive often means to the right or up and negative often means to the left or down. So to say that an
object has a negative acceleration as in Examples C and D is to simply say that its acceleration is to the
left or down (or in whatever direction has been defined as negative). Negative accelerations do not refer
acceleration values that are less than 0. An acceleration of -2 m/s/s is an acceleration with a magnitude of
2 m/s/s that is directed in the negative direction.
Check Your Understanding
To test your understanding of the concept of acceleration, consider the following problems and the
corresponding solutions. Use the equation for acceleration to determine the acceleration for the following
two motions.
Describing Motion with Diagrams
Introduction to Diagrams
Throughout the course, there will be a persistent appeal to your ability to
represent physical concepts in a visual manner. You will quickly notice that
this effort to provide visual representation of physical concepts permeates
much of the discussion in The Physics Classroom Tutorial. The world that we
study in physics is a physical world - a world that we can see. And if we can
see it, we certainly ought to visualize it. And if we seek to understand it, then
that understanding ought to involve visual representations. So as you continue
your pursuit of physics understanding, always be mindful of your ability (or
lack of ability) to visually represent it. Monitor your study and learning
habits, asking if your knowledge has become abstracted to a series of
vocabulary words that have (at least in your own mind) no relation to the physical world which it seeks to
describe. Your understanding of physics should be intimately tied to the physical world as demonstrated
by your visual images.
Like the study of all of physics, our study of 1-dimensional kinematics will be concerned with the
multiple means by which the motion of objects can be represented. Such means include the use of words,
the use of graphs, the use of numbers, the use of equations, and the use of diagrams. Lesson 2 focuses on
the use of diagrams to describe motion. The two most commonly used types of diagrams used to describe
the motion of objects are:


ticker tape diagrams
vector diagrams
Begin cultivating your visualization skills early in the course. Spend some time on the rest of Lesson 2,
seeking to connect the visuals and graphics with the words and the physical reality. And as you proceed
through the remainder of the unit 1 lessons, continue to make these same connections.
Ticker Tape Diagrams
A common way of analyzing the motion of objects in physics labs is to perform a ticker tape analysis. A
long tape is attached to a moving object and threaded through a device that places a tick upon the tape at
regular intervals of time - say every 0.10 second. As the object moves, it drags the tape through the
"ticker," thus leaving a trail of dots. The trail of dots provides a history of the object's motion and
therefore a representation of the object's motion.
The distance between dots on a ticker tape represents the object's position change during that time
interval. A large distance between dots indicates that the object was moving fast during that time interval.
A small distance between dots means the object was moving slow during that time interval. Ticker tapes
for a fast- and slow-moving object are depicted below.
The analysis of a ticker tape diagram will also reveal if the object is moving with a constant velocity or
accelerating. A changing distance between dots indicates a changing velocity and thus an acceleration. A
constant distance between dots represents a constant velocity and therefore no acceleration. Ticker tapes
for objects moving with a constant velocity and with an accelerated motion are shown below.
And so ticker tape diagrams provide one more means of representing various features of the motion of
objects.
Check Your Understanding
Ticker tape diagrams are sometimes referred to as oil drop diagrams. Imagine a car with a leaky engine
that drips oil at a regular rate. As the car travels through town, it would leave a trace of oil on the street.
That trace would reveal information about the motion of the car. Renatta Oyle owns such a car and it
leaves a signature of Renatta's motion wherever she goes. Analyze the three traces of Renatta's ventures
as shown below. Assume Renatta is traveling from left to right. Describe Renatta's motion characteristics
during each section of the diagram. Click the button to check your answers.
1.
2.
3.
Vector Diagrams
Vector diagrams are diagrams that depict the direction and relative magnitude of a vector quantity by a
vector arrow. Vector diagrams can be used to describe the velocity of a moving object during its motion.
For example, a vector diagram could be used to represent the motion of a car moving down the road.
In a vector diagram, the magnitude of a vector quantity is represented by the size of the vector arrow. If
the size of the arrow in each consecutive frame of the vector diagram is the same, then the magnitude of
that vector is constant. The diagrams below depict the velocity of a car during its motion. In the top
diagram, the size of the velocity vector is constant, so the diagram is depicting a motion of constant
velocity. In the bottom diagram, the size of the velocity vector is increasing, so the diagram is depicting a
motion with increasing velocity - i.e., an acceleration.
Vector diagrams can be used to represent any vector quantity. In future studies, vector diagrams will be
used to represent a variety of physical quantities such as acceleration, force, and momentum. Be familiar
with the concept of using a vector arrow to represent the direction and relative size of a quantity. It will
become a very important representation of an object's motion as we proceed further in our studies of the
physics of motion.
Describing Motion with Position(displacement) vs. Time Graphs
Meaning of Shape for a p-t Graph | Meaning of Slope for a p-t Graph
Determining Slope on a p-t Graph
The Meaning of Shape for a p-t Graph
Our study of 1-dimensional kinematics has been concerned with the multiple means by which the motion
of objects can be represented. Such means include the use of words, the use of diagrams, the use of
numbers, the use of equations, and the use of graphs. Lesson 3 focuses on the use of position vs. time
graphs to describe motion. As we will learn, the specific features of the motion of objects are
demonstrated by the shape and the slope of the lines on a position vs. time graph. The first part of this
lesson involves a study of the relationship between the shape of a p-t graph and the motion of the object.
To begin, consider a car moving with a constant, rightward (+) velocity - say of +10 m/s.
If the position-time data for such a car were graphed, then the
resulting graph would look like the graph at the right. Note that
a motion described as a constant, positive velocity results in a
line of constant and positive slope when plotted as a positiontime graph.
Now consider a car moving with a rightward (+), changing
velocity - that is, a car that is moving rightward but speeding up
or accelerating.
If the position-time data for such a car were
graphed, then the resulting graph would look like
the graph at the right. Note that a motion described
as a changing, positive velocity results in a line of
changing and positive slope when plotted as a
position-time graph.
The position vs. time graphs for the two types of
motion - constant velocity and changing velocity
(acceleration) - are depicted as follows.
Constant Velocity
Positive Velocity
Positive Velocity
Changing Velocity (acceleration)
The Importance of Slope
The shapes of the position versus time graphs for these two basic types of motion constant velocity motion and accelerated motion (i.e., changing velocity) - reveal an
important principle. The principle is that the slope of the line on a position-time
graph reveals useful information about the velocity of the object. It is often said,
"As the slope goes, so goes the velocity." Whatever characteristics the velocity has,
the slope will exhibit the same (and vice versa). If the velocity is constant, then the
slope is constant (i.e., a straight line). If the velocity is changing, then the slope is changing (i.e., a curved
line). If the velocity is positive, then the slope is positive (i.e., moving upwards and to the right). This
very principle can be extended to any motion conceivable.
Consider the graphs below as example applications of this principle
concerning the slope of the line on a position versus time graph. The
graph on the left is representative of an object that is moving with a
positive velocity (as denoted by the positive slope), a constant
velocity (as denoted by the constant slope) and a small velocity (as
denoted by the small slope). The graph on the right has similar
features - there is a constant, positive velocity (as denoted by the
constant, positive slope). However, the slope of the graph on the right
is larger than that on the left. This larger slope is indicative of a
larger velocity. The object represented by the graph on the right is traveling faster than the object
represented by the graph on the left. The principle of slope can be used to extract relevant motion
characteristics from a position vs. time graph. As the slope goes, so goes the velocity.
Slow, Rightward(+)
Constant Velocity
Fast, Rightward(+)
Constant Velocity
Consider the graphs below as another application of this
principle of slope. The graph on the left is representative of an
object that is moving with a negative velocity (as denoted by
the negative slope), a constant velocity (as denoted by the
constant slope) and a small velocity (as denoted by the small
slope). The graph on the right has similar features - there is a
constant, negative velocity (as denoted by the constant,
negative slope). However, the slope of the graph on the right is
larger than that on the left. Once more, this larger slope is indicative of a larger velocity. The object
represented by the graph on the right is traveling faster than the object represented by the graph on the
left.
Slow, Leftward(-)
Constant Velocity
Fast, Leftward(-)
Constant Velocity
As a final application of this principle of slope, consider the two graphs below. Both graphs show plotted
points forming a curved line. Curved lines have changing slope; they may start with a very small slope
and begin curving sharply (either upwards or downwards) towards a large slope. In either case, the curved
line of changing slope is a sign of accelerated motion (i.e., changing velocity). Applying the principle of
slope to the graph on the left, one would conclude that the object depicted by the graph is moving with a
negative velocity (since the slope is negative ). Furthermore, the object is starting with a small velocity
(the slope starts out with a small slope) and finishes with a large velocity (the slope becomes large). That
would mean that this object is moving in the negative direction and speeding up (the small velocity turns
into a larger velocity). This is an example of negative acceleration - moving in the negative direction and
speeding up. The graph on the right also depicts an object with negative velocity (since there is a negative
slope). The object begins with a high velocity (the slope is initially large) and finishes with a small
velocity (since the slope becomes smaller). So this object is moving in the negative direction and slowing
down. This is an example of positive acceleration.
Negative (-) Velocity
Slow to Fast
Leftward (-) Velocity
Fast to Slow
The principle of slope is an incredibly useful principle for extracting relevant information about the
motion of objects as described by their position vs. time graph. Once you've practiced the principle a few
times, it becomes a very natural means of analyzing position-time graphs.
Check Your Understanding
Use the principle of slope to describe the motion of the objects depicted by the two plots below. In your
description, be sure to include such information as the direction of the velocity vector (i.e., positive or
negative), whether there is a constant velocity or an acceleration, and whether the object is moving slow,
fast, from slow to fast or from fast to slow. Be complete in your description.
The Meaning of Slope for a p-t Graph
As discussed in the previous part of Lesson 3, the slope of a position vs. time graph reveals pertinent
information about an object's velocity. For example, a small slope means a small velocity; a negative
slope means a negative velocity; a constant slope (straight line) means a constant velocity; a changing
slope (curved line) means a changing velocity. Thus the shape of the line on the graph (straight, curving,
steeply sloped, mildly sloped, etc.) is descriptive of the object's motion. In this part of the lesson, we will
examine how the actual slope value of any straight line on a graph is the velocity of the object.
Consider a car moving with a constant velocity of +10 m/s for 5 seconds. The diagram below depicts such
a motion.
The position-time graph would look like the graph at
the right. Note that during the first 5 seconds, the line
on the graph slopes up 10 m for every 1 second along
the horizontal (time) axis. That is, the slope of the
line is +10 meter/1 second. In this case, the slope of
the line (10 m/s) is obviously equal to the velocity of
the car. We will examine a few other graphs to see if
this a principle that is true of all position vs. time
graphs.
Now consider a car moving at a constant velocity of +5 m/s for 5 seconds, abruptly stopping, and then
remaining at rest (v = 0 m/s) for 5 seconds.
If the position-time data for such a car were graphed,
then the resulting graph would look like the graph at
the right. For the first five seconds the line on the
graph slopes up 5 meters for every 1 second along
the horizontal (time) axis. That is, the line on the
position vs. time graph has a slope of +5 meters/1
second for the first five seconds. Thus, the slope of
the line on the graph equals the velocity of the car.
During the last 5 seconds (5 to 10 seconds), the line
slopes up 0 meters. That is, the slope of the line is 0 m/s - the same as the velocity during this time
interval.
Both of these examples reveal an important principle. The principle
is that the slope of the line on a position-time graph is equal to the
velocity of the object. If the object is moving with a velocity of +4
m/s, then the slope of the line will be +4 m/s. If the object is moving
with a velocity of -8 m/s, then the slope of the line will be -8 m/s. If
the object has a velocity of 0 m/s, then the slope of the line will be 0
m/s.
Determining the Slope on a p-t Graph
It was learned earlier in Lesson 3 that the slope of the line on a position versus time graph is equal to the
velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be
+4 m/s. If the object is moving with a velocity of -8 m/s, then the slope of the line will be -8 m/s. If the
object has a velocity of 0 m/s, then the slope of the line will be 0 m/s. The slope of the line on a position
versus time graph tells it all. Because of its importance, a student of physics must have a good
understanding of how to calculate the slope of a line. In this part of the lesson, the method for determining
the slope of a line on a position-time graph will be discussed.
Let's begin by considering the position versus time graph below.
The line is sloping upwards to the right. But mathematically, by how much does it slope upwards for
every 1 second along the horizontal (time) axis? To answer this question we must use the slope equation.
The slope equation says that the slope of a line is found by determining the amount of rise of the line
between any two points divided by the amount of run of the line between the same two points. In other
words,




Pick two points on the line and determine their coordinates.
Determine the difference in y-coordinates of these two points (rise).
Determine the difference in x-coordinates for these two points (run).
Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).
The diagram below shows this method being applied to determine the slope of the line. Note that three
different calculations are performed for three different sets of two points on the line. In each case, the
result is the same: the slope is 10 m/s.
So that was easy - rise over run is all that is involved.
Now let's attempt a more difficult example. Consider the graph below. Note that the slope is not positive
but rather negative; that is, the line slopes in the downward direction. Note also that the line on the graph
does not pass through the origin. Slope calculations are relatively easy when the line passes through the
origin since one of the points is (0,0). But that is not the case here. Test your understanding of slope
calculations by determining the slope of the line below. Then click the button to check your answer.
Check Your Understanding
1. Determine the velocity (i.e., slope) of the object as portrayed by the graph below. When you believe
you know the answer (and not before), click the button to check it.
Describing Motion with Velocity vs. Time Graphs
Meaning of Shape for a v-t Graph | Meaning of Slope for a v-t Graph
Relating the Shape to the Motion | Determining Slope on a v-t Graph
Determining Area on a v-t Graph
The Meaning of Shape for a v-t Graph
Our study of 1-dimensional kinematics has been concerned with the multiple means by which the motion
of objects can be represented. Such means include the use of words, the use of diagrams, the use of
numbers, the use of equations, and the use of graphs. Lesson 4 focuses on the use of velocity versus time
graphs to describe motion. As we will learn, the specific features of the motion of objects are
demonstrated by the shape and the slope of the lines on a velocity vs. time graph. The first part of this
lesson involves a study of the relationship between the shape of a v-t graph and the motion of the object.
Consider a car moving with a constant, rightward (+) velocity - say of +10 m/s. As learned in an earlier
lesson, a car moving with a constant velocity is a car with zero acceleration.
If the velocity-time data for such a car were
graphed, then the resulting graph would look like
the graph at the right. Note that a motion described
as a constant, positive velocity results in a line of
zero slope (a horizontal line has zero slope) when
plotted as a velocity-time graph. Furthermore, only
positive velocity values are plotted, corresponding
to a motion with positive velocity.
Now consider a car moving with a rightward (+), changing velocity - that is, a car that is moving
rightward but speeding up or accelerating. Since the car is moving in the positive direction and speeding
up, the car is said to have a positive acceleration.
If the velocity-time data for such a car were
graphed, then the resulting graph would look
like the graph at the right. Note that a motion
described as a changing, positive velocity
results in a sloped line when plotted as a
velocity-time graph. The slope of the line is
positive, corresponding to the positive
acceleration. Furthermore, only positive velocity
values are plotted, corresponding to a motion
with positive velocity.
The velocity vs. time graphs for the two types of motion - constant velocity and changing velocity
(acceleration) - can be summarized as follows.
Positive Velocity
Zero Acceleration
The Importance of Slope
Positive Velocity
Positive Acceleration
The shapes of the velocity vs. time graphs for these two
basic types of motion - constant velocity motion and
accelerated motion (i.e., changing velocity) - reveal an
important principle. The principle is that the slope of the
line on a velocity-time graph reveals useful information
about the acceleration of the object. If the acceleration is
zero, then the slope is zero (i.e., a horizontal line). If the
acceleration is positive, then the slope is positive (i.e., an
upward sloping line). If the acceleration is negative, then the slope is negative (i.e., a downward sloping
line). This very principle can be extended to any conceivable motion.
The slope of a velocity-time graph reveals information about an object's acceleration. But how can one
tell whether the object is moving in the positive direction (i.e., positive velocity) or in the negative
direction (i.e., negative velocity)? And how can one tell if the object is speeding up or slowing down?
The answers to these questions hinge on one's ability to read a graph. Since the graph is a velocity-time
graph, the velocity would be positive whenever the line lies in the positive region (above the x-axis) of
the graph. Similarly, the velocity would be negative whenever the line lies in the negative region (below
the x-axis) of the graph. As learned in Lesson 1, a positive velocity means the object is moving in the
positive direction; and a negative velocity means the object is moving in the negative direction. So one
knows an object is moving in the positive direction if the line is located in the positive region of the graph
(whether it is sloping up or sloping down). And one knows that an object is moving in the negative
direction if the line is located in the negative region of the graph (whether it is sloping up or sloping
down). And finally, if a line crosses over the x-axis from the positive region to the negative region of the
graph (or vice versa), then the object has changed directions.
Now how can one tell if the object is speeding up or slowing down? Speeding up means that the
magnitude (or numerical value) of the velocity is getting large. For instance, an object with a velocity
changing from +3 m/s to + 9 m/s is speeding up. Similarly, an object with a velocity changing from -3
m/s to -9 m/s is also speeding up. In each case, the magnitude of the velocity (the number itself, not the
sign or direction) is increasing; the speed is getting bigger. Given this fact, one would believe that an
object is speeding up if the line on a velocity-time graph is changing from near the 0-velocity point to a
location further away from the 0-velocity point. That is, if the line is getting further away from the x-axis
(the 0-velocity point), then the object is speeding up. And conversely, if the line is approaching the x-axis,
then the object is slowing down.
Check Your Understanding
1. Consider the graph at the right. The object whose motion is represented by this graph is
... (include all that are true):
1.
2.
3.
4.
5.
6.
7.
8.
moving in the positive direction.
moving with a constant velocity.
moving with a negative velocity.
slowing down.
changing directions.
speeding up.
moving with a positive acceleration.
moving with a constant acceleration.
The Meaning of Slope for a v-t Graph
As discussed in the previous part of Lesson 4, the shape of a velocity versus time graph reveals pertinent
information about an object's acceleration. For example, if the acceleration is zero, then the velocity-time
graph is a horizontal line (i.e., the slope is zero). If the acceleration is positive, then the line is an upward
sloping line (i.e., the slope is positive). If the acceleration is negative, then the velocity-time graph is a
downward sloping line (i.e., the slope is negative). If the acceleration is great, then the line slopes up
steeply (i.e., the slope is great). This principle can be extended to any motion conceivable. Thus the shape
of the line on the graph (horizontal, sloped, steeply sloped, mildly sloped, etc.) is descriptive of the
object's motion. In this part of the lesson, we will examine how the actual slope value of any straight line
on a velocity-time graph is the acceleration of the object.
Consider a car moving with a constant velocity of +10 m/s. A car moving with a constant velocity has an
acceleration of 0 m/s/s.
The velocity-time data and graph would look like the graph below. Note that the line on the graph is
horizontal. That is the slope of the line is 0 m/s/s. In this case, it is obvious that the slope of the line (0
m/s/s) is the same as the acceleration (0 m/s/s) of the car.
Time
(s)
0
1
2
3
4
5
Velocity
(m/s)
10
10
10
10
10
10
So in this case, the slope of the line is equal to the acceleration of the velocity-time graph. Now we will
examine a few other graphs to see if this is a principle that is true of all velocity versus time graphs.
Now consider a car moving with a changing velocity. A car with a changing velocity will have an
acceleration.
The velocity-time data for this motion show that the car has an acceleration value of 10 m/s/s. (In Lesson
6, we will learn how to relate position-time data such as that in the diagram above to an acceleration
value.) The graph of this velocity-time data would look like the graph below. Note that the line on the
graph is diagonal - that is, it has a slope. The slope of the line can be calculated as 10 m/s/s. It is obvious
once again that the slope of the line (10 m/s/s) is the same as the acceleration (10 m/s/s) of the car.
Time
(s)
0
1
2
3
4
5
Velocity
(m/s)
0
10
20
30
40
50
In both instances above, the slope of the line was equal to the acceleration. As a last illustration, we will
examine a more complex case. Consider the motion of a car that first travels with a constant velocity (a=0
m/s/s) of 2 m/s for four seconds and then accelerates at a rate of +2 m/s/s for four seconds. That is, in the
first four seconds, the car is not changing its velocity (the velocity remains at 2 m/s) and then the car
increases its velocity by 2 m/s per second over the next four seconds. The velocity-time data and graph
are displayed below. Observe the relationship between the slope of the line during each four-second
interval and the corresponding acceleration value.
Time
(s)
0
1
2
3
4
5
6
7
8
Velocity
(m/s)
2
2
2
2
2
4
6
8
10
From 0 s to 4 s: slope = 0 m/s/s
From 4 s to 8 s: slope = 2 m/s/s
A motion such as the one above further illustrates the important principle: the slope of the line on a
velocity-time graph is equal to the acceleration of the object. This principle can be used for all velocitytime in order to determine the numerical value of the acceleration. A single example is given below in the
Check Your Understanding section.
Check Your Understanding
The velocity-time graph for a two-stage rocket is shown below. Use the graph and your understanding of
slope calculations to determine the acceleration of the rocket during the listed time intervals. When
finished, click the buttons to see the answers. (Help with Slope Calculations)
1. t = 0 - 1 second
2. t = 1 - 4 second
3. t = 4 - 12 second
Relating the Shape to the Motion
As discussed in a previous part of Lesson 4, the shape of a velocity vs. time graph reveals pertinent
information about an object's acceleration. For example, if the acceleration is zero, then the velocity-time
graph is a horizontal line - having a slope of zero. If the acceleration is positive, then the line is an upward
sloping line - having a positive slope. If the acceleration is negative, then the velocity-time graph is a
downward sloping line - having a negative slope. If the acceleration is great, then the line slopes up
steeply - having a large slope. The shape of the line on the graph (horizontal, sloped, steeply sloped,
mildly sloped, etc.) is descriptive of the object's motion. This principle can be extended to any motion
conceivable. In this part of the lesson, we will examine how the principle applies to a variety of types of
motion. In each diagram below, a short verbal description of a motion is given (e.g., "constant, rightward
velocity") and an accompanying ticker tape diagram is shown. Finally, the corresponding velocity-time
graph is sketched and an explanation is given. Near the end of this page, a few practice problems are
given.
Check Your Understanding
Describe the motion depicted by the following velocity-time graphs. In your descriptions, make reference
to the direction of motion (+ or - direction), the velocity and acceleration and any changes in speed
(speeding up or slowing down) during the various time intervals (e.g., intervals A, B, and C). When
finished, click the button to see the answers.
Determining the Slope on a v-t Graph
It was learned earlier in Lesson 4 that the slope of the line on a velocity versus time graph is equal to the
acceleration of the object. If the object is moving with an acceleration of +4 m/s/s (i.e., changing its
velocity by 4 m/s per second), then the slope of the line will be +4 m/s/s. If the object is moving with an
acceleration of -8 m/s/s, then the slope of the line will be -8 m/s/s. If the object has a velocity of 0 m/s,
then the slope of the line will be 0 m/s. Because of its importance, a student of physics must have a good
understanding of how to calculate the slope of a line. In this part of the lesson, the method for determining
the slope of a line on a velocity-time graph will be discussed.
Let's begin by considering the velocity versus time graph below.
The line is sloping upwards to the right. But mathematically, by how much does it slope upwards for
every 1 second along the horizontal (time) axis? To answer this question we must use the slope equation.
The slope equation says that the slope of a line is found by determining the amount of rise of the line
between any two points divided by the amount of run of the line between the same two points. A method
for carrying out the calculation is
1.
2.
3.
4.
Pick two points on the line and determine their coordinates.
Determine the difference in y-coordinates for these two points (rise).
Determine the difference in x-coordinates for these two points (run).
Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).
The diagram below shows this method being applied to determine the slope of the line. Note that three
different calculations are performed for three different sets of two points on the line. In each case, the
result is the same: the slope is 10 m/s/s.
Observe that regardless of which two points on the line are chosen for the slope calculation, the result
remains the same - 10 m/s/s.
Check Your Understanding
Consider the velocity-time graph below. Determine the acceleration (i.e., slope) of the object as portrayed
by the graph. Click the button to check your answer.
Determining the Area on a v-t Graph
As learned in an earlier part of this lesson, a plot of velocity-time can be used to determine the
acceleration of an object (the slope). In this part of the lesson, we will learn how a plot of velocity versus
time can also be used to determine the displacement of an object. For velocity versus time graphs, the area
bound by the line and the axes represents the displacement. The diagram below shows three different
velocity-time graphs; the shaded regions between the line and the time-axis represent the displacement
during the stated time interval.
The shaded area is representative of the displacement during from 0 seconds
to 6 seconds. This area takes on the shape of a rectangle can be calculated
using the appropriate equation.
The shaded area is representative of the displacement during from 0 seconds
to 4 seconds. This area takes on the shape of a triangle can be calculated
using the appropriate equation.
The shaded area is representative of the displacement during from 2 seconds
to 5 seconds. This area takes on the shape of a trapezoid can be calculated
using the appropriate equation.
The method used to find the area under a line on a velocity-time graph depends upon whether the section
bound by the line and the axes is a rectangle, a triangle or a trapezoid. Area formulas for each shape are
given below.
Calculating the Area of a Rectangle
Now we will look at a few example computations of the area for each of the above geometric shapes. First
consider the calculation of the area for a few rectangles. The solution for finding the area is shown for the
first example below. The shaded rectangle on the velocity-time graph has a base of 6 s and a height of 30
m/s. Since the area of a rectangle is found by using the formula A = b x h, the area is 180 m (6 s x 30
m/s). That is, the object was displaced 180 meters during the first 6 seconds of motion.
Area = b * h
Area = (6 s) * (30 m/s)
Area = 180 m
Now try the following two practice problems as a check of your understanding. Determine the
displacement (i.e., the area) of the object during the first 4 seconds (Practice A) and from 3 to 6 seconds
(Practice B).
Calculating the Area of a Triangle
Now we will look at a few example computations of the area for a few triangles. The solution for finding
the area is shown for the first example below. The shaded triangle on the velocity-time graph has a base
of 4 seconds and a height of 40 m/s. Since the area of triangle is found by using the formula A = 0.5 * b *
h, the area is (.5) * (4 s) * (40 m/s) = 80 m. That is, the object was displaced 80 meters during the four
seconds of motion.
Area = 0.5 * b * h
Area = (0.5) * (4 s) * (40 m/s)
Area = 80 m
Now try the following two practice problems as a check of your understanding. Determine the
displacement of the object during the first second (Practice A) and during the first 3 seconds (Practice B).
Calculating the Area of a Trapezoid
Finally we will look at a few example computations of the area for a few trapezoids. The solution for
finding the area is shown for the first example below. The shaded trapezoid on the velocity-time graph
has a base of 2 seconds and heights of 10 m/s (on the left side) and 30 m/s (on the right side). Since the
area of trapezoid is found by using the formula A = (0.5) * (b) * (h1 + h2), the area is 40 m [(0.5) * (2 s) *
(10 m/s + 30 m/s)]. That is, the object was displaced 40 meters during the time interval from 1 second to 3
seconds.
Area = 0.5 * b * (h1 + h2)
Area = (0.5) * (2 s) * (10 m/s + 30 m/s)
Area = 40 m
Now try the following two practice problems as a check of your understanding. Determine the
displacement of the object during the time interval from 2 to 3 seconds (Practice A) and during the first 2
seconds (Practice B).
Alternative Method for Trapezoids
An alternative means of determining the area of a trapezoid involves breaking the trapezoid into a triangle
and a rectangle. The areas of the triangle and rectangle can be computed individually; the area of the
trapezoid is then the sum of the areas of the triangle and the rectangle. This method is illustrated in the
graphic below.
Triangle: Area = (0.5) * (2 s) * (20 m/s) = 20 m
Rectangle: Area = (2 s) * (10 m/s) = 20 m
Total Area = 20 m + 20 m = 40 m
It has been learned in this lesson that the area bounded by the line and the axes of a velocity-time graph is
equal to the displacement of an object during that particular time period. The area can be identified as a
rectangle, triangle, or trapezoid. The area can be subsequently determined using the appropriate formula.
Once calculated, this area represents the displacement of the object.
Free Fall and the Acceleration of Gravity
Introduction | Acceleration of Gravity | Representing Free Fall by Graphs
How Fast? and How Far? | The Big Misconception
Introduction to Free Fall
A free falling object is an object that is falling under the sole influence of gravity. Any object that is being
acted upon only by the force of gravity is said to be in a state of free fall. There are two
important motion characteristics that are true of free-falling objects:


Free-falling objects do not encounter air resistance.
All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s
(often approximated as 10 m/s/s for back-of-the-envelope calculations)
Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker
tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at
the right depicts the acceleration of a free-falling object. The position of the object at
regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels
every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall
from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is
downward.
Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light
demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper.
The dropper drips water and the strobe illuminates the falling droplets at a regular rate - say once every
0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several
consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot
diagram shown in the graphic at the right.
The Acceleration of Gravity
It was learned in the previous part of this lesson that a free-falling object is an object that is falling under
the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth).
This numerical value for the acceleration of a free-falling object is such an important value that it is given
a special name. It is known as the acceleration of gravity - the acceleration for any object moving under
the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an
important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value
for the acceleration of gravity is most accurately known as 9.8 m/s/s. There are slight variations in this
numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will
occasionally use the approximated value of 10 m/s/s in The Physics Classroom Tutorial in order to reduce
the complexity of the many mathematical tasks that we will perform with this number. By so doing, we
will be able to better focus on the conceptual nature of physics without too much of a sacrifice in
numerical accuracy.
g = 9.8 m/s/s, downward
( ~ 10 m/s/s, downward)
Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the
ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s/s
means to change the velocity by 9.8 m/s each second.
If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then
one would note the following pattern.
Time (s)
0
1
2
3
4
5
Velocity (m/s)
0
- 9.8
- 19.6
- 29.4
- 39.2
- 49.0
Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each
consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s/s.
Another way to represent this acceleration of 9.8 m/s/s is to add
numbers to our dot diagram that we saw earlier in this lesson. The
velocity of the ball is seen to increase as depicted in the diagram at the
right. (NOTE: The diagram is not drawn to scale - in two seconds, the
object would drop considerably further than the distance from shoulder
to toes.)
Representing Free Fall by Graphs
Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects.
One such means of describing the motion of objects is through the use of graphs - position versus time
and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be
represented using these two basic types of graphs.
A position versus time graph for a free-falling object is shown below.
Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph
signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it
would be expected that its position-time graph would be curved. A further look at the position-time graph
reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since
the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small
initial slope indicates a small initial velocity and the large final slope indicates a large final velocity.
Finally, the negative slope of the line indicates a negative (i.e., downward) velocity.
A velocity versus time graph for a free-falling object is shown below.
Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a
velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an
acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be
diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as
read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the
negative direction and speeding up. An object that is moving in the negative direction and speeding up is
said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the
slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the
constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph
is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8
m/s/s in the downward direction.
How Fast? and How Far?
Free-falling objects are in a state of acceleration. Specifically, they are
accelerating at a rate of 9.8 m/s/s. This is to say that the velocity of a
free-falling object is changing by 9.8 m/s every second. If dropped from
a position of rest, the object will be traveling 9.8 m/s (approximately 10
m/s) at the end of the first second, 19.6 m/s (approximately 20 m/s) at
the end of the second second, 29.4 m/s (approximately 30 m/s) at the
end of the third second, etc. Thus, the velocity of a free-falling object
that has been dropped from a position of rest is dependent upon the time
that it has fallen. The formula for determining the velocity of a falling
object after a time of t seconds is
vf = g * t
where g is the acceleration of gravity. The value for g on Earth is 9.8
m/s/s. The above equation can be used to calculate the velocity of the
object after any given amount of time when dropped from rest. Example
calculations for the velocity of a free-falling object after six and eight
seconds are shown below.
Example Calculations:
At t = 6 s
vf = (9.8 m/s2) * (6 s) = 58.8 m/s
At t = 8 s
vf = (9.8 m/s2) * (8 s) = 78.4 m/s
The distance that a free-falling object has fallen from a position of rest is also dependent upon the time of
fall. This distance can be computed by use of a formula; the distance fallen after a time of t seconds is
given by the formula.
d = 0.5 * g * t2
where g is the acceleration of gravity (9.8 m/s/s on Earth). Example calculations for the distance fallen by
a free-falling object after one and two seconds are shown below.
Example Calculations:
At t = 1 s
d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m
At t = 2 s
d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m
At t = 5 s
d = (0.5) * (9.8 m/s2) * (5 s)2 = 123 m
(rounded from 122.5 m)
The diagram below (not drawn to scale) shows the results of several distance calculations for a freefalling object dropped from a position of rest.
The Big Misconception
Earlier in this lesson, it was stated that the acceleration of a free-falling object
(on earth) is 9.8 m/s/s. This value (known as the acceleration of gravity) is the
same for all free-falling objects regardless of how long they have been
falling, or whether they were initially dropped from rest or thrown up into the
air. Yet the questions are often asked "doesn't a more massive object
accelerate at a greater rate than a less massive object?" "Wouldn't an elephant
free-fall faster than a mouse?" This question is a reasonable inquiry that is
probably based in part upon personal observations made of falling objects in
the physical world. After all, nearly everyone has observed the difference in the rate of fall of a single
piece of paper (or similar object) and a textbook. The two objects clearly travel to the ground at different
rates - with the more massive book falling faster.
The answer to the question (doesn't a more massive object accelerate at a greater rate than a less massive
object?) is absolutely not! That is, absolutely not if we are considering the specific type of falling motion
known as free-fall. Free-fall is the motion of objects that move under the sole influence of gravity; freefalling objects do not encounter air resistance. More massive objects will only fall faster if there is an
appreciable amount of air resistance present.
The actual explanation of why all objects accelerate at the same rate involves the concepts of force and
mass. The details will be discussed in Unit 2 of The Physics Classroom. At that time, you will learn that
the acceleration of an object is directly proportional to force and inversely proportional to mass.
Increasing force tends to increase acceleration while increasing mass tends to decrease acceleration. Thus,
the greater force on more massive objects is offset by the inverse influence of greater mass. Subsequently,
all objects free fall at the same rate of acceleration, regardless of their mass.
Kinematic Equations and Problem-Solving
Kinematic Equations | Kinematic Equations and Problem Solving
Kinematic Equations and Free Fall | Sample Problems and Solutions
Kinematic Equations and Graphs
The Kinematic Equations
The goal of this first unit of The Physics Classroom has been to investigate the variety of means by which
the motion of objects can be described. The variety of representations that we have investigated includes
verbal representations, pictorial representations, numerical representations, and graphical representations
(position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to
describe and represent the motion of objects. These equations are known as kinematic equations.
There are a variety of quantities associated with the motion of objects - displacement (and distance),
velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive
information about an object's motion. For example, if a car is known to move with a constant velocity of
22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car
is fully described. And if a second car is known to accelerate from a rest position with an eastward
acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an
eastward displacement of 96 meters, then the motion of this car is fully described. These two statements
provide a complete description of the motion of an object. However, such completeness is not always
known. It is often the case that only a few parameters of an object's motion are known, while the rest are
unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22
m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the
displacement that your car would experience if you were to slam on your brakes and skid to a stop; and
you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters
can be determined using physics principles and mathematical equations (the kinematic equations).
The kinematic equations are a set of four equations that can be utilized to predict unknown information
about an object's motion if other information is known. The equations can be utilized for any motion that
can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant
acceleration motion. They can never be used over any time period during which the acceleration is
changing. Each of the kinematic equations include four variables. If the values of three of the four
variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic
equations provide a useful means of predicting information about an object's motion if other information
is known. For example, if the acceleration value and the initial and final velocity values of a skidding car
is known, then the displacement of the car and the time can be predicted using the kinematic equations.
Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of
unknown quantities for an object's motion.
The four kinematic equations that describe an object's motion are:
There are a variety of symbols used in the above equations. Each symbol has its own specific meaning.
The symbol d stands for the displacement of the object. The symbol t stands for the time for which the
object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the
velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial
velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value.
Each of these four equations appropriately describes the mathematical relationship between the
parameters of an object's motion. As such, they can be used to predict unknown information about an
object's motion if other information is known. In the next part of Lesson 6 we will investigate the process
of doing this.
Kinematic Equations and Problem-Solving
The four kinematic equations that describe the mathematical relationship between the parameters that
describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic
equations are:
In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the
time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol
v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the
velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is
the final velocity value.
In this part of Lesson 6 we will investigate the process of using the equations to determine unknown
information about an object's motion. The process involves the use of a problem-solving strategy that will
be used throughout the course. The strategy involves the following steps:
1.
2.
3.
4.
Construct an informative diagram of the physical situation.
Identify and list the given information in variable form.
Identify and list the unknown information in variable form.
Identify and list the equation that will be used to determine unknown information from known
information.
5. Substitute known values into the equation and use appropriate algebraic steps to solve for the
unknown information.
6. Check your answer to insure that it is reasonable and mathematically correct.
The use of this problem-solving strategy in the solution of the following problem is modeled in Examples
A and B below.
Example A
Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and
Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the
displacement of the car during the skidding process. (Note that the direction of the velocity and the
acceleration vectors are denoted by a + and a - sign.)
The solution to this problem begins by the construction of an informative diagram of the physical
situation. This is shown below. The second step involves the identification and listing of known
information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a
stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the
motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay
careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the
listing of the unknown (or desired) information in variable form. In this case, the problem requests
information about the displacement of the car. So d is the unknown quantity. The results of the first three
steps are shown in the table below.
Diagram:
Given:
vi = +30.0 m/s
vf = 0 m/s
a = - 8.00 m/s
Find:
d = ??
2
The next step of the strategy involves identifying a kinematic equation that would allow you to determine
the unknown quantity. There are four kinematic equations to choose from. In general, you will always
choose the equation that contains the three known and the one unknown variable. In this specific case, the
three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an
equation that has these four variables listed in it. An inspection of the four equations above reveals that
the equation on the top right contains all four variables.
Once the equation is identified and written down, the next step of the strategy involves substituting
known values into the equation and using proper algebraic steps to solve for the unknown information.
This step is shown below.
(0 m/s)2 = (30.0 m/s)2 + 2*(-8.00 m/s2)*d
0 m2/s2 = 900 m2/s2 + (-16.0 m/s2)*d
(16.0 m/s2)*d = 900 m2/s2 - 0 m2/s2
(16.0 m/s2)*d = 900 m2/s2
d = (900 m2/s2)/ (16.0 m/s2)
d = (900 m2/s2)/ (16.0 m/s2)
d = 56.3 m
The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded
to the third digit.)
The last step of the problem-solving strategy involves checking the answer to assure that it is both
reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid
from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a
football field, making this a very reasonable skidding distance. Checking for accuracy involves
substituting the calculated value back into the equation for displacement and insuring that the left side of
the equation is equal to the right side of the equation. Indeed it is!
Example B
Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a
6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period.
Once more, the solution to this problem begins by the construction of an informative diagram of the
physical situation. This is shown below. The second step of the strategy involves the identification and
listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since
Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s.
The next step of the strategy involves the listing of the unknown (or desired) information in variable form.
In this case, the problem requests information about the displacement of the car. So d is the unknown
information. The results of the first three steps are shown in the table below.
Diagram:
Given:
vi = 0 m/s
t = 4.10 s
Find:
d = ??
a = 6.00 m/s
2
The next step of the strategy involves identifying a kinematic equation that would allow you to determine
the unknown quantity. There are four kinematic equations to choose from. Again, you will always search
for an equation that contains the three known variables and the one unknown variable. In this specific
case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four
equations above reveals that the equation on the top left contains all four variables.
Once the equation is identified and written down, the next step of the strategy involves substituting
known values into the equation and using proper algebraic steps to solve for the unknown information.
This step is shown below.
d = (0 m/s)*(4.1 s) + 0.5*(6.00 m/s2)*(4.10 s)2
d = (0 m) + 0.5*(6.00 m/s2)*(16.81 s2)
d = 0 m + 50.43 m
d = 50.4 m
The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is
rounded to the third digit.)
The last step of the problem-solving strategy involves checking the answer to assure that it is both
reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s
will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which
such a car would be displaced during this time period would be approximately one-half a football field,
making this a very reasonable distance. Checking for accuracy involves substituting the calculated value
back into the equation for displacement and insuring that the left side of the equation is equal to the right
side of the equation. Indeed it is!
The two example problems above illustrate how the kinematic equations can be combined with a simple
problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three
motion parameters are known, any of the remaining values can be determined. In the next part of Lesson
6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some
practice problems and check your answer against the given solutions.
Kinematic Equations and Free Fall
As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of
gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is
said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s.
Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence
of gravity, then its acceleration value is 9.8 m/s/s.
Like any moving object, the motion of an object in free fall can be described by four kinematic equations.
The kinematic equations that describe any object's motion are:
The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the
symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands
for the initial velocity value; and the symbol vf stands for the final velocity.
There are a few conceptual characteristics of free fall motion that will be of value when using the
equations to analyze free fall motion. These concepts are described as follows:




An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward
acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic
equations is -9.8 m/s/s for any freely falling object.
If an object is merely dropped (as opposed to being thrown) from an elevated height, then the
initial velocity of the object is 0 m/s.
If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises
upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value
can be used as one of the motion parameters in the kinematic equations; for example, the final
velocity (vf) after traveling to the peak would be assigned a value of 0 m/s.
If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is
projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to
the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have
a downward velocity of -30 m/s when it returns to the same height.
These four principles and the four kinematic equations can be combined to solve problems involving the
motion of free falling objects. The two examples below illustrate application of free fall principles to
kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in
this lesson will be utilized.
Example A
Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground.
Determine the time required for the shingles to reach the ground.
The solution to this problem begins by the construction of an informative diagram of the physical
situation. This is shown below. The second step involves the identification and listing of known
information in variable form. You might note that in the statement of the problem, there is only one piece
of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m.
(The - sign indicates that the displacement is downward). The remaining information must be extracted
from the problem statement based upon your understanding of the above principles. For example, the vi
value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And
the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see
note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of
the solution involves the listing of the unknown (or desired) information in variable form. In this case, the
problem requests information about the time of fall. So t is the unknown quantity. The results of the first
three steps are shown in the table below.
Diagram:
Given:
vi = 0.0 m/s
d = -8.52 m
Find:
t = ??
a = - 9.8 m/s2
The next step involves identifying a kinematic equation that allows you to determine the unknown
quantity. There are four kinematic equations to choose from. In general, you will always choose the
equation that contains the three known and the one unknown variable. In this specific case, the three
known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that
has these four variables listed in it. An inspection of the four equations above reveals that the equation on
the top left contains all four variables.
Once the equation is identified and written down, the next step involves substituting known values into
the equation and using proper algebraic steps to solve for the unknown information. This step is shown
below.
-8.52 m = (0 m/s)*(t) + 0.5*(-9.8 m/s2)*(t)2
-8.52 m = (0 m) *(t) + (-4.9 m/s2)*(t)2
-8.52 m = (-4.9 m/s2)*(t)2
(-8.52 m)/(-4.9 m/s2) = t2
1.739 s2 = t2
t = 1.32 s
The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground.
(Note that this value is rounded to the third digit.)
The last step of the problem-solving strategy involves checking the answer to assure that it is both
reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of
approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2
seconds would be highly reasonable. The calculated time easily falls within this range of reasonability.
Checking for accuracy involves substituting the calculated value back into the equation for time and
insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!
Example B
Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s.
Determine the height to which the vase will rise above its initial height.
Once more, the solution to this problem begins by the construction of an informative diagram of the
physical situation. This is shown below. The second step involves the identification and listing of known
information in variable form. You might note that in the statement of the problem, there is only one piece
of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The
+ sign indicates that the initial velocity is an upwards velocity). The remaining information must be
extracted from the problem statement based upon your understanding of the above principles. Note that
the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see
note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the
listing of the unknown (or desired) information in variable form. In this case, the problem requests
information about the displacement of the vase (the height to which it rises above its starting height). So d
is the unknown information. The results of the first three steps are shown in the table below.
Diagram:
Given:
vi = 26.2 m/s
vf = 0 m/s
Find:
d = ??
a = -9.8 m/s2
The next step involves identifying a kinematic equation that would allow you to determine the unknown
quantity. There are four kinematic equations to choose from. Again, you will always search for an
equation that contains the three known variables and the one unknown variable. In this specific case, the
three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four
equations above reveals that the equation on the top right contains all four variables.
Once the equation is identified and written down, the next step involves substituting known values into
the equation and using proper algebraic steps to solve for the unknown information. This step is shown
below.
(0 m/s)2 = (26.2 m/s)2 + 2*(-9.8m/s2)*d
0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2)*d
(-19.6 m/s2)*d = 0 m2/s2 -686.44 m2/s2
(-19.6 m/s2)*d = -686.44 m2/s2
d = (-686.44 m2/s2)/ (-19.6 m/s2)
d = 35.0 m
The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before
reaching its peak. (Note that this value is rounded to the third digit.)
The last step of the problem-solving strategy involves checking the answer to assure that it is both
reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of
approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never
make it further than one football field in height (approximately 100 m), yet will surely make it past the
10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of
reasonability. Checking for accuracy involves substituting the calculated value back into the equation for
displacement and insuring that the left side of the equation is equal to the right side of the equation.
Indeed it is!
Kinematic equations provide a useful means of determining the value of an unknown motion parameter if
three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And
in many cases, another motion parameter can be inferred through a solid knowledge of some basic
kinematic principles. The next part of Lesson 6 provides a wealth of practice problems with answers and
solutions.
Sample Problems and Solutions
Earlier in Lesson 6, four kinematic equations were introduced and discussed. A useful problem-solving
strategy was presented for use with these equations and two examples were given that illustrated the use
of the strategy. Then, the application of the kinematic equations and the problem-solving strategy to freefall motion was discussed and illustrated. In this part of Lesson 6, several sample problems will be
presented. These problems allow any student of physics to test their understanding of the use of the four
kinematic equations to solve problems involving the one-dimensional motion of objects. You are
encouraged to read each problem and practice the use of the strategy in the solution of the problem. Then
click the button to check the answer or use the link to view the solution.
Check Your Understanding
1. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground.
Determine the distance traveled before takeoff. Ans d= 1720 m
2. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110
m. Determine the acceleration of the car. Ans a= 8.1 m/s2
3. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what
will be his final velocity and how far will he fall? Ans v = -25.5 m/s
4. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the
acceleration of the car and the distance traveled. Ans a = 11.2 m/s2 and d = 79.8 m
5. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the
moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon. Ans
t=1.29 s
6. Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered
sled is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the acceleration and what is
the distance that the sled travels? Ans a=247 m/s2 and d = 400 m
7. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m.
Determine the acceleration of the bike. Ans a = 0.712 m/s2
8. An engineer is designing the runway for an airport. Of the planes that will use the airport, the
lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s.
Assuming this minimum acceleration, what is the minimum allowed length for the runway?
Ans d = 704 m
9. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car
(assume uniform acceleration). Ans d= 28.6 m
10. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the
kangaroo. Ans u= 1.17 m/s
11. If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time
(total time to move upwards to the peak and then return to the ground)? Ans u = 5.03 m/s and
hang-time = 1.03 s
12. A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of
the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume
a uniform acceleration). Ans a =
m/s2
13. A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height
to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one-half the
total hang-time.) Ans d = 47.9 m
14. The observation deck of tall skyscraper 370 m above the street. Determine the time required for a
penny to free fall from the deck to the street below. Ans t = 8.69 s
15. A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet
penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into
the clay. (Assume a uniform acceleration.) ans a =
m/s2
16. A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped.
Determine the depth of the well. Ans = -57 m (NOTE: the - sign indicates direction)
17. It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the
Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the
Jaguar before it began to skid. Ans =47.6 m/s
18. A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the
acceleration of the plane and the time required to reach this speed. Ans = 30.8 s and a = 2.86 m/s2
19. A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration
(assume uniform) of the dragster. Ans= 15.8 m/s2
20. With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of
91.5 m (equivalent to one football field)? Assume negligible air resistance. Ans = 94.4 mi/hr
Kinematic Equations and Graphs
Lesson 4 of this unit at The Physics Classroom focused on the use of velocity-time graphs to describe the
motion of objects. In that Lesson, it was emphasized that the slope of the line on a velocity-time graph is
equal to the acceleration of the object and the area between the line and the time axis is equal to the
displacement of the object. Thus, velocity-time graphs can be used to determine numerical values and
relationships between the quantities displacement (d), velocity (v), acceleration (a) and time (t). In Lesson
6, the focus has been upon the use of four kinematic equations to describe the motion of objects and to
predict the numerical values of one of the four motion parameters - displacement (d), velocity (v),
acceleration (a) and time (t). Thus, there are now two methods available to solve problems involving the
numerical relationships between displacement, velocity, acceleration and time. In this part of Lesson 6,
we will investigate the relationships between these two methods.
Consider an object that moves with a constant velocity of +5 m/s for a time period of 5 seconds and then
accelerates to a final velocity of +15 m/s over the next 5 seconds. Such a verbal description of motion can
be represented by a velocity-time graph. The graph is shown below.
The horizontal section of the graph depicts a constant velocity motion, consistent with the verbal
description. The positively sloped (i.e., upward sloped) section of the graph depicts a positive
acceleration, consistent with the verbal description of an object moving in the positive direction and
speeding up from 5 m/s to 15 m/s. The slope of the line can be computed using the rise over run ratio.
Between 5 and 10 seconds, the line rises from 5 m/s to 15 m/s and runs from 5 s to 10 s. This is a total
rise of +10 m/s and a total run of 5 s. Thus, the slope (rise/run ratio) is (10 m/s)/(5 s) = 2 m/s 2. Using the
velocity-time graph, the acceleration of the object is determined to be 2 m/s2 during the last five seconds
of the object's motion. The displacement of the object can also be determined using the velocity-time
graph. The area between the line on the graph and the time-axis is representative of the displacement; this
area assumes the shape of a trapezoid. As discussed in Lesson 4, the area of a trapezoid can be equated to
the area of a triangle lying on top of the area of a rectangle. This is illustrated in the diagram below.
The total area is then the area of the rectangle plus the area of the triangle. The calculation of these areas
is shown below.
Rectangle
Area = base * height
Area = (10 s) * (5 m/s)
Triangle
Area = 0.5 * base * height
Area = 0.5 * (5 s) * (10 m/s)
Area = 50 m
Area = 25 m
The total area (rectangle plus triangle) is equal to 75 m. Thus the displacement of the object is 75 meters
during the 10 seconds of motion.
The above discussion illustrates how a graphical representation of an object's motion can be used to
extract numerical information about the object's acceleration and displacement. Once constructed, the
velocity-time graph can be used to determine the velocity of the object at any given instant during the 10
seconds of motion. For example, the velocity of the object at 7 seconds can be determined by reading the
y-coordinate value at the x-coordinate of 7 s. Thus, velocity-time graphs can be used to reveal (or
determine) numerical values and relationships between the quantities displacement (d), velocity (v),
acceleration (a) and time (t) for any given motion.
Now let's consider the same verbal description and the corresponding analysis using kinematic equations.
The verbal description of the motion was:
An object that moves with a constant velocity of +5 m/s for a time period of 5 seconds and then
accelerates to a final velocity of +15 m/s over the next 5 seconds
Kinematic equations can be applied to any motion for which the acceleration is constant. Since this
motion has two separate acceleration stages, any kinematic analysis requires that the motion parameters
for the first 5 seconds not be mixed with the motion parameters for the last 5 seconds. The table below
lists the given motion parameters.
t = 0 s - 5 s t = 5 s - 10 s
vi = 5 m/s
vf = 5 m/s vi = 5 m/s
vf = 15 m/s
t=5s
t=5s
a = 0 m/s2
Note that the acceleration during the first 5 seconds is listed as 0 m/s2 despite the fact that it is not
explicitly stated. The phrase constant velocity indicates a motion with a 0 acceleration. The acceleration
of the object during the last 5 seconds can be calculated using the following kinematic equation.
vf = vi + a*t
The substitution and algebra are shown here.
15 m/s = 5 m/s + a*(5 s)
15 m/s - 5 m/s = a*(5 s)
10 m/s = a*(5 s)
(10 m/s)/(5 s) = a
a = 2 m/s2
This value for the acceleration of the object during the time from 5 s to 10 s is consistent with the value
determined from the slope of the line on the velocity-time graph.
The displacement of the object during the entire 10 seconds can also be calculated using kinematic
equations. Since these 10 seconds include two distinctly different acceleration intervals, the calculations
for each interval must be done separately. This is shown below.
t=0s-5s
t = 5 s - 10 s
d = vi*t + 0.5*a*t2
d = ((vi + vf)/2)*t
d = (5 m/s)*(5 s) +0.5*(0 m/s2)*(5 s)2 d = ((5 m/s + 15 m/s)/2)*(5 s)
d = 25 m + 0 m
d = (10 m/s)*(5 s)
d = 25 m
d = 50 m
The total displacement during the first 10 seconds of motion is 75 meters, consistent with the value
determined from the area under the line on the velocity-time graph.
The analysis of this simple motion illustrates the value of these two representations of motion - velocitytime graph and kinematic equations. Each representation can be utilized to extract numerical information
about unknown motion quantities for any given motion. The examples below provide useful opportunity
for those requiring additional practice.
Check Your Understanding
1. Rennata Gas is driving through town at 25.0 m/s and begins to accelerate at a constant rate of -1.0
m/s2. Eventually Rennata comes to a complete stop.
a) Represent Rennata's accelerated motion by sketching a velocity-time graph. Use the
velocity-time graph to determine this distance.
b) Use kinematic equations to calculate the distance that Rennata travels while
decelerating.
2. Otto Emissions is driving his car at 25.0 m/s. Otto accelerates at 2.0 m/s2 for 5 seconds. Otto then
maintains a constant velocity for 10.0 more seconds.
a) Represent the 15 seconds of Otto Emission's motion by sketching a velocitytime graph. Use the graph to determine the distance that Otto traveled during the
entire 15 seconds.
b) Finally, break the motion into its two segments and use kinematic equations to
calculate the total distance traveled during the entire 15 seconds.
3. Luke Autbeloe, a human cannonball artist, is shot off the edge of a cliff with an initial upward
velocity of +40.0 m/s. Luke accelerates with a constant downward acceleration of -10.0 m/s2 (an
approximate value of the acceleration of gravity).
a) Sketch a velocity-time graph for the first 8 seconds of Luke's motion.
b) Use kinematic equations to determine the time required for Luke Autbeloe to
drop back to the original height of the cliff. Indicate this time on the graph.
4. Chuck Wagon travels with a constant velocity of 0.5 mile/minute for 10 minutes. Chuck then
decelerates at -.25 mile/min2 for 2 minutes.
a) Sketch a velocity-time graph for Chuck Wagon's motion. Use the velocity-time
graph to determine the total distance traveled by Chuck Wagon during the 12
minutes of motion.
b) b) Finally, break the motion into its two segments and use kinematic equations
to determine the total distance traveled by Chuck Wagon.
5. Vera Side is speeding down the interstate at 45.0 m/s. Vera looks ahead and observes an accident
that results in a pileup in the middle of the road. By the time Vera slams on the breaks, she is 50.0 m
from the pileup. She slows down at a rate of -10.0 m/s2.
a) Construct a velocity-time plot for Vera Side's motion. Use the plot to determine the distance that
Vera would travel prior to reaching a complete stop (if she did not collide with the pileup).
b) Use kinematic equations to determine the distance that Vera Side would travel prior to reaching a
complete stop (if she did not collide with the pileup). Will Vera hit the cars in the pileup? That is,
will Vera travel more than 50.0 meters?
Earl E. Bird travels 30.0 m/s for 10.0 seconds. He then accelerates at 3.00 m/s2 for 5.00 seconds.
a)
Construct a velocity-time graph for Earl E. Bird's motion. Use the plot to determine
the total distance traveled.
b) Divide the motion of the Earl E. Bird into the two time segments and use kinematic
equations to calculate the total displacement.
Solutions to Above Questions
1. Solution to Question 1
1. a. The velocity-time graph for the motion is:
The distance traveled can be found by a calculation of the area between the line on the
graph and the time axis.
Area = 0.5*b*h = 0.5*(25.0 s)*(25.0 m/s)
Area = 313 m
b. The distance traveled can be calculated using a kinematic equation. The solution is
shown here.
Given:
vi = 25.0 m/s
Find:
vf = 0.0 m/s
a = -1.0 m/s2
d = ??
vf2 = vi2 + 2*a*d
(0 m/s)2 = (25.0 m/s)2 + 2 * (-1.0 m/s2)*d
0.0 m2/s2 = 625.0 m2/s2 + (-2.0 m/s2)*d
0.0 m2/s2 - 625.0 m2/s2 = (-2.0 m/s2)*d
(-625.0 m2/s2)/(-2.0 m/s2) =d
313 m = d
2. Return to Question 1
3.
4.
5. Solution to Question 2
1. a. The velocity-time graph for the motion is:
The distance traveled can be found by a calculation of the area between the line on the
graph and the time axis. This area would be the area of the triangle plus the area of
rectangle 1 plus the area of rectangle 2.
Area = 0.5*btri*htri + brect1*hrect1 + brect2*hrect2
Area = 0.5*(5.0 s)*(10.0 m/s) + (5.0 s)*(25.0 m/s) + (10.0 s)*(35.0 m/s)
Area = 25 m + 125 m + 350 m
Area = 500 m
b. The distance traveled can be calculated using a kinematic equation. The solution is
shown here.
First find the d for the first 5 seconds:
Given:
vi = 25.0 m/s
Find:
d = ??
a = 2.0 m/s2
t = 5.0 s
d = vi*t + 0.5*a*t2
d = (25.0 m/s)*(5.0 s) + 0.5*(2.0 m/s2)*(5.0 s)2
d = 125 m + 25.0 m
d = 150 m
Now find the d for the last 10 seconds:
Given:
vi = 35.0 m/s
Find:
t = 10.0 s
a = 0.0 m/s2
d = ??
(Note: the velocity at the 5 second mark can be found from knowing that the car
accelerates from 25.0 m/s at +2.0 m/s2 for 5 seconds. This results in a velocity change of
a*t = 10 m/s, and thus a velocity of 35.0 m/s.)
d = vi*t + 0.5*a*t2
d = (35.0 m/s)*(10.0 s) + 0.5*(0.0 m/s2)*(10.0 s)2
d = 350 m + 0 m
d =350 m
The total distance for the 15 seconds of motion is the sum of these two distance
calculations (150 m + 350 m):
distance = 500 m
6. Return to Question 2
7.
8.
9. Solution to Question 3
1. a. The velocity-time graph for the motion is:
b. The time to rise up and fall back down to the original height is twice the time to rise up
to the peak. So the solution involves finding the time to rise up to the peak and then
doubling it.
Given:
vi = 40.0 m/s
Find:
vf = 0.0 m/s
a = -10.0 m/s2
tup = ??
2*tup = ??
vf = vi + a*tup
0 m/s = 40 m/s + (-10 m/s2)*tup
(10 m/s2)*tup = 40 m/s
tup = (40 m/s)/(10 m/s2)
tup = 4.0 s
2*tup = 8.0 s
Return to Question 3
Solution to Question 4
2. a. The velocity-time graph for the motion is:
The distance traveled can be found by a calculation of the area between the line on the
graph and the time axis. This area would be the area of the rectangle plus the area of the
triangle.
Area = brect*hrect + 0.5*btri*htri
Area = (10.0 min)*(0.50 mi/min) + 0.5*(2.0 min)*(0.50 mi/min)
Area = 5 mi + 0.5 mi
Area = 5.5 mi
b. The distance traveled can be calculated using a kinematic equation. The solution is
shown here.
First find the d for the first 10 minutes:
Given:
Find:
vi = 0.50 mi/min
a = 0.0 mi/min2
t = 10.0 min
d = ??
d = vi*t + 0.5*a*t2
d = (0.50 mi/min)*(10.0 min) + 0.5*(0.0 mi/min2)*(10.0 min)2
d = 5.0 mi + 0 mi
d = 5.0 mi
Now find the d for the last 2 minutes:
Given:
Find:
vi = 0.50 mi/min
t = 2.0 min
a = -0.25 mi/min2
d = ??
d = vi*t + 0.5*a*t2
d = (0.50 mi/min)*(2.0 min) + 0.5*(-0.25 m/s2)*(2.0 min)2
d = 1.0 mi + (-0.5 mi)
d = 0.5 mi
The total distance for the 12 minutes of motion is the sum of these two distance
calculations (5.0 mi + 0.5 mi):
distance = 5.5 mi
Solution to Question 5
3. a. The velocity-time graph for the motion is:
The distance traveled can be found by a calculation of the area between the line on the
graph and the time axis.
Area = 0.5*b*h = 0.5*(4.5 s)*(45.0 m/s)
Area = 101 m
b.
Given:
vi = 45.0 m/s
Find:
vf = 0.0 m/s
a = -10.0 m/s2
d = ??
vf2 = vi2 + 2*a*d
(0 m/s)2 = (45.0 m/s)2 + 2 * (-10.0 m/s2)*d
0.0 m2/s2 = 2025.0 m2/s2 + (-20.0 m/s2)*d
0.0 m2/s2 - 2025.0 m2/s2 = (-20.0 m/s2)*d
(-2025.0 m2/s2)/(-20.0 m/s2) =d
101 m =d
Since the accident pileup is less than 101 m from Vera, she will indeed hit the pileup
before completely stopping (unless she veers aside).
Solution to Question 6
4. a. The velocity-time graph for the motion is:
The distance traveled can be found by a calculation of the area between the line on the
graph and the time axis. This area would be the area of the triangle plus the area of
rectangle 1 plus the area of rectangle 2.
Area = 0.5*btri*htri + b1*h1 + b2*h2
Area = 0.5*(5.0 s)*(15.0 m/s) + (10.0 s)*(30.0 m/s) + (5.0 s)*(30.0 m/s)
Area = 37.5 m + 300 m + 150 m
Area = 488 m
b. The distance traveled can be calculated using a kinematic equation. The solution is
shown here.
First find the d for the first 10 seconds:
Given:
vi = 30.0 m/s
Find:
t = 10.0 s
a = 0.0 m/s2
d = ??
d = vi*t + 0.5*a*t2
d = (30.0 m/s)*(10.0 s) + 0.5*(0.0 m/s2)*(10.0 s)2
d = 300 m + 0 m
d =300 m
Now find the d for the last 5 seconds:
Given:
vi = 30.0 m/s
Find:
t = 5.0 s
a = 3.0 m/s2
d = ??
d = vi*t + 0.5*a*t2
d = (30.0 m/s)*(5.0 s) + 0.5*(3.0 m/s2)*(5.0 s)2
d = 150 m + 37.5 m
d = 187.5 m
The total distance for the 15 seconds of motion is the sum of these two distance
calculations (300 m + 187.5 m):
distance = 488 m
DYNAMICS
Newton's Laws - Lesson 1
Newton's First Law of Motion
Newton's First Law | Inertia and Mass | State of Motion | Balanced and Unbalanced Forces
Newton's First Law
In a previous chapter of study, the variety of ways by which motion can be described (words, graphs,
diagrams, numbers, etc.) was discussed. In this unit (Newton's Laws of Motion), the ways in which
motion can be explained will be discussed. Isaac Newton (a 17th century scientist) put forth a variety of
laws that explain why objects move (or don't move) as they do. These three laws have become known as
Newton's three laws of motion. The focus of Lesson 1 is Newton's first law of motion - sometimes
referred to as the law of inertia.
Newton's first law of motion is often stated as
An object at rest stays at rest and an object in motion stays in motion with the same speed and in the
same direction unless acted upon by an unbalanced force.
There are two parts to this statement - one that predicts the behavior of stationary objects and the other
that predicts the behavior of moving objects. The two parts are summarized in the following diagram.
The behavior of all objects can be described by saying that
objects tend to "keep on doing what they're doing" (unless acted
upon by an unbalanced force). If at rest, they will continue in this
same state of rest. If in motion with an eastward velocity of 5
m/s, they will continue in this same state of motion (5 m/s, East).
If in motion with a leftward velocity of 2 m/s, they will continue
in this same state of motion (2 m/s, left). The state of motion of
an object is maintained as long as the object is not acted upon by an unbalanced force. All objects resist
changes in their state of motion - they tend to "keep on doing what they're doing."
Suppose that you filled a baking dish to the rim with water and walked around an oval track making an
attempt to complete a lap in the least amount of time. The water would have a tendency to spill from the
container during specific locations on the track. In general the water spilled when:



the container was at rest and you attempted to move it
the container was in motion and you attempted to stop it
the container was moving in one direction and you
attempted to change its direction.
The water spills whenever the state of motion of the container is
changed. The water resisted this change in its own state of
motion. The water tended to "keep on doing what it was doing."
The container was moved from rest to a high speed at the starting
line; the water remained at rest and spilled onto the table. The
container was stopped near the finish line; the water kept moving and spilled over container's leading
edge. The container was forced to move in a different direction to make it around a curve; the water kept
moving in the same direction and spilled over its edge. The behavior of the water during the lap around
the track can be explained by Newton's first law of motion.
Everyday Applications of Newton's First Law
There are many applications of Newton's first law of motion. Consider some of your experiences in an
automobile. Have you ever observed the behavior of coffee in a coffee cup filled to the rim while starting
a car from rest or while bringing a car to rest from a state of motion? Coffee "keeps on doing what it is
doing." When you accelerate a car from rest, the road provides an unbalanced force on the spinning
wheels to push the car forward; yet the coffee (that was at rest) wants to stay at rest. While the car
accelerates forward, the coffee remains in the same position; subsequently, the car accelerates out from
under the coffee and the coffee spills in your lap. On the other hand, when braking from a state of motion
the coffee continues forward with the same speed and in the same direction, ultimately hitting the
windshield or the dash. Coffee in motion stays in motion.
Have you ever experienced inertia (resisting changes in your
state of motion) in an automobile while it is braking to a
stop? The force of the road on the locked wheels provides the
unbalanced force to change the car's state of motion, yet there
is no unbalanced force to change your own state of motion.
Thus, you continue in motion, sliding along the seat in
forward motion. A person in motion stays in motion with the
same speed and in the same direction ...
unless acted upon by the unbalanced
force of a seat belt. Yes! Seat belts are
used to provide safety for passengers whose
motion is governed by Newton's laws.
The seat belt provides the unbalanced force
that brings you from a state of motion
to a state of rest. Perhaps you could
speculate what would occur when no
seat belt is used.
There are many more applications of
applications are listed below. Perhaps
provide explanations for each






Newton's first law of motion. Several
you could think about the law of inertia and
application.
Blood rushes from your head
to your feet while quickly stopping when
riding on a descending
elevator.
The head of a hammer can be
tightened onto the wooden handle by
banging the bottom of the handle against a hard surface.
A brick is painlessly broken over the hand of a physics teacher by slamming it with a hammer.
(CAUTION: do not attempt this at home!)
To dislodge ketchup from the bottom of a ketchup bottle, it is often turned upside down and
thrusted downward at high speeds and then abruptly halted.
Headrests are placed in cars to prevent whiplash injuries during rear-end collisions.
While riding a skateboard (or wagon or bicycle), you fly forward off the board when hitting a
curb or rock or other object that abruptly halts the motion of the skateboard.
Try This At Home
Acquire a metal coat hanger for which you have
permission to destroy. Pull the coat hanger apart. Using
duct tape, attach two tennis balls to opposite ends of the
coat hanger as shown in the diagram at the right. Bend
the hanger so that there is a flat part that balances on the
head of a person. The ends of the hanger with the tennis
balls should hang low (below the balancing point). Place
the hanger on your head and balance it. Then quickly spin in a circle. What do the
tennis balls do?
Inertia and Mass
Newton's first law of motion states that "An object at rest stays at
rest and an object in motion stays in motion with the same speed
and in the same direction unless acted upon by an unbalanced
force." Objects tend to "keep on doing what they're doing." In
fact, it is the natural tendency of objects to resist changes in their state of motion. This tendency to resist
changes in their state of motion is described as inertia.
Inertia: the resistance an object has to a change in its state of motion.
Newton's conception of inertia stood in direct opposition to more popular conceptions about motion. The
dominant thought prior to Newton's day was that it was the natural tendency of objects to come to a rest
position. Moving objects, so it was believed, would eventually stop moving; a force was necessary to
keep an object moving. But if left to itself, a moving object would eventually come to rest and an object at
rest would stay at rest; thus, the idea that dominated people's thinking for nearly 2000 years prior to
Newton was that it was the natural tendency of all objects to assume a rest position.
Galileo and the Concept of Inertia
Galileo, a premier scientist in the seventeenth century, developed the concept of inertia. Galileo reasoned
that moving objects eventually stop because of a force called friction. In experiments using a pair of
inclined planes facing each other, Galileo observed that a ball would roll down one plane and up the
opposite plane to approximately the same height. If smoother planes were used, the ball would roll up the
opposite plane even closer to the original height. Galileo reasoned that any difference between initial and
final heights was due to the presence of friction. Galileo postulated that if friction could be entirely
eliminated, then the ball would reach exactly the same height.
Galileo further observed that regardless of the angle at which the planes were oriented, the final height
was almost always equal to the initial height. If the slope of the opposite incline were reduced, then the
ball would roll a further distance in order to reach that original height.
Galileo's reasoning continued - if the opposite incline were elevated at nearly a 0-degree angle, then the
ball would roll almost forever in an effort to reach the original height. And if the opposing incline was not
even inclined at all (that is, if it were oriented along the horizontal), then ... an object in motion would
continue in motion... .
Forces Don't Keep Objects Moving
Isaac Newton built on Galileo's thoughts about motion.
Newton's first law of motion declares that a force is not needed
to keep an object in motion. Slide a book across a table and
watch it slide to a rest position. The book in motion on the table
top does not come to a rest position because of the absence of a
force; rather it is the presence of a force - that force being the
force of friction - that brings the book to a rest position. In the
absence of a force of friction, the book would continue in motion with the same speed and direction forever! (Or at least to the end of the table top.) A force is not required to keep a moving book in motion.
In actuality, it is a force that brings the book to rest.
Mass as a Measure of the Amount of Inertia
All objects resist changes in their state of motion. All objects have
this tendency - they have inertia. But do some objects have more of
a tendency to resist changes than others? Absolutely yes! The
tendency of an object to resist changes in its state of motion varies
with mass. Mass is that quantity that is solely dependent upon the
inertia of an object. The more inertia that an object has, the more
mass that it has. A more massive object has a greater tendency to
resist changes in its state of motion.
Suppose that there are two seemingly identical bricks at rest on the
physics lecture table. Yet one brick consists of mortar and the other brick consists of Styrofoam. Without
lifting the bricks, how could you tell which brick was the Styrofoam brick? You could give the bricks an
identical push in an effort to change their state of motion. The brick that offers the least resistance is the
brick with the least inertia - and therefore the brick with the least mass (i.e., the Styrofoam brick).
A common physics demonstration relies on this principle that the more massive
the object, the more that object resist changes in its state of motion. The
demonstration goes as follows: several massive books are placed upon a
teacher's head. A wooden board is placed on top of the books and a hammer is
used to drive a nail into the board. Due to the large mass of the books, the force
of the hammer is sufficiently resisted (inertia). This is demonstrated by the fact
that the teacher does not feel the hammer blow. (Of course, this story may
explain many of the observations that you previously have made concerning
your "weird physics teacher.") A common variation of this demonstration
involves breaking a brick over the teacher's hand using the swift blow of a
hammer. The massive bricks resist the force and the hand is not hurt.
(CAUTION: do not try these demonstrations at home.)
Check Your Understanding
1. Imagine a place in the cosmos far from all gravitational and frictional
influences. Suppose that you visit that place (just suppose) and throw a rock.
The rock will
a) gradually stop.
b) continue in motion in the same direction at constant speed.
2. A 2-kg object is moving horizontally with a speed of 4 m/s. How much net force is required to keep the
object moving at this speed and in this direction?
3. Mac and Tosh are arguing in the cafeteria. Mac says that if he flings the Jell-O
with a greater speed it will have a greater inertia. Tosh argues that inertia does not
depend upon speed, but rather upon mass. Who do you agree with? Explain why.
4. Supposing you were in space in a weightless environment, would it require a
force to set an object in motion?
5. Fred spends most Sunday afternoons at rest on the sofa, watching pro football games and consuming
large quantities of food. What effect (if any) does this practice have upon his inertia? Explain.
6. Ben Tooclose is being chased through the woods by a bull moose that he was attempting to
photograph. The enormous mass of the bull moose is extremely intimidating. Yet, if Ben makes a zigzag
pattern through the woods, he will be able to use the large mass of the moose to his own advantage.
Explain this in terms of inertia and Newton's first law of motion.
7. Two bricks are resting on edge of the lab table. Shirley Sheshort stands on her toes and spots the two
bricks. She acquires an intense desire to know which of the two bricks are most massive. Since Shirley is
vertically challenged, she is unable to reach high enough and lift the bricks; she can however reach high
enough to give the bricks a push. Discuss how the process of pushing the bricks will allow Shirley to
determine which of the two bricks is most massive. What difference will Shirley observe and how can this
observation lead to the necessary conclusion?
State of Motion
Inertia is the tendency of an object to resist changes in its state of motion. But what is meant by the phrase
state of motion? The state of motion of an object is defined by its velocity - the speed with a direction.
Thus, inertia could be redefined as follows:
Inertia: tendency of an object to resist changes in its velocity.
An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero
velocity. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an
unbalanced force. An object in motion with a velocity of 2 m/s, East will (in the absence of an unbalanced
force) remain in motion with a velocity of 2 m/s, East. Such an object will not change its state of motion
(i.e., velocity) unless acted upon by an unbalanced force. Objects resist changes in their velocity.
As learned in an earlier unit, an object that is not changing its velocity is said to have an acceleration of 0
m/s/s. Thus, we could provide an alternative means of defining inertia:
Inertia: tendency of an object to resist accelerations.
Check Your Understanding
1. A group of physics teachers is taking some time off for a little putt-putt golf.
The 15th hole at the Hole-In-One Putt-Putt Golf Course has a large metal rim
that putters must use to guide their ball towards the hole. Mr. S guides a golf
ball around the metal rim When the ball leaves the rim, which path (1, 2, or 3)
will the golf ball follow?
2. A 4.0-kg object is moving across a friction-free surface with a constant
velocity of 2 m/s. Which one of the following horizontal forces is necessary to
maintain this state of motion?
a. 0 N
b. 0.5 N
e. depends on the speed.
c. 2.0 N
d. 8.0 N
Balanced and Unbalanced Forces
Newton's first law of motion has been frequently stated throughout this lesson.
An object at rest stays at rest and an object in motion stays in motion with the same speed and in the
same direction unless acted upon by an unbalanced force.
But what exactly is meant by the phrase unbalanced force? What is an unbalanced force? In pursuit of an
answer, we will first consider a physics book at rest on a tabletop. There are two forces acting upon the
book. One force - the Earth's gravitational pull - exerts a downward force. The other force - the push of
the table on the book (sometimes referred to as a normal force) - pushes upward on the book.
Since these two forces are of equal magnitude and in opposite directions, they balance each other. The
book is said to be at equilibrium. There is no unbalanced force acting upon the book and thus the book
maintains its state of motion. When all the forces acting upon an object balance each other, the object will
be at equilibrium; it will not accelerate. (Note: diagrams such as the one above are known as free-body
diagrams and will be discussed in detail in Lesson 2.)
Consider another example involving balanced forces - a person standing upon the ground. There are two
forces acting upon the person. The force of gravity exerts a downward force. The floor of the floor exerts
an upward force.
Since these two forces are of equal magnitude and in opposite directions, they balance each other. The
person is at equilibrium. There is no unbalanced force acting upon the person and thus the person
maintains its state of motion. (Note: diagrams such as the one above are known as free-body diagrams and
will be discussed in detail in Lesson 2.)
Now consider a book sliding from left to right across a tabletop. Sometime in the prior history of the
book, it may have been given a shove and set in motion from a rest position. Or perhaps it acquired its
motion by sliding down an incline from an elevated position. Whatever the case, our focus is not upon the
history of the book but rather upon the current situation of a book sliding to the right across a tabletop.
The book is in motion and at the moment there is no one pushing it to the right. (Remember: a force is not
needed to keep a moving object moving to the right.) The forces acting upon the book are shown below.
The force of gravity pulling downward and the force of the table pushing upwards on the book are of
equal magnitude and opposite directions. These two forces balance each other. Yet there is no force
present to balance the force of friction. As the book moves to the right, friction acts to the left to slow the
book down. There is an unbalanced force; and as such, the book changes its state of motion. The book is
not at equilibrium and subsequently accelerates. Unbalanced forces cause accelerations. In this case, the
unbalanced force is directed opposite the book's motion and will cause it to slow down. (Note: diagrams
such as the one above are known as free-body diagrams and will be discussed in detail in Lesson 2.)
To determine if the forces acting upon an object are balanced or unbalanced, an analysis must first be
conducted to determine what forces are acting upon the object and in what direction. If two individual
forces are of equal magnitude and opposite direction, then the forces are said to be balanced. An object is
said to be acted upon by an unbalanced force only when there is an individual force that is not being
balanced by a force of equal magnitude and in the opposite direction. Such analyses are discussed in
Lesson 2 of this unit and applied in Lesson 3.
Check Your Understanding
Luke Autbeloe drops an approximately 5.0 kg fat cat (weight = 50.0 N) off the roof of his house into the
swimming pool below. Upon encountering the pool, the cat encounters a 50.0 N upward resistance force
(assumed to be constant). Use this description to answer the following questions. Click the button to view
the correct answers.
1. Which one of the velocity-time graphs best describes the motion of the cat? Support your answer with
sound reasoning.
2. Which one of the following dot diagrams best describes the motion of the falling cat from the time that
they are dropped to the time that they hit the bottom of the pool? The arrows on the diagram represent the
point at which the cat hits the water. Support your answer with sound reasoning.
3. Several of Luke's friends were watching the motion of the falling cat. Being "physics types", they
began discussing the motion and made the following comments. Indicate whether each of the comments is
correct or incorrect? Support your answers.
a. Once the cat hits the water, the forces are balanced and the cat will stop.
b. Upon hitting the water, the cat will accelerate upwards because the water applies an upward force.
c. Upon hitting the water, the cat will bounce upwards due to the upward force.
4. If the forces acting upon an object are balanced, then the object
a. must not be moving.
b. must be moving with a constant velocity.
c. must not be accelerating.
d. none of these
Force and Its Representation
The Meaning of Force | Types of Forces | Drawing Free-Body Diagrams
Determining the Net Force
The Meaning of Force
A force is a push or pull upon an object resulting from the object's interaction with another object.
Whenever there is an interaction between two objects, there is a force upon each of the objects. When the
interaction ceases, the two objects no longer experience the force. Forces only exist as a result of an
interaction.
For simplicity sake, all forces (interactions) between objects can be placed into two broad categories:


contact forces, and
forces resulting from action-at-a-distance
Contact forces are those types of forces that result when the two interacting objects are perceived to be
physically contacting each other. Examples of contact forces include frictional forces, tensional forces,
normal forces, air resistance forces, and applied forces. These specific forces will be discussed in more
detail later in Lesson 2 as well as in other lessons.
Action-at-a-distance forces are those types of forces that result even when the two interacting objects are
not in physical contact with each other, yet are able to exert a push or pull despite their physical
separation. Examples of action-at-a-distance forces include gravitational forces. For example, the sun and
planets exert a gravitational pull on each other despite their large spatial separation. Even when your feet
leave the earth and you are no longer in physical contact with the earth, there is a gravitational pull
between you and the Earth. Electric forces are action-at-a-distance forces. For example, the protons in the
nucleus of an atom and the electrons outside the nucleus experience an electrical pull towards each other
despite their small spatial separation. And magnetic forces are action-at-a-distance forces. For example,
two magnets can exert a magnetic pull on each other even when separated by a distance of a few
centimeters. These specific forces will be discussed in more detail later in Lesson 2 as well as in other
lessons.
Force is a quantity that is measured using the standard metric unit known as the Newton. A Newton is
abbreviated by an "N." To say "10.0 N" means 10.0 Newton of force. One Newton is the amount of force
required to give a 1-kg mass an acceleration of 1 m/s/s. Thus, the following unit equivalency can be
stated:
A force is a vector quantity. As learned in an earlier unit, a vector quantity is a quantity that has both
magnitude and direction. To fully describe the force acting upon an object, you must describe both the
magnitude (size or numerical value) and the direction. Thus, 10 Newton is not a full description of the
force acting upon an object. In contrast, 10 Newton, downward is a complete description of the force
acting upon an object; both the magnitude (10 Newton) and the direction (downward) are given.
Because a force is a vector that has a direction, it is common to represent
forces using diagrams in which a force is represented by an arrow. Such
vector diagrams were introduced in an earlier unit and are used throughout
the study of physics. The size of the arrow is reflective of the magnitude of
the force and the direction of the arrow reveals the direction that the force is
acting. (Such diagrams are known as free-body diagrams and are discussed
later in this lesson.) Furthermore, because forces are vectors, the effect of an
individual force upon an object is often canceled by the effect of another
force. For example, the effect of a 20-Newton upward force acting upon a
book is canceled by the effect of a 20-Newton downward force acting upon the book. In such instances, it
is said that the two individual forces balance each other; there would be no unbalanced force acting upon
the book
Other situations could be imagined in which two of the individual vector
forces cancel each other ("balance"), yet a third individual force exists
that is not balanced by another force. For example, imagine a book
sliding across the rough surface of a table from left to right. The downward force of gravity and the
upward force of the table supporting the book act in opposite directions and thus balance each other.
However, the force of friction acts leftwards, and there is no rightward force to balance it. In this case, an
unbalanced force acts upon the book to change its state of motion.
The exact details of drawing free-body diagrams are discussed later. For now, the emphasis is upon the
fact that a force is a vector quantity that has a direction. The importance of this fact will become clear as
we analyze the individual forces acting upon an object later in this lesson.
Force and Its Representation
The Meaning of Force | Types of Forces | Drawing Free-Body Diagrams
Determining the Net Force
Types of Forces
A force is a push or pull acting upon an object as a result of its interaction with another object. There are a
variety of types of forces. Previously in this lesson, a variety of force types were placed into two broad
category headings on the basis of whether the force resulted from the contact or non-contact of the two
interacting objects.
Contact Forces
Frictional Force
Action-at-a-Distance Forces
Gravitational Force
Tension Force
Electrical Force
Normal Force
Magnetic Force
Air Resistance Force
Applied Force
Spring Force
These types of individual forces will now be discussed in more detail. To read about each force listed
above, continue scrolling through this page. Or to read about an individual force, click on its name from
the list below
Type of Force
Description of Force
(and Symbol)
Fapp
An applied force is a force that is applied to an object by a person or
another object. If a person is pushing a desk across the room, then there
is an applied force acting upon the object. The applied force is the force
exerted on the desk by the person.
Gravity Force
The force of gravity is the force with which the earth, moon, or other
massively large object attracts another object towards itself. By
definition, this is the weight of the object. All objects upon earth
experience a force of gravity that is directed "downward" towards the
center of the earth. The force of gravity on earth is always equal to the
weight of the object as found by the equation:
Applied Force
(also known as Weight)
Fgrav = m * g
Fgrav
where g = 9.8 N/kg (on Earth)
and m = mass (in kg)
Normal Force
Fnorm
Friction Force
Ffrict
The normal force is the support force exerted upon an object that is in
contact with another stable object. For example, if a book is resting upon
a surface, then the surface is exerting an upward force upon the book in
order to support the weight of the book. On occasions, a normal force is
exerted horizontally between two objects that are in contact with each
other. For instance, if a person leans against a wall, the wall pushes
horizontally on the person.
The friction force is the force exerted by a surface as an object moves
across it or makes an effort to move across it. There are at least two
types of friction force - sliding and static friction. Thought it is not
always the case, the friction force often opposes the motion of an object.
For example, if a book slides across the surface of a desk, then the desk
exerts a friction force in the opposite direction of its motion. Friction
results from the two surfaces being pressed together closely, causing
intermolecular attractive forces between molecules of different surfaces.
As such, friction depends upon the nature of the two surfaces and upon
the degree to which they are pressed together. The maximum amount of
friction force that a surface can exert upon an object can be calculated
using the formula below:
Ffrict = µ • Fnorm
The friction force is discussed in more detail later on this page.
Air Resistance Force
Fair
The air resistance is a special type of frictional force that acts upon
objects as they travel through the air. The force of air resistance is often
observed to oppose the motion of an object. This force will frequently be
neglected due to its negligible magnitude (and due to the fact that it is
mathematically difficult to predict its value). It is most noticeable for
objects that travel at high speeds (e.g., a skydiver or a downhill skier) or
for objects with large surface areas. Air resistance will be discussed in
more detail in Lesson 3.
Tension Force
Ftens
Spring Force
Fspring
The tension force is the force that is transmitted through a string, rope,
cable or wire when it is pulled tight by forces acting from opposite ends.
The tension force is directed along the length of the wire and pulls
equally on the objects on the opposite ends of the wire.
The spring force is the force exerted by a compressed or stretched spring
upon any object that is attached to it. An object that compresses or
stretches a spring is always acted upon by a force that restores the object
to its rest or equilibrium position. For most springs (specifically, for
those that are said to obey "Hooke's Law"), the magnitude of the force is
directly proportional to the amount of stretch or compression of the
spring.
Confusion of Mass and Weight
A few further comments should be added about the single force that is a source of much confusion to
many students of physics - the force of gravity. As mentioned above, the force of gravity acting upon an
object is sometimes referred to as the weight of the object. Many students of physics confuse weight with
mass. The mass of an object refers to the amount of matter that is contained by the object; the weight of
an object is the force of gravity acting upon that object. Mass is related to how much stuff is there and
weight is related to the pull of the Earth (or any other planet) upon that stuff. The mass of an object
(measured in kg) will be the same no matter where in the universe that object is located. Mass is never
altered by location, the pull of gravity, speed or even the existence of other forces. For example, a 2-kg
object will have a mass of 2 kg whether it is located on Earth, the moon, or Jupiter; its mass will be 2 kg
whether it is moving or not (at least for purposes of our study); and its mass will be 2 kg whether it is
being pushed upon or not.
On the other hand, the weight of an object (measured in Newton) will vary according to where in the
universe the object is. Weight depends upon which planet is exerting the force and the distance the object
is from the planet. Weight, being equivalent to the force of gravity, is dependent upon the value of g - the
gravitational field strength. On earth's surface g is 9.8 N/kg (often approximated as 10 N/kg). On the
moon's surface, g is 1.7 N/kg. Go to another planet, and there will be another g value. Furthermore, the g
value is inversely proportional to the distance from the center of the planet. So if we were to measure g at
a distance of 400 km above the earth's surface, then we would find the g value to be less than 9.8 N/kg.
(The nature of the force of gravity will be discussed in more detail in a later unit of The Physics
Classroom.) Always be cautious of the distinction between mass and weight. It is the source of much
confusion for many students of physics.
Sliding versus Static Friction
As mentioned above, the friction force is the force exerted by a surface as an object moves across it or
makes an effort to move across it. For the purpose of our study of physics at The Physics Classroom,
there are two types of friction force - static friction and sliding friction. Sliding friction results when an
object slides across a surface. As an example, consider pushing a box across a floor. The floor surface
offers resistance to the movement of the box. We often say that the floor exerts a friction force upon the
box. This is an example of a sliding friction force since it results from the sliding motion of the box. If a
car slams on its brakes and skids to a stop (without antilock brakes), there is a sliding friction force
exerted upon the car tires by the roadway surface. This friction force is also a sliding friction force
because the car is sliding across the road surface. Sliding friction forces can be calculated from
knowledge of the coefficient of friction and the normal force exerted upon the object by the surface it is
sliding across. The formula is:
Sliding Ffrict = μ • Fnorm
The symbol
represents the coefficient of sliding friction between the two surfaces. The
coefficient value is dependent primarily upon the nature of the surfaces that are in contact with each other.
For most surface combinations, the friction coefficients show little dependence upon other variables such
as area of contact, temperature, etc. Values of
have been experimentally determined for a variety
of surface combinations and are often tabulated in technical manuals and handbooks. The values of μ
provide a measure of the relative amount of adhesion or attraction of the two surfaces for each other. The
more that surface molecules tend to adhere to each other, the greater the coefficient values and the greater
the friction force.
Friction forces can also exist when the two surfaces are not sliding across each other. Such friction forces
are referred to as static friction. Static friction results when the surfaces of two objects are at rest relative
to one another and a force exists on one of the objects to set it into motion relative to the other object.
Suppose you were to push with 5-Newton of force on a large box to move it across the floor. The box
might remain in place. A static friction force exists between the surfaces of the floor and the box to
prevent the box from being set into motion. The static friction force balances the force that you exert on
the box such that the stationary box remains at rest. When exerting 5 Newton of applied force on the box,
the static friction force has a magnitude of 5 Newton. Suppose that you were to push with 25 Newton of
force on the large box and the box were to still remain in place. Static friction now has a magnitude of 25
Newton. Then suppose that you were to increase the force to 26 Newton and the box finally budged from
its resting position and was set into motion across the floor. The box-floor surfaces were able to provide
up to 25 Newton of static friction force to match your applied force. Yet the two surfaces were not able to
provide 26 Newton of static friction force. The amount of static friction resulting from the adhesion of
any two surfaces has an upper limit. In this case, the static friction force spans the range from 0 Newton
(if there is no force upon the box) to 25 Newton (if you push on the box with 25 Newton of force). This
relationship is often expressed as follows:
Ffrict-static ≤ μfrict-static• Fnorm
The symbol μfrict-static represents the coefficient of static friction between the two surfaces. Like the
coefficient of sliding friction, this coefficient is dependent upon the types of surfaces that are attempting
to move across each other. In general, values of static friction coefficients are greater than the values of
sliding friction coefficients for the same two surfaces. Thus, it typically takes more force to budge an
object into motion than it does to maintain the motion once it has been started.
The meaning of each of these forces listed in the table above will have to be thoroughly understood to be
successful during this unit. Ultimately, you must be able to read a verbal description of a physical
situation and know enough about these forces to recognize their presence (or absence) and to construct a
free-body diagram that illustrates their relative magnitude and direction.
Check Your Understanding
1. Complete the following table showing the relationship between mass and weight.
3. When a person diets, is their goal to lose mass or to lose weight? Explain.
Drawing Free-Body Diagrams
Free-body diagrams are diagrams used to show the relative magnitude
and direction of all forces acting upon an object in a given situation. A
free-body diagram is a special example of the vector diagrams that were
discussed in an earlier unit. These diagrams will be used throughout our
study of physics. The size of the arrow in a free-body diagram reflects
the magnitude of the force. The direction of the arrow shows the
direction that the force is acting. Each force arrow in the diagram is
labeled to indicate the exact type of force. It is generally customary in a
free-body diagram to represent the object by a box and to draw the
force arrow from the center of the box outward in the direction that the
force is acting. An example of a free-body diagram is shown at the
right.
The free-body diagram above depicts four forces acting upon the object. Objects do not necessarily
always have four forces acting upon them. There will be cases in which the number of forces depicted by
a free-body diagram will be one, two, or three. There is no hard and fast rule about the number of forces
that must be drawn in a free-body diagram. The only rule for drawing free-body diagrams is to depict all
the forces that exist for that object in the given situation. Thus, to construct free-body diagrams, it is
extremely important to know the various types of forces. If given a description of a physical situation,
begin by using your understanding of the force types to identify which forces are present. Then determine
the direction in which each force is acting. Finally, draw a box and add arrows for each existing force in
the appropriate direction; label each force arrow according to its type. If necessary, refer to the list of
forces and their description in order to understand the various force types and their appropriate symbols
Newton's Second Law of Motion
Newton's Second Law | The Big Misconception | Finding Acceleration
Finding Individual Forces | Free Fall and Air Resistance | Double Trouble
Newton's Second Law
Newton's first law of motion predicts the behavior of objects for which all existing forces are balanced.
The first law - sometimes referred to as the law of inertia - states that if the forces acting upon an object
are balanced, then the acceleration of that object will be 0 m/s/s. Objects at equilibrium (the condition in
which all forces balance) will not accelerate. According to Newton, an object will only accelerate if there
is a net or unbalanced force acting upon it. The presence of an unbalanced force will accelerate an object changing its speed, its direction, or both its speed and direction.
Newton's second law of motion pertains to the behavior of objects for which all existing forces are not
balanced. The second law states that the acceleration of an object is dependent upon two variables - the
net force acting upon the object and the mass of the object. The acceleration of an object depends directly
upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting
upon an object is increased, the acceleration of the object is increased. As the mass of an object is
increased, the acceleration of the object is decreased.
Newton's second law of motion can be formally stated as follows:
The acceleration of an object as produced by a net force is directly proportional to the magnitude of the
net force, in the same direction as the net force, and inversely proportional to the mass of the object.
This verbal statement can be expressed in equation form as follows:
a = Fnet / m
The above equation is often rearranged to a more familiar form as shown below. The net force is equated
to the product of the mass times the acceleration.
Fnet = m * a
In this entire discussion, the emphasis has been on the net force. The acceleration is
directly proportional to the net force; the net force equals mass times acceleration; the
acceleration in the same direction as the net force; an acceleration is produced by a net
force. The NET FORCE. It is important to remember this distinction. Do not use the
value of merely "any 'ole force" in the above equation. It is the net force that is related
to acceleration. As discussed in an earlier lesson, the net force is the vector sum of all
the forces. If all the individual forces acting upon an object are known, then the net
force can be determined. If necessary, review this principle by returning to the practice
questions in Lesson 2.
Consistent with the above equation, a unit of force is equal to a unit of mass times a unit of acceleration.
By substituting standard metric units for force, mass, and acceleration into the above equation, the
following unit equivalency can be written.
The definition of the standard metric unit of force is stated by the above equation. One Newton is defined
as the amount of force required to give a 1-kg mass an acceleration of 1 m/s/s.
The Fnet = m • a equation is often used in algebraic problem solving. The table below can be filled
by substituting into the equation and solving for the unknown quantity. Try it yourself and then use the
click on the buttons to view the answers.
Net Force
Mass
Acceleration
(N)
(kg)
(m/s/s)
1.
10
2
2.
20
2
The numerical information in the table above demonstrates some important qualitative relationships
between force, mass, and acceleration. Comparing the values in rows 1 and 2, it can be seen that a
doubling of the net force results in a doubling of the acceleration (if mass is held constant). Similarly,
comparing the values in rows 2 and 4 demonstrates that a halving of the net force results in a halving of
the acceleration (if mass is held constant). Acceleration is directly proportional to net force.
Furthermore, the qualitative relationship between mass and acceleration can be seen by a comparison of
the numerical values in the above table. Observe from rows 2 and 3 that a doubling of the mass results in
a halving of the acceleration (if force is held constant). And similarly, rows 4 and 5 show that a halving of
the mass results in a doubling of the acceleration (if force is held constant). Acceleration is inversely
proportional to mass.
The analysis of the table data illustrates that an equation such as Fnet = m*a can be a guide to thinking
about how a variation in one quantity might effect another quantity. Whatever alteration is made of the
net force, the same change will occur with the acceleration. Double, triple or quadruple the net force, and
the acceleration will do the same. On the other hand, whatever alteration is made of the mass, the opposite
or inverse change will occur with the acceleration. Double, triple or quadruple the mass, and the
acceleration will be one-half, one-third or one-fourth its original value.
As stated above, the direction of the net force is in the same direction as the acceleration. Thus, if
the direction of the acceleration is known, then the direction of the net force is also known. Consider the
two oil drop diagrams below for an acceleration of a car. From the diagram, determine the direction of the
net force that is acting upon the car. Then click the buttons to view the answers. (If necessary, review
acceleration from the previous unit.)
In conclusion, Newton's second law provides the explanation for the behavior of objects upon which the
forces do not balance. The law states that unbalanced forces cause objects to accelerate with an
acceleration that is directly proportional to the net force and inversely proportional to the mass.
Rocket Science!
NASA rockets (and others) accelerate upward off the launch pad as they burn a tremendous amount of
fuel. As the fuel is burned and exhausted to propel the rocket, the mass of the rocket changes. As such,
the same propulsion force can result in increasing acceleration values over time. Use the Rocket Science
widget below to explore this effect.
Check Your Understanding
1. Determine the accelerations that result when a 12-N net force is applied to a 3-kg object and then to a
6-kg object.
2. A net force of 15 N is exerted on an encyclopedia to cause it to accelerate at a rate of 5 m/s2. Determine
the mass of the encyclopedia.
3. Suppose that a sled is accelerating at a rate of 2 m/s2. If the net force is tripled and the mass is doubled,
then what is the new acceleration of the sled?
4. Suppose that a sled is accelerating at a rate of 2 m/s2. If the net force is tripled and the mass is halved,
then what is the new acceleration of the sled?
The Big Misconception
So what's the big deal? Many people have known Newton's first law since eighth
grade (or earlier). And if prompted with the first few words, most people could
probably recite the law word for word. And what is so terribly difficult about
remembering that F = ma? It seems to be a simple algebraic statement for solving
story problems. The big deal however is not the ability to recite the first law nor
to use the second law to solve problems; but rather the ability to understand their
meaning and to believe their implications. While most people know what
Newton's laws say, many people do not know what they mean (or simply do not
believe what they mean).
Cognitive scientists (scientists who study how people learn) have shown that
physics students come into physics class with a set of beliefs that they are unwilling
(or not easily willing) to discard despite evidence to the contrary. These beliefs
about motion (known as misconceptions) hinder further learning. The task of
overcoming misconceptions involves becoming aware of the misconceptions,
considering alternative conceptions or explanations, making a personal evaluation
of the two competing ideas and adopting a new conception that is more reasonable
than the previously held-misconception. This process involves self-reflection (to
ponder your own belief systems), critical thinking (to analyze the reasonableness of two competing ideas),
and evaluation (to select the most reasonable and harmonious model that explains the world of motion).
Self-reflection, critical thinking, and evaluation. While this process may seem terribly complicated, it is
simply a matter of using your noodle (that's your brain).
The most common misconception is one that dates back for ages; it is the
idea that sustaining motion requires a continued force. The misconception
has already been discussed in a previous lesson, but will now be discussed
in more detail. This misconception sticks out its ugly head in a number of
different ways and at a number of different times. As your read through
the following discussion, give careful attention to your own belief
systems. View physics as a system of thinking about the world rather than
information that can be dumped into your brain without evaluating its
consistency with your own belief systems.
Newton's laws declare loudly that a net force (an unbalanced force) causes
an acceleration; the acceleration is in the same direction as the net force. To test your own belief system,
consider the following question and its answer as seen by clicking the button.
Are You Infected with the Misconception?
Two students are discussing their physics homework prior to class. They are
discussing an object that is being acted upon by two individual forces (both in a
vertical direction); the free-body diagram for the particular object is shown at the
right. During the discussion, Anna Litical suggests to Noah Formula that the object
under discussion could be moving. In fact, Anna suggests that if friction and air
resistance could be ignored (because of their negligible size), the object could be
moving in a horizontal direction. According to Anna, an object experiencing forces as
described at the right could be experiencing a horizontal motion as described below.
Noah Formula objects, arguing that the object could not have any horizontal motion if there are only
vertical forces acting upon it. Noah claims that the object must be at rest, perhaps on a table or floor.
After all, says Noah, an object experiencing a balance of forces will be at rest. Who do you agree with?
Remember last winter when you went sledding down the hill and across the level surface at the local
park? (Apologies are extended to those who live in warmer winter climates.)
Imagine a the moment that there was no friction along the level surface from
point B to point C and that there was no air resistance to impede your motion.
How far would your sled travel? And what would its motion be like? Most
students I've talked to quickly answer: the sled would travel forever at constant
speed. Without friction or air resistance to slow it down, the sled would
continue in motion with the same speed and in the same direction. The forces
acting upon the sled from point B to point C would be the normal force (the
snow pushes up on the sled) and the gravity force (see diagram at right). These
forces are balanced and since the sled is already in motion at point B it will
continue in motion with the same speed and direction. So, as in the case of the
sled and as in the case of the object that Noah and Anna are discussing, an
object can be moving to the right even if the only forces acting upon the object
are vertical forces. Forces do not cause motion; forces cause accelerations.
Newton's first law of motion declares that a force is not needed to keep an object in motion. Slide a book
across a table and watch it slide to a rest position. The book in motion on the table top does not come to a
rest position because of the absence of a force; rather it is the presence of a force - that force being the
force of friction - that brings the book to a rest position. In the absence of a force of friction, the book
would continue in motion with the same speed and direction - forever (or at least to the end of the table
top)! A force is not required to keep a moving book in motion; and a force is not required to keep a
moving sled in motion; and a force is not required to keep any object horizontally moving object in
motion. To read more about this misconception, return to an earlier lesson.
Finding Acceleration
As learned earlier in Lesson 3 (as well as in Lesson 2), the net force is the vector sum of all the individual
forces. In Lesson 2, we learned how to determine the net force if the magnitudes of all the individual
forces are known. In this lesson, we will learn how to determine the acceleration of an object if the
magnitudes of all the individual forces are known. The three major equations that will be useful are the
equation for net force (Fnet = m•a), the equation for gravitational force (Fgrav = m•g), and the equation for
frictional force (Ffrict = μ•Fnorm).
The process of determining the acceleration of an object demands that the mass and the net force are
known. If mass (m) and net force (Fnet) are known, then the acceleration is determined by use of the
equation.
Thus, the task involves using the above equations, the given information, and your understanding of
Newton's laws to determine the acceleration. To gain a feel for how this method is applied, try the
following practice problems. Once you have solved the problems, click the button to check your answers.
Practice #1
An applied force of 50 N is used to accelerate an object to the right across a frictional surface. The object
encounters 10 N of friction. Use the diagram to determine the normal force, the net force, the mass, and
the acceleration of the object. (Neglect air resistance.)
Practice #2
An applied force of 20 N is used to accelerate an object to the right across a frictional surface. The object
encounters 10 N of friction. Use the diagram to determine the normal force, the net force, the coefficient
of friction (μ) between the object and the surface, the mass, and the acceleration of the object. (Neglect air
resistance.)
Practice #3
A 5-kg object is sliding to the right and encountering a friction force that slows it down. The coefficient
of friction (μ) between the object and the surface is 0.1. Determine the force of gravity, the normal force,
the force of friction, the net force, and the acceleration. (Neglect air resistance.)
A couple more practice problems are provided below. You should make an effort to solve as
many problems as you can without the assistance of notes, solutions, teachers, and other
students. Commit yourself to individually solving the problems. In the meantime, an
important caution is worth mentioning:
Avoid forcing a problem into the form of a previously solved problem. Problems in physics will seldom
look the same. Instead of solving problems by rote or by mimicry of a previously solved problem, utilize
your conceptual understanding of Newton's laws to work towards solutions to problems. Use your
understanding of weight and mass to find the m or the Fgrav in a problem. Use your conceptual
understanding of net force (vector sum of all the forces) to find the value of Fnet or the value of an
individual force. Do not divorce the solving of physics problems from your understanding of physics
concepts. If you are unable to solve physics problems like those above, it is does not necessarily mean
that you are having math difficulties. It is likely that you are having a physics concepts difficulty.
Check Your Understanding
1. Edwardo applies a 4.25-N rightward force to a 0.765-kg book to accelerate it across a tabletop. The
coefficient of friction between the book and the tabletop is 0.410. Determine the acceleration of the book.
2. In a physics lab, Kate and Rob use a hanging mass and pulley system to exert a 2.45 N rightward force
on a 0.500-kg cart to accelerate it across a low-friction track. If the total resistance force to the motion of
the cart is 0.72 N, then what is the cart's acceleration?
Free Fall and Air Resistance
In a previous unit, it was stated that all objects (regardless of their mass) free fall with the same
acceleration - 9.8 m/s/s. This particular acceleration value is so important in physics that it has its own
peculiar name - the acceleration of gravity - and its own peculiar symbol - g. But why do all objects free
fall at the same rate of acceleration regardless of their mass? Is it because they all weigh the same? ...
because they all have the same gravity? ... because the air resistance is the same for each? Why? These
questions will be explored in this section of Lesson 3.
In addition to an exploration of free fall, the motion of objects that encounter air resistance will also be
analyzed. In particular, two questions will be explored:


Why do objects that encounter air resistance ultimately reach a terminal velocity?
In situations in which there is air resistance, why do more massive objects fall faster than less
massive objects?
To answer the above questions, Newton's second law of motion (Fnet = m•a) will be applied to analyze the
motion of objects that are falling under the sole influence of gravity (free fall) and under the dual
influence of gravity and air resistance.
Free Fall Motion
As learned in an earlier unit, free fall is a special type of motion in which the only force acting upon an
object is gravity. Objects that are said to be undergoing free fall, are not encountering a significant force
of air resistance; they are falling under the sole influence of gravity. Under such conditions, all objects
will fall with the same rate of acceleration, regardless of their mass. But why? Consider the free-falling
motion of a 1000-kg baby elephant and a 1-kg overgrown mouse.
If Newton's second law were applied to their falling motion, and if a free-body diagram
were constructed, then it would be seen that the 1000-kg baby elephant would
experiences a greater force of gravity. This greater force of gravity would have a direct
affect upon the elephant's acceleration; thus, based on force alone, it might be thought
that the 1000-kg baby elephant would accelerate faster. But acceleration depends upon
two factors: force and mass. The 1000-kg baby elephant obviously has more mass (or
inertia). This increased mass has an inverse affect upon the elephant's acceleration. And
thus, the direct affect of greater force on the 1000-kg elephant is offset by the inverse
affect of the greater mass of the 1000-kg elephant; and so each object accelerates at the same rate approximately 10 m/s/s. The ratio of force to mass (Fnet/m) is the same for the elephant and the mouse
under situations involving free fall.
This ratio (Fnet/m) is sometimes called the gravitational field strength and is expressed as 9.8 N/kg (for a
location upon Earth's surface). The gravitational field strength is a property of the location within Earth's
gravitational field and not a property of the baby elephant nor the mouse. All objects placed upon Earth's
surface will experience this amount of force (9.8 N) upon every 1 kilogram of mass within the object.
Being a property of the location within Earth's gravitational field and not a property of the free falling
object itself, all objects on Earth's surface will experience this amount of force per mass. As such, all
objects free fall at the same rate regardless of their mass. Because the 9.8 N/kg gravitational field at
Earth's surface causes a 9.8 m/s/s acceleration of any object placed there, we often call this ratio the
acceleration of gravity.
Falling with Air Resistance
As an object falls through air, it usually encounters some degree of air resistance.
Air resistance is the result of collisions of the object's leading surface with air
molecules. The actual amount of air resistance encountered by the object is
dependent upon a variety of factors. To keep the topic simple, it can be said that
the two most common factors that have a direct affect upon the amount of air
resistance are the speed of the object and the cross-sectional area of the object.
Increased speeds result in an increased amount of air resistance. Increased crosssectional areas result in an increased amount of air resistance.
Why does an object that encounters air resistance eventually reach a terminal velocity? To answer this
questions, Newton's second law will be applied to the motion of a falling skydiver.
In the diagrams below, free-body diagrams showing the forces acting upon an 85-kg skydiver
(equipment included) are shown. For each case, use the diagrams to determine the net force and
acceleration of the skydiver at each instant in time. Then use the button to view the answers.
The diagrams above illustrate a key principle. As an object falls, it picks up speed. The increase in speed
leads to an increase in the amount of air resistance. Eventually, the force of air resistance becomes large
enough to balances the force of gravity. At this instant in time, the net force is 0 Newton; the object will
stop accelerating. The object is said to have reached a terminal velocity. The change in velocity
terminates as a result of the balance of forces. The velocity at which this happens is called the terminal
velocity.
In situations in which there is air resistance, more massive objects fall faster than less massive objects.
But why? To answer the why question, it is necessary to consider the free-body diagrams for objects of
different mass. Consider the falling motion of two skydivers: one with a mass of 100 kg (skydiver plus
parachute) and the other with a mass of 150 kg (skydiver plus parachute). The free-body diagrams are
shown below for the instant in time in which they have reached terminal velocity.
As learned above, the amount of air resistance depends upon the speed of the object. A falling object will
continue to accelerate to higher speeds until they encounter an amount of air resistance that is equal to
their weight. Since the 150-kg skydiver weighs more (experiences a greater force of gravity), it will
accelerate to higher speeds before reaching a terminal velocity. Thus, more massive objects fall faster
than less massive objects because they are acted upon by a larger force of gravity; for this reason, they
accelerate to higher speeds until the air resistance force equals the gravity force.
Newton's Third Law
A force is a push or a pull upon an object that results from its interaction with another object. Forces
result from interactions! As discussed in Lesson 2, some forces result from contact interactions (normal,
frictional, tensional, and applied forces are examples of contact forces) and other forces are the result of
action-at-a-distance interactions (gravitational, electrical, and magnetic forces). According to Newton,
whenever objects A and B interact with each other, they exert forces upon each other. When you sit in
your chair, your body exerts a downward force on the chair and the chair exerts an upward force on your
body. There are two forces resulting from this interaction - a force on the chair and a force on your body.
These two forces are called action and reaction forces and are the subject of Newton's third law of
motion. Formally stated, Newton's third law is:
For every action, there is an equal and opposite reaction.
The statement means that in every interaction, there is a pair of forces acting on the two
interacting objects. The size of the forces on the first object equals the size of the force
on the second object. The direction of the force on the first object is opposite to the
direction of the force on the second object. Forces always come in pairs - equal and
opposite action-reaction force pairs.
A variety of action-reaction force pairs are evident in nature. Consider the propulsion of
a fish through the water. A fish uses its fins to push water backwards. But a push on the water will only
serve to accelerate the water. Since forces result from mutual interactions, the water must also be pushing
the fish forwards, propelling the fish through the water. The size of the force on the water equals the size
of the force on the fish; the direction of the force on the water (backwards) is opposite the direction of the
force on the fish (forwards). For every action, there is an equal (in size) and opposite (in direction)
reaction force. Action-reaction force pairs make it possible for fish to swim.
Consider the flying motion of birds. A bird flies by use of its wings. The wings
of a bird push air downwards. Since forces result from mutual interactions, the
air must also be pushing the bird upwards. The size of the force on the air
equals the size of the force on the bird; the direction of the force on the air
(downwards) is opposite the direction of the force on the bird (upwards). For
every action, there is an equal (in size) and opposite (in direction) reaction.
Action-reaction force pairs make it possible for birds to fly.
Consider the motion of a car on the way to school. A car is equipped with
wheels that spin. As the wheels spin, they grip the road and push the road
backwards. Since forces result from mutual interactions, the road must also be pushing the wheels
forward. The size of the force on the road equals the size of the force on the wheels (or car); the direction
of the force on the road (backwards) is opposite the direction of the force on the wheels (forwards). For
every action, there is an equal (in size) and opposite (in direction) reaction. Action-reaction force pairs
make it possible for cars to move along a roadway surface.
Check Your Understanding
1. While driving down the road, a firefly strikes the windshield of a bus
and makes a quite obvious mess in front of the face of the driver. This is
a clear case of Newton's third law of motion. The firefly hit the bus and
the bus hits the firefly. Which of the two forces is greater: the force on
the firefly or the force on the bus?
2. For years, space travel was believed to be impossible because there
was nothing that rockets could push off of in space in order to provide
the propulsion necessary to accelerate. This inability of a rocket to provide propulsion is because ...
a. ... space is void of air so the rockets have nothing to push off of.
b. ... gravity is absent in space.
c. ... space is void of air and so there is no air resistance in space.
d. ... nonsense! Rockets do accelerate in space and have been able to do so for a
long time.
3. Many people are familiar with the fact that a rifle recoils when fired. This
recoil is the result of action-reaction force pairs. A gunpowder explosion creates
hot gases that expand outward allowing the rifle to push forward on the bullet.
Consistent with Newton's third law of motion, the bullet pushes backwards upon
the rifle. The acceleration of the recoiling rifle is ...
a. greater than the acceleration of the bullet.
b. smaller than the acceleration of the bullet.
c. the same size as the acceleration of the bullet.
4. In the top picture (below), Kent Budgett is pulling upon a rope that is attached to a wall. In the bottom
picture, the Kent is pulling upon a rope that is attached to an elephant. In each case, the force scale reads
500 Newton. Kent is pulling ...
a. with more force when the rope is attached to the wall.
b. with more force when the rope is attached to the elephant.
c. the same force in each case.
Identifying Action and Reaction Force Pairs
According to Newton's third law, for every action force there is an equal (in size) and opposite (in
direction) reaction force. Forces always come in pairs - known as "action-reaction force pairs."
Identifying and describing action-reaction force pairs is a simple matter of identifying the two interacting
objects and making two statements describing who is pushing on whom and in what direction. For
example, consider the interaction between a baseball bat and a baseball.
The baseball forces the bat to the left; the bat forces the ball to the right. Together, these two forces
exerted upon two different objects form the action-reaction force pair. Note that in the description of the
two forces, the nouns in the sentence describing the forces simply switch places.
Consider the following three examples. One of the forces in the mutual interaction is described; describe
the other force in the action-reaction force pair. Click the button to view the answer.
Baseball pushes glove leftwards.
Bowling ball pushes pin leftwards.
Enclosed air particles push balloon wall outwards.
Check Your Understanding
1. Consider the interaction depicted below between foot A, ball B, and foot C. The three objects interact
simultaneously (at the same time). Identify the two pairs of action-reaction forces. Use the notation "foot
A", "foot C", and "ball B" in your statements. Click the button to view the answer.
2. Identify at least six pairs of action-reaction force pairs in the following diagram.
Center of mass
The centre of mass is the point where we can assume all the mass of the object is concentrated.
As the gravity only acts at a single point in the object (this point is the center off mass) , so a single arrow on
diagram can represent the weight of the object.
W
The centre of mass for regularly shaped objects is in the centre.
For irregular shaped objects, we can find the centre of mass by following steps.
1.
2.
3.
4.
5.
6.
Hang up the object.
Suspend a plumb line from the same place.
Mark the position of the thread.
The centre of mass is along the line of thread.
Repeat the above steps with object suspended from different places.
The centre of mass is where these lines cross.
Centre of mass links stability:
It is important to know where the centre of mass of a body is, as its position determines the stability of the
body. A body is stable if, when it is tilted slightly, the line of action of its weight passes through its base.
Hook's law
The law states at "the extension of a spring (or any elastic body) is directly proportional to the applied load
(stretching force), provided that elastic limit is not exceeded.
The graph of load and extension is a straight line which passes through the origin.The gradient of the line is the
measure of the stiffness of the spring, which is called spring constant.
If the applied force or load is removed the spring returns to its original length. This behavior is
called elastic behavior.This behavior is up to a particular point or limit called elastic limit.If the applied
force or load is removed the spring does not return to its original length. This behavior is called plastic
behavior.This behavior is after a particular point or limit called elastic limit.
Elastic limit:
If you stretch the spring too far, the line (Force-extension graph) no longer straightens, and hook’s law is
no longer true.
The point at the end of the straight line is known as the limit of proportionality or elastic limit. If the
material (spring) is stretched beyond the elastic limit, there is a permanent change in its shape.After
elastic limit different materials behave differently.
Effect of applied force on the materials:
Whenever force is applied on the material, It bring the change either in its length, area or volume. Hook's
law explains the the effect of applied force on the length of the material.
Hooke's Law, elastic and plastic behaviour
(Note: Plastic behaviour is sometimes called inelastic behaviour.)
An elastic material is one that will return to its original shape when the force applied to it is taken away.A
plastic (or inelastic) material is one that stays deformed after you have taken the force away. If you apply
too big a force a material will lose its elasticity.
Hooke discovered that the amount a spring stretches is proportional to the amount of force applied to it.
This means if you double the force its extension will double, if you triple the force the extension will
triple and so on.
Gradient = spring constant
Area under the graph = elastic potential energy
The elastic limit can be seen on the graph. This is where the graph stops being a straight line. If you
stretch the spring beyond this point it will not return to its original shape.
You can write Hooke's law as an equation:
F = kx
where:



F is the applied force (in newtons, N),
x is the extension (in metres, m) and
k is the spring constant (in N/m).
The spring constant measures how stiff the spring is. The larger the spring constant the stiffer the spring.
You may be able to see this by looking at the graphs below:
Elastic behaviour is very important in car safety, as car seatbelts are made from elastic materials.
However, after a crash they must be replaced as they will go past their elastic limit.
Why have seat belts that are elastic?
Why not just have very rigid seatbelts that would keep you firmly in place?
The reason for this, is that it would be very dangerous and cause large injuries. This is because it would
slow your body down too quickly. The quicker a collision, the bigger the force that is produced.
This can be seen very plainly by comparing the effect of kicking a football, which squashes as you kick it
giving a big collision time, followed by kicking a brick. The brick doesn't squash, giving a very quick
collision time and a very painful foot.
Moment of force
The turning effect of force is called moment of force.
The moment of force depends on the following factors.


The size (magnitude) of the force
The perpendicular distance between the line of action of the force and the turning point which is called the
pivot.
We calculate the moment of force by using the following formula
Moment of force = force * perpendicular distance from pivot to the line of action of the force
Moment=F * d
d
Moment is measured in newton meters(Nm).
d
Priciple of moment of force
This principle states that "if a system of force is not turning (or if the system is in equilibrium) then the
algebaric sum of the all clockwise moments is equal to the algebaric the sum of all anticlockwise moments
about any point".
Sum of clockwise moments = sum of anticlockwise moments
Equilibrium
If a body under the action of a number of forces is at rest or moving with uniform velocity, it is said to be in
equilibrium.
There are two conditions of equilibrium.
First condition of equilibrium:Vector sum of all the forces acting on the object is zero.
Second condition of equilibrium:The algebaric sum of all the moment of forces acting on an object iz
zero. i.e. Sum of anticlock wise moment of forces = Sum of clock wise moment of forces
WORK, ENERGY AND POWER
Basic Terminology and Concepts
Definition and Mathematics of Work | Calculating the Amount of Work Done by Forces
Potential Energy | Kinetic Energy | Mechanical Energy | Power
Definition and Mathematics of Work
In the first three units of The Physics Classroom, we utilized Newton's laws to analyze the motion of
objects. Force and mass information were used to determine the acceleration of an object. Acceleration
information was subsequently used to determine information about the velocity or
displacement of an object after a given period of time. In this manner, Newton's laws
serve as a useful model for analyzing motion and making predictions about the final
state of an object's motion. In this unit, an entirely different model will be used to
analyze the motion of objects. Motion will be approached from the perspective of
work and energy. The affect that work has upon the energy of an object (or system
of objects) will be investigated; the resulting velocity and/or height of the object can
then be predicted from energy information. In order to understand this work-energy
approach to the analysis of motion, it is important to first have a solid understanding of a few basic terms.
Thus, Lesson 1 of this unit will focus on the definitions and meanings of such terms as work, mechanical
energy, potential energy, kinetic energy, and power.
When a force acts upon an object to cause a displacement of the object, it is said that work was done
upon the object. There are three key ingredients to work - force, displacement, and cause. In order for a
force to qualify as having done work on an object, there must be a displacement and the force must cause
the displacement. There are several good examples of work that can be observed in everyday life - a horse
pulling a plow through the field, a father pushing a grocery cart down the aisle of a grocery store, a
freshman lifting a backpack full of books upon her shoulder, a weightlifter lifting a barbell above his
head, an Olympian launching the shot-put, etc. In each case described here there is a force exerted upon
an object to cause that object to be displaced.
Read the following five statements and determine whether or not they represent examples of work.
Then click on the See Answer button to view the answer.
Statement
A teacher applies a force to a wall and becomes exhausted.
Answer with
Explanation
A book falls off a table and free falls to the ground.
A waiter carries a tray full of meals above his head by one arm straight across the
room at constant speed. (Careful! This is a very difficult question that will be
discussed in more detail later.)
A rocket accelerates through space.
Mathematically, work can be expressed by the following equation.
where F is the force, d is the displacement, and the angle (theta) is defined as the angle between the force
and the displacement vector. Perhaps the most difficult aspect of the above equation is the angle "theta."
The angle is not just any 'ole angle, but rather a very specific angle. The angle measure is defined as the
angle between the force and the displacement. To gather an idea of it's meaning, consider the following
three scenarios.



Scenario A: A force acts rightward
upon an object as it is displaced
rightward. In such an instance, the
force vector and the displacement
vector are in the same direction.
Thus, the angle between F and d is 0
degrees.
Scenario B: A force acts leftward
upon an object that is displaced
rightward. In such an instance, the
force vector and the displacement vector are in the opposite direction. Thus, the angle between
F and d is 180 degrees.
Scenario C: A force acts upward on an object as it is displaced rightward. In such an instance, the
force vector and the displacement vector are at right angles to each other. Thus, the angle
between F and d is 90 degrees.
To Do Work, Forces Must Cause Displacements
Let's consider Scenario C above in more detail. Scenario C
involves a situation similar to the waiter who carried a tray full of
meals above his head by one arm straight across the room at
constant speed. It was mentioned earlier that the waiter does not
do work upon the tray as he carries it across the room. The force
supplied by the waiter on the tray is an upward force and the
displacement of the tray is a horizontal displacement. As such,
the angle between the force and the displacement is 90 degrees.
If the work done by the waiter on the tray were to be calculated,
then the results would be 0. Regardless of the magnitude of the force and displacement, F*d*cosine 90
degrees is 0 (since the cosine of 90 degrees is 0). A vertical force can never cause a horizontal
displacement; thus, a vertical force does not do work on a horizontally displaced object!!
It can be accurately noted that the waiter's hand did push forward on the tray for a brief period of time to
accelerate it from rest to a final walking speed. But once up to speed, the tray will stay in its straight-line
motion at a constant speed without a forward force. And if the only force exerted upon the tray during the
constant speed stage of its motion is upward, then no work is done upon the tray. Again, a vertical force
does not do work on a horizontally displaced object.
The equation for work lists three variables - each variable is associated with one
of the three key words mentioned in the definition of work (force, displacement,
and cause). The angle theta in the equation is associated with the amount of force
that causes a displacement. As mentioned in a previous unit, when a force is
exerted on an object at an angle to the horizontal, only a part of the force
contributes to (or causes) a horizontal displacement. Let's consider the force of a
chain pulling upwards and rightwards upon Fido in order to drag Fido to the
right. It is only the horizontal component of the tension force in the chain that
causes Fido to be displaced to the right. The horizontal component is found by
multiplying the force F by the cosine of the angle between F and d. In this sense,
the cosine theta in the work equation relates to the cause factor - it selects the
portion of the force that actually causes a displacement.
The Meaning of Negative Work
On occasion, a force acts upon a moving object to hinder a displacement. Examples might include a car
skidding to a stop on a roadway surface or a baseball runner sliding to a stop on the infield dirt. In such
instances, the force acts in the direction opposite the objects motion in order to slow it down. The force
doesn't cause the displacement but rather hinders it. These situations involve what is commonly called
negative work. The negative of negative work refers to the numerical value that results when values of F,
d and theta are substituted into the work equation. Since the force vector is directly opposite the
displacement vector, theta is 180 degrees. The cosine(180 degrees) is -1 and so a negative value results
for the amount of work done upon the object. Negative work will become important (and more
meaningful) in Lesson 2 as we begin to discuss the relationship between work and energy.
Units of Work
Whenever a new quantity is introduced in physics, the standard metric units associated with that quantity
are discussed. In the case of work (and also energy), the standard metric unit is the Joule (abbreviated J).
One Joule is equivalent to one Newton of force causing a displacement of one meter. In other words,
The Joule is the unit of work.
1 Joule = 1 Newton * 1 meter
1J=1N*m
In fact, any unit of force times any unit of displacement is equivalent to a unit of work. Some nonstandard
units for work are shown below. Notice that when analyzed, each set of units is equivalent to a force unit
times a displacement unit.
In summary, work is done when a force acts upon an object to cause a displacement. Three quantities
must be known in order to calculate the amount of work. Those three quantities are force, displacement
and the angle between the force and the displacement.
Potential Energy
An object can store energy as the result of its position. For example, the heavy ball of a demolition
machine is storing energy when it is held at an elevated position. This stored energy of position is referred
to as potential energy. Similarly, a drawn bow is able to store energy as the result of its position. When
assuming its usual position (i.e., when not drawn), there is no energy stored in the bow. Yet when its
position is altered from its usual equilibrium position, the bow is able to store energy by virtue of its
position. This stored energy of position is referred to as potential energy. Potential energy is the stored
energy of position possessed by an object.
Gravitational Potential Energy
The two examples above illustrate the two forms of potential energy to be discussed in this course gravitational potential energy and elastic potential energy.
Gravitational potential energy is the energy stored in an object
as the result of its vertical position or height. The energy is
stored as the result of the gravitational attraction of the Earth for
the object. The gravitational potential energy of the massive ball
of a demolition machine is dependent on two variables - the
mass of the ball and the height to which it is raised. There is a direct relation between gravitational
potential energy and the mass of an object. More massive objects have greater gravitational potential
energy. There is also a direct relation between gravitational potential energy and the height of an object.
The higher that an object is elevated, the greater the gravitational potential energy. These relationships are
expressed by the following equation:
PEgrav = mass • g • height
PEgrav = m *• g • h
In the above equation, m represents the mass of the object, h represents the height of the object and g
represents the gravitational field strength (9.8 N/kg on Earth) - sometimes referred to as the acceleration
of gravity.
To determine the gravitational potential energy of an object, a
zero height position must first be arbitrarily assigned.
Typically, the ground is considered to be a position of zero
height. But this is merely an arbitrarily assigned position that
most people agree upon. Since many of our labs are done on
tabletops, it is often customary to assign the tabletop to be the
zero height position. Again this is merely arbitrary. If the
tabletop is the zero position, then the potential energy of an
object is based upon its height relative to the tabletop. For
example, a pendulum bob swinging to and from above the
tabletop has a potential energy that can be measured based on
its height above the tabletop. By measuring the mass of the bob and the height of the bob above the
tabletop, the potential energy of the bob can be determined.
Since the gravitational potential energy of an object is directly proportional to its height above the zero
position, a doubling of the height will result in a doubling of the gravitational potential energy. A tripling
of the height will result in a tripling of the gravitational potential energy.
Use this principle to determine the blanks in the following diagram. Knowing that the potential
energy at the top of the tall platform is 50 J, what is the potential energy at the other positions
shown on the stair steps and the incline?
Elastic Potential Energy
The second form of potential energy that we will discuss is
elastic potential energy. Elastic potential energy is the
energy stored in elastic materials as the result of their
stretching or compressing. Elastic potential energy can be
stored in rubber bands, bungee chords, trampolines,
springs, an arrow drawn into a bow, etc. The amount of
elastic potential energy stored in such a device is related to
the amount of stretch of the device - the more stretch, the
more stored energy.
Springs are a special instance of a device that can store elastic potential energy due to either compression
or stretching. A force is required to compress a spring; the more compression there is, the more force that
is required to compress it further. For certain springs, the amount of force is directly proportional to the
amount of stretch or compression (x); the constant of proportionality is known as the spring constant (k).
Such springs are said to follow Hooke's Law. If a spring is not stretched or compressed, then there is no
elastic potential energy stored in it. The spring is said to be at its equilibrium position. The equilibrium
position is the position that the spring naturally assumes when there is no force applied to it. In terms of
potential energy, the equilibrium position could be called the zero-potential energy position. There is a
special equation for springs that relates the amount of elastic potential energy to the amount of stretch (or
compression) and the spring constant. The equation is
To summarize, potential energy is the energy that is stored in an object due to its position relative to some
zero position. An object possesses gravitational potential energy if it is positioned at a height above (or
below) the zero height. An object possesses elastic potential energy if it is at a position on an elastic
medium other than the equilibrium position.
Check Your Understanding
Check your understanding of the concept of potential energy by answering the following questions. When
finished, click the button to view the answers.
1. A cart is loaded with a brick and pulled at constant speed along
an inclined plane to the height of a seat-top. If the mass of the
loaded cart is 3.0 kg and the height of the seat top is 0.45 meters,
then what is the potential energy of the loaded cart at the height of
the seat-top?
2. If a force of 14.7 N is used to drag the loaded cart (from previous
question) along the incline for a distance of 0.90 meters, then how much work is done on the loaded cart?
Note that the work done to lift the loaded cart up the inclined plane at constant speed is equal to the
potential energy change of the cart. This is not coincidental! The reason for the relation between the
potential energy change of the cart and the work done upon it is the subject of Lesson 2
Kinetic Energy
Kinetic energy is the energy of motion. An object that has motion - whether it is vertical or horizontal
motion - has kinetic energy. There are many forms of kinetic energy - vibrational (the energy due to
vibrational motion), rotational (the energy due to rotational motion), and translational (the energy due to
motion from one location to another). To keep matters simple, we will focus upon translational kinetic
energy. The amount of translational kinetic energy (from here on, the phrase kinetic energy will refer to
translational kinetic energy) that an object has depends upon two variables: the mass (m) of the object and
the speed (v) of the object. The following equation is used to represent the kinetic energy (KE) of an
object.
where m = mass of object AND v = speed of object
This equation reveals that the kinetic energy of an object is directly
proportional to the square of its speed. That means that for a twofold
increase in speed, the kinetic energy will increase by a factor of four. For a
threefold increase in speed, the kinetic energy will increase by a factor of
nine. And for a fourfold increase in speed, the kinetic energy will increase
by a factor of sixteen. The kinetic energy is dependent upon the square of
the speed. As it is often said, an equation is not merely a recipe for algebraic problem solving, but also a
guide to thinking about the relationship between quantities.
Kinetic energy is a scalar quantity; it does not have a direction. Unlike velocity, acceleration, force, and
momentum, the kinetic energy of an object is completely described by magnitude alone. Like work and
potential energy, the standard metric unit of measurement for kinetic energy is the Joule. As might be
implied by the above equation, 1 Joule is equivalent to 1 kg*(m/s)^2.
Check Your Understanding
Use your understanding of kinetic energy to answer the following questions. Then click the button to
view the answers.
1. Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s.
2. If the roller coaster car in the above problem were moving with twice the speed, then what would be its
new kinetic energy?
3. Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12
000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then what is her speed?
4. A 900-kg compact car moving at 60 mi/hr has approximately 320 000 Joules of kinetic energy.
Estimate its new kinetic energy if it is moving at 30 mi/hr. (HINT: use the kinetic energy equation as a
"guide to thinking.")
Mechanical Energy
In a previous part of Lesson 1, it was said that work is done upon an
object whenever a force acts upon it to cause it to be displaced.
Work involves a force acting upon an object to cause a
displacement. In all instances in which work is done, there is an
object that supplies the force in order to do the work. If a World
Civilization book is lifted to the top shelf of a student locker, then
the student supplies the force to do the work on the book. If a plow
is displaced across a field, then some form of farm equipment
(usually a tractor or a horse) supplies the force to do the work on the
plow. If a pitcher winds up and accelerates a baseball towards home
plate, then the pitcher supplies the force to do the work on the baseball. If a roller coaster car is displaced
from ground level to the top of the first drop of a roller coaster ride, then a chain driven by a motor
supplies the force to do the work on the car. If a barbell is displaced from ground level to a height above a
weightlifter's head, then the weightlifter is supplying a force to do work on the barbell. In all instances, an
object that possesses some form of energy supplies the force to do the work. In the instances described
here, the objects doing the work (a student, a tractor, a pitcher, a motor/chain) possess chemical potential
energy stored in food or fuel that is transformed into work. In the process of doing work, the object that is
doing the work exchanges energy with the object upon which the work is done. When the work is done
upon the object, that object gains energy. The energy acquired by the objects upon which work is done is
known as mechanical energy.
Mechanical energy is the energy that is possessed by an object due to
its motion or due to its position. Mechanical energy can be either
kinetic energy (energy of motion) or potential energy (stored energy of
position). Objects have mechanical energy if they are in motion and/or
if they are at some position relative to a zero potential energy position
(for example, a brick held at a vertical position above the ground or
zero height position). A moving car possesses mechanical energy due
to its motion (kinetic energy). A moving baseball possesses
mechanical energy due to both its high speed (kinetic energy) and its
vertical position above the ground (gravitational potential energy). A
World Civilization book at rest on the top shelf of a locker possesses
mechanical energy due to its vertical position above the ground (gravitational potential energy). A barbell
lifted high above a weightlifter's head possesses mechanical energy due to its vertical position above the
ground (gravitational potential energy). A drawn bow possesses mechanical energy due to its stretched
position (elastic potential energy).
Mechanical Energy as the Ability to Do Work
An object that possesses mechanical energy is able to do work. In
fact, mechanical energy is often defined as the ability to do work.
Any object that possesses mechanical energy - whether it is in the
form of potential energy or kinetic energy - is able to do work. That
is, its mechanical energy enables that object to apply a force to
another object in order to cause it to be displaced.
Numerous examples can be given of how an object with mechanical energy can harness that energy in
order to apply a force to cause another object to be displaced. A classic example involves the massive
wrecking ball of a demolition machine. The wrecking ball is a massive object that is swung backwards to
a high position and allowed to swing forward into building structure or other object in order to demolish
it. Upon hitting the structure, the wrecking ball applies a force to it in order to cause the wall of the
structure to be displaced. The diagram below depicts the process by which the mechanical energy of a
wrecking ball can be used to do work.
A hammer is a tool that utilizes mechanical energy to do work. The
mechanical energy of a hammer gives the hammer its ability to apply a
force to a nail in order to cause it to be displaced. Because the hammer has
mechanical energy (in the form of kinetic energy), it is able to do work on
the nail. Mechanical energy is the ability to do work.
Another example that illustrates how mechanical
energy is the ability of an object to do work can be
seen any evening at your local bowling alley. The
mechanical energy of a bowling ball gives the ball the ability to apply a force to a
bowling pin in order to cause it to be displaced. Because the massive ball has
mechanical energy (in the form of kinetic energy), it is able to do work on the pin.
Mechanical energy is the ability to do work.
A dart gun is still another example of how mechanical energy
of an object can do work on another object. When a dart gun is
loaded and the springs are compressed, it possesses mechanical
energy. The mechanical energy of the compressed springs gives
the springs the ability to apply a force to the dart in order to
cause it to be displaced. Because of the springs have
mechanical energy (in the form of elastic potential energy), it is able to do work on the dart. Mechanical
energy is the ability to do work.
A common scene in some parts of the countryside is a "wind
farm." High-speed winds are used to do work on the blades of a
turbine at the so-called wind farm. The mechanical energy of the
moving air gives the air particles the ability to apply a force and
cause a displacement of the blades. As the blades spin, their
energy is subsequently converted into electrical energy (a nonmechanical form of energy) and supplied to homes and
industries in order to run electrical appliances. Because the moving wind has mechanical energy (in the
form of kinetic energy), it is able to do work on the blades. Once more, mechanical energy is the ability to
do work.
The Total Mechanical Energy
As already mentioned, the mechanical energy of an object can be the result of its motion (i.e., kinetic
energy) and/or the result of its stored energy of position (i.e., potential energy). The total amount of
mechanical energy is merely the sum of the potential energy and the kinetic energy. This sum is simply
referred to as the total mechanical energy (abbreviated TME).
TME = PE + KE
As discussed earlier, there are two forms of potential energy discussed in our course - gravitational
potential energy and elastic potential energy. Given this fact, the above equation can be rewritten:
TME = PEgrav + PEspring + KE
The diagram below depicts the motion of Li Ping Phar (esteemed Chinese ski jumper) as she glides down
the hill and makes one of her record-setting jumps.
The total mechanical energy of Li Ping Phar is the sum of the potential and kinetic energies. The two
forms of energy sum up to 50 000 Joules. Notice also that the total mechanical energy of Li Ping Phar is a
constant value throughout her motion. There are conditions under which the total mechanical energy will
be a constant value and conditions under which it will be a changing value. This is the subject of Lesson 2
- the work-energy relationship. For now, merely remember that total mechanical energy is the energy
possessed by an object due to either its motion or its stored energy of position. The total amount of
mechanical energy is merely the sum of these two forms of energy. And finally, an object with
mechanical energy is able to do work on another object.
Power
The quantity work has to do with a force causing a displacement. Work has
nothing to do with the amount of time that this force acts to cause the
displacement. Sometimes, the work is done very quickly and other times the
work is done rather slowly. For example, a rock climber takes an abnormally long time to elevate her
body up a few meters along the side of a cliff. On the other hand, a trail hiker (who selects the easier path
up the mountain) might elevate her body a few meters in a short amount of time. The two people might do
the same amount of work, yet the hiker does the work in considerably less time than the rock climber. The
quantity that has to do with the rate at which a certain amount of work is done is known as the power. The
hiker has a greater power rating than the rock climber.
Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using
the following equation.
The standard metric unit of power is the Watt. As is implied by the
equation for power, a unit of power is equivalent to a unit of work
divided by a unit of time. Thus, a Watt is equivalent to a
Joule/second. For historical reasons, the horsepower is occasionally
used to describe the power delivered by a machine. One horsepower
is equivalent to approximately 750 Watts.
Most machines are designed and built to do work on objects. All machines are typically described by a
power rating. The power rating indicates the rate at which that machine can do work upon other objects.
Thus, the power of a machine is the work/time ratio for that particular
machine. A car engine is an example of a machine that is given a power
rating. The power rating relates to how rapidly the car can accelerate the car.
Suppose that a 40-horsepower engine could accelerate the car from 0 mi/hr
to 60 mi/hr in 16 seconds. If this were the case, then a car with four times
the horsepower could do the same amount of work in one-fourth the time. That is, a 160-horsepower
engine could accelerate the same car from 0 mi/hr to 60 mi/hr in 4 seconds. The point is that for the same
amount of work, power and time are inversely proportional. The power equation suggests that a more
powerful engine can do the same amount of work in less time.
A person is also a machine that has a power rating. Some people are more power-full than others. That is,
some people are capable of doing the same amount of work in less time or more work in the same amount
of time. A common physics lab involves quickly climbing a flight of stairs and using mass, height and
time information to determine a student's personal power. Despite the diagonal motion along the staircase,
it is often assumed that the horizontal motion is constant and all the force from the steps is used to elevate
the student upward at a constant speed. Thus, the weight of the student is equal to the force that does the
work on the student and the height of the staircase is the upward
displacement. Suppose that Ben Pumpiniron elevates his 80-kg body up the
2.0-meter stairwell in 1.8 seconds. If this were the case, then we could
calculate Ben's power rating. It can be assumed that Ben must apply an 800Newton downward force upon the stairs to elevate his body. By so doing, the
stairs would push upward on Ben's body with just enough force to lift his
body up the stairs. It can also be assumed that the angle between the force of
the stairs on Ben and Ben's displacement is 0 degrees. With these two approximations, Ben's power rating
could be determined as shown below.
Ben's power rating is 871 Watts. He is quite a horse.
The expression for power is work/time. And since the expression for work is force*displacement, the
expression for power can be rewritten as (force*displacement)/time. Since the expression for velocity is
displacement/time, the expression for power can be rewritten once more as force*velocity. This is shown
below.
This new equation for power reveals that a powerful machine is
both strong (big force) and fast (big velocity). A powerful car
engine is strong and fast. A powerful piece of farm equipment is
strong and fast. A powerful weightlifter is strong and fast. A
powerful lineman on a football team is strong and fast. A machine that is strong enough to apply a big
force to cause a displacement in a small mount of time (i.e., a big velocity) is a powerful machine
Check Your Understanding
Use your understanding of work and power to answer the following questions. When finished, click the
button to view the answers.
1. Two physics students, Will N. Andable and Ben Pumpiniron, are in the
weightlifting room. Will lifts the 100-pound barbell over his head 10 times in
one minute; Ben lifts the 100-pound barbell over his head 10 times in 10
seconds. Which student does the most work? ______________ Which student
delivers the most power? ______________ Explain your answers.
2. During a physics lab, Jack and Jill ran up a hill. Jack is twice as massive as Jill; yet Jill ascends the
same distance in half the time. Who did the most work? ______________ Who delivered the most
power? ______________ Explain your answers.
3. A tired squirrel (mass of approximately 1 kg) does
push-ups by applying a force to elevate its center-of-mass
by 5 cm in order to do a mere 0.50 Joule of work. If the
tired squirrel does all this work in 2 seconds, then
determine its power.
4. When doing a chin-up, a physics student lifts her 42.0-kg body a distance
of 0.25 meters in 2 seconds. What is the power delivered by the student's
biceps?
5. Your household's monthly electric bill is often expressed in kilowatthours. One kilowatt-hour is the amount of energy delivered by the flow of l
kilowatt of electricity for one hour. Use conversion factors to show how
many joules of energy you get when you buy 1 kilowatt-hour of electricity.
6. An escalator is used to move 20 passengers every minute from the first floor of a department store to
the second. The second floor is located 5.20 meters above the first floor. The average passenger's mass is
54.9 kg. Determine the power requirement of the escalator in order to move this number of passengers in
this amount of time.
THERMAL PHYSICS
Simple Kinetic Molecular Model of Matter
Solids
1. In solids the molecules are closely packed by the forces between them.
2. They vibrate constantly about the fixed positions.
3. Solids have definite shape and volume.
Liquids
1. In liquids the molecules are slightly farther apart than in solids.
2. Molecules vibrate, at the same move rapidly over short distances, slipping past each other in all
directions.
3. Liquids don’t have definite shape but have definite volume
Gases
1. The molecules in gases are much farther apart than in solids or liquids.
2. Molecules move randomly at very high speed in all the space available. They hit each other and
collide with the wall of the container.
3. Gases don’t have definite volume and shape
Evaporation
When particles break away from the surface of a liquid and form vapours, this process is known as
evaporation.
Evaporation depends upon the following factors




Surface area
Temperature
Air blowing
Humidity
Difference between boiling and evaporation
Boiling
Evaporation
Occurs at only one
temperature, which
depends on the external
pressure.
Occurs at all temperature.
Takes place at surface
and within the
liquid(bubbles are
Takes place from surface only.
formed).
Rate of boiling depends
on the rate of supply of
heat.
Rate of evaporation depends on
the temperature, the area and the
surface and the state of the air
above the surface.
Thermal Properties
Thermal Expansion of Solids , Liquids and Gases
When matter is heated it expands (increases in size/volume) and when cooled it contracts.According to
the kinetic theory the molecules of solid and liquids and gases are in constant vibration. When heated,
they vibrate faster and force each other a little further apart. Expansion results , and this is greater for
liquid , gases expand even more.
Precautions against thermal expansion
1) Gaps must be left between lengths of railway track to allow the expansion of the steel rails in the hot
seasons.
2) Metal bridges must be constructed to allow for expansion. One end of the bridge is usually fixed and the
other end rests on rollers to allow movement due to expansion
Uses of thermal expansion
If equal lengths of two different metals, e.g. copper and iron , are fixed together so that they cannot move
separately , they form a bimetallic strip.When the strip is heated it bends , copper expands more than iron
and to allow this the strip bends with copper on the out side.
Bimetallic strips have many uses, few of them are mentioned here.
1) Fire alarm:- Heat from the fire makes the bimetallic strip bend and complete the electrical circuit , so
ringing the alarm bell.
2) Thermostat:- A thermostat keeps the temperature of a room or an appliance constant. Bimetallic strip
can be used in thermostat to control the temperature in a heating system such as an electric iron.
Unusual expansion of water
As water is cooled to 4oC it contracts as we would expect. However between 4oC – 0oC, it expands
surprisingly. Water has a maximum density at 4oC.The unusual expansion of water between 4oC –
0oC explains why fish survives in a frozen pond.
Temperature and Thermometers
We all have a feel for what temperature is. We even have a shared language that we use to qualitatively
describe temperature. The water in the shower or bathtub feels hot or cold or warm. The weather
outside is chilly or steamy. We certainly have a good feel for how one temperature is qualitatively
different than another temperature. We may not always agree on whether the room temperature is too
hot or too cold or just right. But we will likely all agree that we possess built-in thermometers for
making qualitative judgments about relative temperatures.
What is Temperature?
Despite our built-in feel for temperature, it remains one of those concepts in science that is difficult to
define. It seems that a tutorial page exploring the topic of temperature and thermometers should begin
with a simple definition of temperature. But it is at this point that I'm stumped. So I turn to that familiar
resource, Dictionary.com ... where I find definitions that vary from the simple-yet-not-too-enlightening
to the too-complex-to-be-enlightening. At the risk of doing a belly flop in the pool of enlightenment, I
will list some of those definitions here:





The degree of hotness or coldness of a body or environment.
A measure of the warmth or coldness of an object or substance with reference to some standard value.
A measure of the average kinetic energy of the particles in a sample of matter, expressed in terms of
units or degrees designated on a standard scale.
A measure of the ability of a substance, or more generally of any physical system, to transfer heat
energy to another physical system.
Any of various standardized numerical measures of this ability, such as the Kelvin, Fahrenheit, and
Celsius scale.
For certain, we are comfortable with the first two definitions the degree or measure of how hot or cold and object is. But
our understanding of temperature is not furthered by such
definitions. The third and the fourth definitions that reference
the kinetic energy of particles and the ability of a substance to
transfer heat are scientifically accurate. However, these
definitions are far too sophisticated to serve as good starting
points for a discussion of temperature. So we will resign to a
definition similar to the fifth one that is listed - temperature
can be defined as the reading on a thermometer. Admittedly, this definition lacks the power that is
needed for eliciting the much-desired Aha! Now I Understand! moment. Nonetheless it serves as a great
starting point for this lesson and heat and temperature. Temperatureis what the thermometer reads.
Whatever it is that temperature is a measure of, it is reflected by the reading on a thermometer. So
exactly how does a thermometer work? How does it reliably meter whatever it is that temperature is a
measure of?
How a Thermometer Works
Today, there are a variety of types of thermometers. The type that most of us are familiar with from
science class is the type that consists of a liquid encased in a narrow glass column. Older thermometers
of this type used liquid mercury. In response to our understanding of the health concerns associated
with mercury exposure, these types of thermometers usually use some type of liquid alcohol. These
liquid thermometers are based on the principal of thermal expansion. When a substance gets hotter, it
expands to a greater volume. Nearly all substances exhibit this behavior of thermal expansion. It is the
basis of the design and operation of thermometers.
As the temperature of the liquid in a thermometer increases, its volume increases. The liquid is enclosed
in a tall, narrow glass (or plastic) column with a constant cross-sectional area. The increase in volume is
thus due to a change in height of the liquid within the column. The increase in volume, and thus in the
height of the liquid column, is proportional to the increase in temperature. Suppose that a 10-degree
increase in temperature causes a 1-cm increase in the column's height. Then a 20-degree increase in
temperature will cause a 2-cm increase in the column's height. And a 30-degree increase in temperature
will cause s 3-cm increase in the column's height. The relationship between the temperature and the
column's height is linear over the small temperature range for which the thermometer is used. This
linear relationship makes the calibration of a thermometer a relatively easy task.
The calibration of any measuring tool involves the placement of divisions or marks upon the tool to
measure a quantity accurately in comparison to known standards. Any measuring tool - even a meter
stick - must be calibrated. The tool needs divisions or
markings; for instance, a meter stick typically has markings
every 1-cm apart or every 1-mm apart. These markings must
be accurately placed and the accuracy of their placement can
only be judged when comparing it to another object known
to have an accurate length.
A thermometer is calibrated by using two objects of known
temperatures. The typical process involves using the freezing
point and the boiling point of water. Water is known to freeze at 0°C and to boil at 100°C at an
atmospheric pressure of 1 atm. By placing a thermometer in mixture of ice water and allowing the
thermometer liquid to reach a stable height, the 0-degree mark can be placed upon the thermometer.
Similarly, by placing the thermometer in boiling water (at 1 atm of pressure) and allowing the liquid
level to reach a stable height, the 100-degree mark can be placed upon the thermometer. With these two
markings placed upon the thermometer, 100 equally spaced divisions can be placed between them to
represent the 1-degree marks. Since there is a linear relationship between the temperature and the height
of the liquid, the divisions between 0 degree and 100 degree can be equally spaced. With a calibrated
thermometer, accurate measurements can be made of the temperature of any object within the
temperature range for which it has been calibrated.
What is Heat?
Earlier in this lesson, five dictionary style definitions of temperature were given. They were:





The degree of hotness or coldness of a body or environment.
A measure of the warmth or coldness of an object or substance with reference to some standard value.
A measure of the average kinetic energy of the particles in a sample of matter, expressed in terms of
units or degrees designated on a standard scale.
A measure of the ability of a substance, or more generally of any physical system, to transfer heat
energy to another physical system.
Any of various standardized numerical measures of this ability, such as the Kelvin, Fahrenheit, and
Celsius scale
As mentioned, the first two bullet points have rather obvious meanings. The third bullet point was the
topic of the previous page in this lesson. The fifth bullet point was the definition that we started with as
we discussed temperature and the operation of thermometers; it was the topic of the second page in this
lesson. That leaves us with the fourth bullet point - defining temperature in terms of the ability of a
substance to transfer heat to another substance. This part of Lesson 1 is devoted to understanding how
the relative temperature of two objects effects the direction that heat is transferred between the two
objects.
What is Heat?
Consider a very hot mug of coffee on the countertop of your kitchen. For discussion purposes, we will
say that the cup of coffee has a temperature of 80°C and that the surroundings (countertop, air in the
kitchen, etc.) has a temperature of 26°C. What do you suppose will happen in this situation? I suspect
that you know that the cup of coffee will gradually cool down over time. At 80°C, you wouldn't dare
drink the coffee. Even the coffee mug will likely be too hot to touch. But over time, both the coffee
mug and the coffee will cool down. Soon it will be at a drinkable temperature. And if you resist the
temptation to drink the coffee, it will eventually reach room temperature. The coffee cools from 80°C to
about 26°C. So what is happening over the course of time to cause the coffee to cool down? The answer
to this question can be both macroscopic and particulate in nature.
On the macroscopic level, we would say that the coffee and the
mug are transferring heat to the surroundings. This transfer of
heat occurs from the hot coffee and hot mug to the surrounding
air. The fact that the coffee lowers its temperature is a sign that
the average kinetic energy of its particles is decreasing. The
coffee is losing energy. The mug is also lowering its temperature;
the average kinetic energy of its particles is also decreasing. The
mug is also losing energy. The energy that is lost by the coffee
and the mug is being transferred to the colder surroundings. We
refer to this transfer of energy from the coffee and the mug to the surrounding air and countertop as
heat. In this sense, heat is simply the transfer of energy from a hot object to a colder object.
Now let's consider a different scenario - that of a cold can of pop placed on the same kitchen counter.
For discussion purposes, we will say that the pop and the can which contains it has a temperature of
5°C and that the surroundings (countertop, air in the kitchen, etc.) has a temperature of 26°C. What will
happen to the cold can of pop over the course of time? Once more, I suspect that you know the answer.
The cold pop and the container will both warm up to room temperature. But what is happening to cause
these colder-than-room-temperature objects to increase their temperature? Is the cold escaping from the
pop and its container? No! There is no such thing as the cold escaping or leaking. Rather, our
explanation is very similar to the explanation used to explain why the coffee cools down. There is a
heat transfer.
Over time, the pop and the container increase their
temperature. The temperature rises from 5°C to nearly 26°C.
This increase in temperature is a sign that the average kinetic
energy of the particles within the pop and the container is
increasing. In order for the particles within the pop and the
container to increase their kinetic energy, they must be
gaining energy from somewhere. But from where? Energy is
being transferred from the surroundings (countertop, air in the kitchen, etc.) in the form of heat. Just as
in the case of the cooling coffee mug, energy is being transferred from the higher temperature objects to
the lower temperature object. Once more, this is known as heat - the transfer of energy from the higher
temperature object to a lower temperature object.
Both of these scenarios could be summarized by two simple statements. An object
decreases its temperature by releasing energy in the form of heat to its surroundings.
And an object increases its temperature by gaining energy in the form of heat from
its surroundings. Both the warming up and the cooling down of objects works in the
same way - by heat transfer from the higher temperature object to the lower
temperature object. So now we can meaningfully re-state the definition of temperature. Temperature is
a measure of the ability of a substance, or more generally of any physical system, to transfer heat
energy to another physical system. The higher the temperature of an object is, the greater the tendency
of that object to transfer heat. The lower the temperature of an object is, the greater the tendency of that
object to be on the receiving end of the heat transfer.
But perhaps you have been asking: what happens to the temperature of surroundings? Do the
countertop and the air in the kitchen increase their temperature when the mug and the coffee cool
down? And do the countertop and the air in the kitchen decrease its temperature when the can and its
pop warm up? The answer is a resounding Yes! The proof? Just touch the countertop - it should feel
cooler or warmer than before the coffee mug or pop can were placed on the countertop. But what about
the air in the kitchen? Now that's a little more difficult to present a convincing proof of. The fact that
the volume of air in the room is so large and that the energy quickly diffuses away from the surface of
the mug and of the means that the temperature change of the air in the kitchen will be abnormally small.
In fact, it will be negligibly small. There would have to be a lot more heat transfer before there is a
noticeable temperature change.
Thermal Equilibrium
In the discussion of the cooling of the coffee mug, the countertop and the air in the kitchen were
referred to as the surroundings. It is common in physics discussions of this type to use a mental
framework of a system and the surroundings. The coffee mug (and the coffee) would be regarded as the
system and everything else in the universe would be regarded as the surroundings. To keep it simple,
we often narrow the scope of the surroundings from the rest of the universe to simply those objects that
are immediately surrounding the system. This approach of analyzing a situation in terms of system and
surroundings is so useful that we will adopt the approach for the rest of this chapter and the next.
Now let's imagine a third situation. Suppose that a small metal cup of hot water is placed inside of a
larger Styrofoam cup of cold water. Let's suppose that the temperature of the hot water is initially 70°C
and that the temperature of the cold water in the outer cup is initially 5°C. And let's suppose that both
cups are equipped with thermometers (or temperature probes) that measure the temperature of the water
in each cup over the course of time. What do you suppose will happen? Before you read on, think about
the question and commit to some form of answer. When the cold water is done warming and the hot
water is done cooling, will their temperatures be the same or different? Will the cold water warm up to
a lower temperature than the temperature that the hot water cools down to? Or as the warming and
cooling occurs, will their temperatures cross each other?
Fortunately, this is an experiment that can be done and in fact has been done on many occasions. The
graph below is a typical representation of the results.
As you can see from the graph, the hot water cooled down to approximately 30°C and the cold water
warmed up to approximately the same temperature. Heat is transferred from the high temperature object
(inner can of hot water) to the low temperature object (outer can of cold water). If we designate the
inner cup of hot water as the system, then we can say that there is a flow of water from the system to the
surroundings. As long as there is a temperature difference between the system and the surroundings,
there is a heat flow between them. The heat flow is more rapid at first as depicted by the steeper slopes
of the lines. Over time, the temperature difference between system and surroundings decreases and the
rate of heat transfer decreases. This is denoted by the gentler slope of the two lines. (Detailed
information about rates of heat transfer will be discussed later in this lesson.) Eventually, the system
and the surroundings reach the same temperature and the heat transfer ceases. It is at this point, that the
two objects are said to have reached thermal equilibrium.
In our chapter on electric circuits, we learned that a difference in electric potential between two
locations causes a flow of charge along a conducting path between those locations. As long as an
electric potential difference is maintained, a flow of charge will exist. Now in this chapter we learn a
similar principle related to the flow of heat. A temperature difference between two locations will cause
a flow of heat along a (thermally) conducting path between those two locations. As long as the
temperature difference is maintained, a flow of heat will occur. This flow of heat continues until the
two objects reach the same temperature. Once their temperatures become equal, they are said to be at
thermal equilibrium and the flow of heat no longer takes place.
This principle is sometimes referred to as the zeroeth law of thermodynamics. This principle became
formalized into a law after the first, second and third laws of thermodynamics had already been
discovered. But because the law seemed more fundamental than the previously discovered three, it was
titled the zeroeth law. All objects are governed by this law - this tendency towards thermal equilibrium.
It represents a daily challenge for those who wish to control the temperature of their bodies, their food,
their drinks and their homes. We use ice and insulation to try to keep our cold drinks cold and we use
insulation and ongoing pulses of microwave energy to keep our hot drinks hot. We equip our vehicles,
our homes and our office buildings equipped with air conditioners and fans in order to keep them cool
during the warm summer months. And we equip these same vehicles and buildings with furnaces and
heaters in order to keep them warm during the cold winter months. Whenever any of these systems are
at a different temperature as the surroundings and not perfectly insulated from the surroundings (an
ideal situation), heat will flow. This heat flow will continue until the system and surroundings have
achieved equal temperatures. Because these systems have a considerably smaller volume than the
surroundings, there will be a more noticeable and substantial change in temperature of these systems.
Methods of Heat Transfer
If you have been following along since the beginning of this lesson, then you have been developing a
progressively sophisticated understanding of temperature and heat. You should be developing a model
of matter as consisting of particles which vibrate (wiggle about a fixed position), translate (move from
one location to another) and even rotate (revolve about an imaginary axis). These motions give the
particles kinetic energy. Temperature is a measure of the average amount of kinetic energy possessed
by the particles in a sample of matter. The more the particles vibrate, translate and rotate, the greater the
temperature of the object. You have hopefully adopted an understanding of heat as a flow of energy
from a higher temperature object to a lower temperature object. It is the temperature difference between
the two neighboring objects that causes this heat transfer. The heat transfer continues until the two
objects have reached thermal equilibrium and are at the same temperature. The discussion of heat
transfer has been structured around some everyday examples such as the cooling of a hot mug of coffee
and the warming of a cold can of pop. Finally, we have explored a thought experiment in which a metal
can containing hot water is placed within a Styrofoam cup containing cold water. Heat is transferred
from the hot water to the cold water until both samples have the same
temperature.
Now we should probe some of the following questions:




What is happening at the particle level when energy is being
transferred between two objects?
Why is thermal equilibrium always established when two objects
transfer heat?
How does heat transfer work within the bulk of an object?
Is there more than one method of heat transfer? If so, then how are they similar and different than one
another?
Conduction - A Particulate View
Let's begin our discussion by returning to our thought experiment in which a metal can containing hot
water was placed within a Styrofoam cup containing cold water. Heat is transferred from the hot water
to the cold water until both samples have the same temperature. In this instance, the transfer of heat
from the hot water through the metal can to the cold water is sometimes referred to as conduction.
Conductive heat flow involves the transfer of heat from one location to another in the absence of any
material flow. There is nothing physical or material moving from the hot water to the cold water. Only
energy is transferred from the hot water to the cold water. Other than the loss of energy, there is nothing
else escaping from the hot water. And other than the gain of energy, there is nothing else entering the
cold water. How does this happen? What is the mechanism that makes conductive heat flow possible?
A question like this is a particle-level question. To understand the
answer, we have to think about matter as consisting of tiny particles
atoms, molecules and ions. These particles are in constant motion;
this gives them kinetic energy. As mentioned previously in this
lesson, these particles move throughout the space of a container,
colliding with each other and with the walls of their container. This is
known as translational kinetic energy and is the main form of kinetic
energy for gases and liquids. But these particles can also vibrate
about a fixed position. This gives the particles vibrational kinetic
energy and is the main form of kinetic energy for solids. To put it
more simply, matter consists of little wigglers and little bangers. The
wigglers are those particles vibrating about a fixed position. They
possess vibrational kinetic energy. The bangers are those particles that move through the container with
translational kinetic energy and collide with the container walls.
The container walls represent the perimeters of a sample of matter. Just as the perimeter of your
property (as in real estate property) is the furthest extension of the property, so the perimeter of an
object is the furthest extension of the particles within a sample of matter. At the perimeter, the little
bangers are colliding with particles of another substance - the particles of the container or even the
surrounding air. Even the wigglers that are fixed in a position along the perimeter are doing some
banging. Being at the perimeter, their wiggling results in collisions with the particles that are next to
them; these are the particles of the container or of the surrounding air.
At this perimeter or boundary, the collisions of the little bangers and wigglers
are elastic collisions in which the total amount of kinetic energy of all
colliding particles is conserved. The net effect of these elastic collisions is that
there is a transfer of kinetic energy across the boundary to the particles on the
opposite side. The more energetic particles will lose a little kinetic energy and
the less energetic particles will gain a little kinetic energy. Temperature is a measure of the average
amount of kinetic energy possessed by the particles in a sample of matter. So on average, there are
more particles in the higher temperature object with greater kinetic energy than there are in the lower
temperature object. So when we average all the collisions together and apply the principles associated
with elastic collisions to the particles within a sample of matter, it is logical to conclude that the higher
temperature object will lose some kinetic energy and the lower temperature object will gain some
kinetic energy. The collisions of our little bangers and wigglers will continue to transfer energy until
the temperatures of the two objects are identical. When this state of thermal equilibrium has been
reached, the average kinetic energy of both objects' particles is equal. At thermal equilibrium, there are
an equal number of collisions resulting in an energy gain as there are collisions resulting in an energy
loss. On average, there is no net energy transfer resulting from the collisions of particles at the
perimeter.
At the macroscopic level, heat is the transfer of energy from the
high temperature object to the low temperature object. At the
particulate level, heat flow can be explained in terms of the net
effect of the collisions of a whole bunch of little bangers.
Warming and cooling is the macroscopic result of this particle-
level phenomenon. Now let's apply this particle view to the scenario of the metal can with the hot water
positioned inside of a Styrofoam cup containing cold water. On average, the particles with the greatest
kinetic energy are the particles of the hot water. Being a fluid, those particles move about with
translational kinetic energy and bang upon the particles of the metal can. As the hot water particles
bang upon the particles of the metal can, they transfer energy to the metal can. This warms the metal
can up. Most metals are good thermal conductors so they warm up quite quickly throughout the bulk of
the can. The can assumes nearly the same temperature as the hot water. Being a solid, the metal can
consists of little wigglers. The wigglers at the outer perimeter of the metal can bang upon particles in
the cold water. The collisions between the particles of the metal can and the particles of the cold water
result in the transfer of energy to the cold water. This slowly warms the cold water up. The interaction
between the particles of the hot water, the metal can and the cold water results in a transfer of energy
outward from the hot water to the cold water. The average kinetic energy of the hot water particles
gradually decreases; the average kinetic energy of the cold-water particles gradually increases; and
eventually, thermal equilibrium would be reached at the point that the particles of the hot water and the
cold water have the same average kinetic energy. At the macroscopic level, one would observe a
decrease in temperature of the hot water and an increase in temperature of the cold water.
The mechanism in which heat is transferred from one object to another object through particle
collisions is known as conduction. In conduction, there is no net transfer of physical stuff between the
objects. Nothing material moves across the boundary. The changes in temperature are wholly explained
as the result of the gains and losses of kinetic energy during collisions.
Conduction Through The Bulk of an Object
We have discussed how heat transfers from one object to another through conduction. But how does it
transfer through the bulk of an object? For instance, suppose we pull a ceramic coffee mug out of the
cupboard and place it on the countertop. The mug is at room temperature - maybe at 26°C. Then
suppose we fill the ceramic coffee mug with hot coffee at a temperature of 80°C. The mug quickly
warms up. Energy first flows into the particles at the boundary between the hot coffee and the ceramic
mug. But then it flows through the bulk of the ceramic to all parts of the ceramic mug. How does heat
conduction occur in the ceramic itself?
The mechanism of heat transfer through the bulk of the ceramic mug is described in a similar manner as
it before. The ceramic mug consists of a collection of orderly arranged wigglers. These are particles that
wiggle about a fixed position. As the ceramic particles at the boundary between the hot coffee and the
mug warm up, they attain a kinetic energy that is much higher than their neighbors. As they wiggle
more vigorously, they bang into their neighbors and increase their vibrational kinetic energy. These
particles in turn begin to wiggle more vigorously and their collisions with their neighbors increase their
vibrational kinetic energy. The process of energy transfer by means of the little bangers continues from
the particles at the inside of the mug (in contact with the coffee particles) to the outside of the mug (in
contact with the surrounding air). Soon the entire coffee mug is warm and your hand feels it.
This mechanism of conduction by particle-to-particle interaction is very common in ceramic materials
such as a coffee mug. Does it work the same in metal objects? For instance, you likely have noticed the
high temperatures attained by the metal handle of a skillet when placed upon a stovetop. The burners on
the stove transfer heat to the metal skillet. If the handle of the skillet is metallic, it too attains a high
temperature, certainly high enough to cause a bad burn. The transfer of heat from the skillet to the
skillet handle occurs by conduction. But in metals, the conduction mechanism is slightly more
complicated. In a manner similar to electrical conductivity, thermal conductivity in metals occurs by the
movement of free electrons. Outer shell electrons of metal atoms are shared among atoms and are free
to move throughout the bulk of the metal. These electrons carry the energy from the skillet to the skillet
handle. The details of this mechanism of thermal conduction in metals are considerably more complex
than the discussion given here. The main point to grasp is that heat transfer through metals occurs
without any movement of atoms from the skillet to the skillet handle. This qualifies the heat transfer as
being categorized as thermal conduction.
Heat Transfer by Convection
Is conduction the only means of heat transfer? Can heat be transferred
across through the bulk of an object in methods other than conduction?
The answer is yes. The model of heat transfer through the ceramic coffee
mug and the metal skillet involved conduction. The ceramic of the coffee
mug and the metal of the skillet are both solids. Heat transfer through solids occurs by conduction. This
is primarily due to the fact that solids have orderly arrangements of particles that are fixed in place.
Liquids and gases are not very good conductors of heat. In fact, they are considered good thermal
insulators. Heat typically does not flow through liquids and gases by means of conduction. Liquids and
gases are fluids; their particles are not fixed in place; they move about the bulk of the sample of matter.
The model used for explaining heat transfer through the bulk of liquids and gases involves convection.
Convection is the process of heat transfer from one location to the next by the movement of fluids. The
moving fluid carries energy with it. The fluid flows from a high temperature location to a low
temperature location.
To understand convection in fluids, let's consider the heat transfer
through the water that is being heated in a pot on a stove. Of
course the source of the heat is the stove burner. The metal pot
that holds the water is heated by the stove burner. As the metal
becomes hot, it begins to conduct heat to the water. The water at
the boundary with the metal pan becomes hot. Fluids expand
when heated and become less dense. So as the water at the
bottom of the pot becomes hot, its density decreases. Differences
in water density between the bottom of the pot and the top of the
pot results in the gradual formation of circulation currents. Hot
water begins to rise to the top of the pot displacing the colder
water that was originally there. And the colder water that was
present at the top of the pot moves towards the bottom of the pot where it is heated and begins to rise.
These circulation currents slowly develop over time, providing the pathway for heated water to transfer
energy from the bottom of the pot to the surface.
Convection also explains how an electric heater placed on the
floor of a cold room warms up the air in the room. Air present
near the coils of the heater warm up. As the air warms up, it
expands, becomes less dense and begins to rise. As the hot air
rises, it pushes some of the cold air near the top of the room out
of the way. The cold air moves towards the bottom of the room
to replace the hot air that has risen. As the colder air approaches
the heater at the bottom of the room, it becomes warmed by the
heater and begins to rise. Once more, convection currents are
slowly formed. Air travels along these pathways, carrying
energy with it from the heater throughout the room.
Convection is the main method of heat transfer in fluids such as
water and air. It is often said that heat rises in these situations.
The more appropriate explanation is to say that heated fluid rises. For instance, as the heated air rises
from the heater on a floor, it carries more energetic particles with it. As the more energetic particles of
the heated air mix with the cooler air near the ceiling, the average kinetic energy of the air near the top
of the room increases. This increase in the average kinetic energy corresponds to an increase in
temperature. The net result of the rising hot fluid is the transfer of heat from one location to another
location. The convection method of heat transfer always involves the transfer of heat by the movement
of matter. This is not to be confused with the caloric theory discussed earlier in this lesson. In caloric
theory, heat was the fluid and the fluid that moved was the heat. Our model of convection considers
heat to be energy transfer that is simply the result of the movement of more energetic particles.
The two examples of convection discussed here - heating water in a pot and heating air in a room - are
examples of natural convection. The driving force of the circulation of fluid is natural - differences in
density between two locations as the result of fluid being heated at some source. (Some sources
introduce the concept of buoyant forces to explain why the heated fluids rise. We will not pursue such
explanations here.) Natural convection is common in nature. The earth's oceans and atmosphere are
heated by natural convection. In contrast to natural convection, forced convection involves fluid being
forced from one location to another by fans, pumps and other devices. Many home heating systems
involve force air heating. Air is heated at a furnace and blown by fans through ductwork and released
into rooms at vent locations. This is an example of forced convection. The movement of the fluid from
the hot location (near the furnace) to the cool location (the rooms throughout the house) is driven or
forced by a fan. Some ovens are forced convection ovens; they have fans that blow heated air from a
heat source into the oven. Some fireplaces enhance the heating ability of the fire by blowing heated air
from the fireplace unit into the adjacent room. This is another example of forced convection.
Heat Transfer by Radiation
A final method of heat transfer involves radiation. Radiation is the transfer of heat by means of
electromagnetic waves. To radiate means to send out or spread from a central location. Whether it is
light, sound, waves, rays, flower petals, wheel spokes or pain, if something radiates then it protrudes or
spreads outward from an origin. The transfer of heat by radiation involves the carrying of energy from
an origin to the space surrounding it. The energy is carried by electromagnetic waves and does not
involve the movement or the interaction of matter. Thermal radiation can occur through matter or
through a region of space that is void of matter (i.e., a vacuum). In fact, the heat received on Earth from
the sun is the result of electromagnetic waves traveling through the void of space between the Earth and
the sun.
All objects radiate energy in the form of electromagnetic waves. The rate at which this energy is
released is proportional to the Kelvin temperature (T) raised to the fourth power.
Radiation rate = k•T4
The hotter the object, the more it radiates. The sun obviously radiates off more energy than a hot mug
of coffee. The temperature also affects the wavelength and frequency of the radiated waves. Objects at
typical room temperatures radiate energy as infrared waves. Being invisible to the human eye, we do
not see this form of radiation. An infrared camera is capable of detecting such radiation. Perhaps you
have seen thermal photographs or videos of the radiation surrounding a person or animal or a hot mug
of coffee or the Earth. The energy radiated from an object is usually a collection or range of
wavelengths. This is usually referred to as an emission spectrum. As the temperature of an object
increases, the wavelengths within the spectra of the emitted radiation also decrease. Hotter objects tend
to emit shorter wavelength, higher frequency radiation. The coils of an electric toaster are considerably
hotter than room temperature and emit electromagnetic radiation in the visible spectrum. Fortunately,
this provides a convenient warning to its users that the coils are hot. The tungsten filament of an
incandescent light bulb emits electromagnetic radiation in the visible (and beyond) range. This radiation
not only allows us to see, it also warms the glass bulb that contains the filament. Put your hand near the
bulb (without touching it) and you will feel the radiation from the bulb as well.
Thermal radiation is a form of heat transfer because the
electromagnetic radiation emitted from the source carries energy
away from the source to surrounding (or distant) objects. This energy
is absorbed by those objects, causing the average kinetic energy of
their particles to increase and causing the temperatures to rise. In this
sense, energy is transferred from one location to another by means of
electromagnetic radiation. The image at the right was taken by a
thermal imaging camera. The camera detects the radiation emitted by
objects and represents it by means of a color photograph. The hotter colors represent areas of objects
that are emitting thermal radiation at a more intense rate. (Images courtesy Peter Lewis and Chris West
of Standford's SLAC.)
Difference between boiling and melting
The process due to which a solid changes into liquid state by absorbing heat energy is called melting or fusion
and the constant temperature at which a solid changes into liquid state by absorbing heat energy is called melting
point.
Again, the process due to which a liquid changes into gaseous state by absorbing heat energy is called boiling or
vaporisation and the constant temperature at which a liquid rapidly changes into gaseous state by absorbing heat
energy is called boiling point.
WAVES
The Nature of a Wave
Waves and Wavelike Motion | What is a Wave? | Categories of Waves
Waves and Wavelike Motion
Waves are everywhere. Whether we recognize it or not, we encounter waves on a
daily basis. Sound waves, visible light waves, radio waves, microwaves, water
waves, sine waves, cosine waves, stadium waves, earthquake waves, waves on a
string, and slinky waves and are just a few of the examples of our daily
encounters with waves. In addition to waves, there are a variety of phenomena in
our physical world that resemble waves so closely that we can describe such
phenomenon as being wavelike. The motion of a pendulum, the motion of a mass
suspended by a spring, the motion of a child on a swing, and the "Hello, Good
Morning!" wave of the hand can be thought of as wavelike phenomena. Waves
(and wavelike phenomena) are everywhere!
We study the physics of waves because it provides a rich glimpse into the physical world that we seek to
understand and describe as students of physics. Before beginning a formal discussion of the nature of
waves, it is often useful to ponder the various encounters and exposures that we have of waves. Where do
we see waves or examples of wavelike motion? What experiences do we already have that will help us in
understanding the physics of waves?
For many people, the first thought concerning waves conjures up a
picture of a wave moving across the surface of an ocean, lake, pond or
other body of water. The waves are created by some form of a
disturbance, such as a rock thrown into the water, a duck shaking its tail
in the water or a boat moving through the water. The water wave has a
crest and a trough and travels from one location to another. One crest is
often followed by a second crest that is often followed by a third crest.
Every crest is separated by a trough to create an alternating pattern of
crests and troughs. A duck or gull at rest on the surface of the water is
observed to bob up-and-down at rather regular time intervals as the
wave passes by. The waves may appear to be plane waves that travel
together as a front in a straight-line direction, perhaps towards a sandy
shore. Or the waves may be circular waves that originate from the point where the disturbances occur;
such circular waves travel across the surface of the water in all directions. These mental pictures of water
waves are useful for understanding the nature of a wave and will be revisited later when we begin our
formal discussion of the topic.
The thought of waves often brings to mind a recent encounter at the baseball or football stadium when the
crowd enthusiastically engaged in doing the wave. When performed with reasonably good timing, a
noticeable ripple is produced that travels around the circular stadium or back and forth across a section of
bleachers. The observable ripple results when a group of enthusiastic fans rise up from their seats, swing
their arms up high, and then sit back down. Beginning in Section 1, the first row of fans abruptly rise up
to begin the wave; as they sit back down, row 2 begins its motion; as row 2 sits back down, row 3 begins
its motion. The process continues, as each consecutive row becomes involved by a momentary standing
up and sitting back down. The wave is passed from row to row as each individual member of the row
becomes temporarily displaced out of his or her seat, only to return to it as the wave passes by. This
mental picture of a stadium wave will also provide a useful context for the discussion of the physics of
wave motion.
Another picture of waves involves the movement of a
slinky or similar set of coils. If a slinky is stretched out
from end to end, a wave can be introduced into the slinky
by either vibrating the first coil up and down vertically or
back and forth horizontally. A wave will subsequently be
seen traveling from one end of the slinky to the other. As
the wave moves along the slinky, each individual coil is
seen to move out of place and then return to its original
position. The coils always move in the same direction that
the first coil was vibrated. A continued vibration of the
first coil results in a continued back and forth motion of
the other coils. If looked at closely, one notices that the wave does not stop when it reaches the end of the
slinky; rather it seems to bounce off the end and head back from where it started. A slinky wave provides
an excellent mental picture of a wave and will be used in discussions and demonstrations throughout this
unit.
We likely have memories from childhood of holding a long jump rope with a friend and vibrating an end
up and down. The up and down vibration of the end of the rope created a disturbance of the rope that
subsequently moved towards the other end. Upon reaching the opposite end, the disturbance often
bounced back to return to the end we were holding. A single disturbance could be created by the single
vibration of one end of the rope. On the other hand, a repeated disturbance would result in a repeated and
regular vibration of the rope. The shape of the pattern formed in the rope was influenced by the frequency
at which we vibrated it. If we vibrated the rope rapidly, then a short wave was created. And if we vibrated
the rope less frequently (not as often), a long wave was created. While we were likely unaware of it as
children, we were entering the world of the physics of waves as we contentedly played with the rope.
Then there is the "Hello, Good Morning!" wave. Whether encountered in the driveway as you begin your
trip to school, on the street on the way to school, in the parking lot upon arrival to school, or in the
hallway on the way to your first class, the "Hello, Good Morning!" wave provides a simple (yet excellent)
example of physics in action. The simple back and forth motion of the hand is called a wave. When Mom
commands us to "wave to Mr. Smith," she is telling us to raise our hand and to temporarily or even
repeatedly vibrate it back and forth. The hand is raised, moved to the left, and then back to the far right
and finally returns to its original position. Energy is put into the hand and the hand begins its back-andforth vibrational motion. And we call the process of doing it "waving." Soon we will see how this simple
act is representative of the nature of a physical wave.
Finally, we are familiar with microwaves and visible light waves. While we have never seen them, we
believe that they exist because we have witnessed how they carry energy from one location to another.
And similarly, we are familiar with radio waves and sound waves. Like microwaves, we have never seen
them. Yet we believe they exist because we have witnessed the signals that they carry from one location
to another and we have even learned how to tune into those signals through use of our ears or a tuner on a
television or radio. Waves, as we will learn, carry energy from one location to another. And if the
frequency of those waves can be changed, then we can also carry a complex signal that is capable of
transmitting an idea or thought from one location to another. Perhaps this is one of the most important
aspects of waves and will become a focus of our study in later units.
Waves are everywhere in nature. Our understanding of the physical world is not complete until we
understand the nature, properties and behaviors of waves. The goal of this unit is to develop mental
models of waves and ultimately apply those models to an understanding of the two most common types of
waves - sound waves and light waves.
What is a Wave?
So waves are everywhere. But what makes a wave a wave? What characteristics, properties, or behaviors
are shared by the phenomena that we typically characterize as being a wave? How can waves be
described in a manner that allows us to understand their basic nature and qualities?
A wave can be described as a disturbance that travels through a medium from one location to another
location. Consider a slinky wave as an example of a wave. When the slinky is stretched from end to end
and is held at rest, it assumes a natural position known as the equilibrium or rest position. The coils of
the slinky naturally assume this position, spaced equally far apart. To introduce a wave into the slinky, the
first particle is displaced or moved from its equilibrium or rest position. The particle might be moved
upwards or downwards, forwards or backwards; but once moved, it is returned to its original equilibrium
or rest position. The act of moving the first coil of the slinky in a given direction and then returning it to
its equilibrium position creates a disturbance in the slinky. We can then observe this disturbance moving
through the slinky from one end to the other. If the first coil of the slinky is given a single back-and-forth
vibration, then we call the observed motion of the disturbance through the slinky a slinky pulse. A pulse
is a single disturbance moving through a medium from one location to another location. However, if the
first coil of the slinky is continuously and periodically vibrated in a back-and-forth manner, we would
observe a repeating disturbance moving within the slinky that endures over some prolonged period of
time. The repeating and periodic disturbance that moves through a medium from one location to another
is referred to as a wave.
What is a Medium?
But what is meant by the word medium? A medium is a substance or material that carries the wave. You
have perhaps heard of the phrase news media. The news media refers to the various institutions
(newspaper offices, television stations, radio stations, etc.) within our society that carry the news from
one location to another. The news moves through the media. The media doesn't make the news and the
media isn't the same as the news. The news media is merely the thing that carries the news from its source
to various locations. In a similar manner, a wave medium is the substance that carries a wave (or
disturbance) from one location to another. The wave medium is not the wave and it doesn't make the
wave; it merely carries or transports the wave from its source to other locations. In the case of our slinky
wave, the medium through that the wave travels is the slinky coils. In the case of a water wave in the
ocean, the medium through which the wave travels is the ocean water. In the case of a sound wave
moving from the church choir to the pews, the medium through which the sound wave travels is the air in
the room. And in the case of the stadium wave, the medium through which the stadium wave travels is the
fans that are in the stadium.
Particle-to-Particle Interaction
To fully understand the nature of a wave, it is important to consider the medium as a collection of
interacting particles. In other words, the medium is composed of parts that are capable of interacting with
each other. The interactions of one particle of the medium with the next adjacent particle allow the
disturbance to travel through the medium. In the case of the slinky wave, the particles or interacting parts
of the medium are the individual coils of the slinky. In the case of a sound wave in air, the particles or
interacting parts of the medium are the individual molecules of air. And in the case of a stadium wave, the
particles or interacting parts of the medium are the fans in the stadium.
Consider the presence of a wave in a slinky. The first coil becomes disturbed and begins to push or pull
on the second coil; this push or pull on the second coil will
displace the second coil from its equilibrium position. As the
second coil becomes displaced, it begins to push or pull on the
third coil; the push or pull on the third coil displaces it from its
equilibrium position. As the third coil becomes displaced, it
begins to push or pull on the fourth coil. This process
continues in consecutive fashion, with each individual particle
acting to displace the adjacent particle. Subsequently, the
disturbance travels through the medium. The medium can be
pictured as a series of particles connected by springs. As one
particle moves, the spring connecting it to the next particle begins to stretch and apply a force to its
adjacent neighbor. As this neighbor begins to move, the spring attaching this neighbor to its neighbor
begins to stretch and apply a force on its adjacent neighbor.
A Wave Transports Energy and Not Matter
When a wave is present in a medium (that is, when there is a disturbance moving through a medium), the
individual particles of the medium are only temporarily displaced from their rest position. There is always
a force acting upon the particles that restores them to their original position. In a slinky wave, each coil of
the slinky ultimately returns to its original position. In a water wave, each molecule of the water
ultimately returns to its original position. And in a stadium wave, each fan in the bleacher ultimately
returns to its original position. It is for this reason, that a wave is said to involve the movement of a
disturbance without the movement of matter. The particles of the medium (water molecules, slinky coils,
stadium fans) simply vibrate about a fixed position as the pattern of the disturbance moves from one
location to another location.
Waves are said to be an energy transport phenomenon. As a disturbance moves through a medium from
one particle to its adjacent particle, energy is being transported from one end of the medium to the other.
In a slinky wave, a person imparts energy to the first coil by doing work upon it. The first coil receives a
large amount of energy that it subsequently transfers to the second coil. When the first coil returns to its
original position, it possesses the same amount of energy as it had before it was displaced. The first coil
transferred its energy to the second coil. The second coil then has a large amount of energy that it
subsequently transfers to the third coil. When the second coil returns to its original position, it possesses
the same amount of energy as it had before it was displaced. The third coil has received the energy of the
second coil. This process of energy transfer continues as each coil interacts with its neighbor. In this
manner, energy is transported from one end of the slinky to the other, from its source to another location.
This characteristic of a wave as an energy transport phenomenon distinguishes waves from other types of
phenomenon. Consider a common phenomenon observed at a softball game - the collision of a bat with a
ball. A batter is able to transport energy from her to the softball by means of a bat. The batter applies a
force to the bat, thus imparting energy to the bat in the form of kinetic energy. The bat then carries this
energy to the softball and transports the energy to the softball upon collision. In this example, a bat is
used to transport energy from the player to the softball. However, unlike wave phenomena, this
phenomenon involves the transport of matter. The bat must move from its starting location to the contact
location in order to transport energy. In a wave phenomenon, energy can move from one location to
another, yet the particles of matter in the medium return to their fixed position. A wave transports its
energy without transporting matter.
Waves are seen to move through an ocean or lake; yet the water always returns to its rest position. Energy
is transported through the medium, yet the water molecules are not transported. Proof of this is the fact
that there is still water in the middle of the ocean. The water has not moved from the middle of the ocean
to the shore. If we were to observe a gull or duck at rest on the water, it would merely bob up-and-down
in a somewhat circular fashion as the disturbance moves through the water. The gull or duck always
returns to its original position. The gull or duck is not transported to the shore because the water on which
it rests is not transported to the shore. In a water wave, energy is transported without the transport of
water.
The same thing can be said about a stadium wave. In a stadium wave, the fans do not get out of their seats
and walk around the stadium. We all recognize that it would be silly (and embarrassing) for any fan to
even contemplate such a thought. In a stadium wave, each fan rises up and returns to the original seat. The
disturbance moves through the stadium, yet the fans are not transported. Waves involve the transport of
energy without the transport of matter.
In conclusion, a wave can be described as a disturbance that travels through a medium, transporting
energy from one location (its source) to another location without transporting matter. Each individual
particle of the medium is temporarily displaced and then returns to its original equilibrium positioned.
Check Your Understanding
1. TRUE or FALSE:
In order for John to hear Jill, air molecules must move from the lips of Jill to the ears of John.
2. Curly and Moe are conducting a wave experiment using a slinky. Curly introduces a disturbance into
the slinky by giving it a quick back and forth jerk. Moe places his cheek (facial) at the opposite end of the
slinky. Using the terminology of this unit, describe what Moe experiences as the pulse reaches the other
end of the slinky.
3. Mac and Tosh are experimenting with pulses on a rope. They vibrate an end up and down to create the
pulse and observe it moving from end to end. How does the position of a point on the rope, before the
pulse comes, compare to the position after the pulse has passed?
4. Minute after minute, hour after hour, day after day, ocean waves continue to splash onto the shore.
Explain why the beach is not completely submerged and why the middle of the ocean has not yet been
depleted of its water supply.
5. A medium is able to transport a wave from one location to another because the particles of the medium
are ____.
a. frictionless
b. isolated from one another
c. able to interact
d. very light
Categories of Waves
Waves come in many shapes and forms. While all waves share some basic characteristic properties and
behaviors, some waves can be distinguished from others based on some observable (and some nonobservable) characteristics. It is common to categorize waves based on these distinguishing
characteristics.
Longitudinal versus Transverse Waves versus Surface Waves
One way to categorize waves is on the basis of the direction of movement of the individual particles of
the medium relative to the direction that the waves travel. Categorizing waves on this basis leads to three
notable categories: transverse waves, longitudinal waves, and surface waves.
A transverse wave is a wave in which particles of the medium move in a direction perpendicular to the
direction that the wave moves. Suppose that a slinky is stretched out in a horizontal direction across the
classroom and that a pulse is introduced into the slinky on the left end by vibrating the first coil up and
down. Energy will begin to be transported through the slinky from left to right. As the energy is
transported from left to right, the individual coils of the medium will be displaced upwards and
downwards. In this case, the particles of the medium move perpendicular to the direction that the pulse
moves. This type of wave is a transverse wave. Transverse waves are always characterized by particle
motion being perpendicular to wave motion.
A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the
direction that the wave moves. Suppose that a slinky is stretched out in a horizontal direction across the
classroom and that a pulse is introduced into the slinky on the left end by vibrating the first coil left and
right. Energy will begin to be transported through the slinky from left to right. As the energy is
transported from left to right, the individual coils of the medium will be displaced leftwards and
rightwards. In this case, the particles of the medium move parallel to the direction that the pulse moves.
This type of wave is a longitudinal wave. Longitudinal waves are always characterized by particle motion
being parallel to wave motion.
A sound wave traveling through air is a classic example of a longitudinal wave. As a sound wave moves
from the lips of a speaker to the ear of a listener, particles of air vibrate back and forth in the same
direction and the opposite direction of energy transport. Each individual particle pushes on its
neighboring particle so as to push it forward. The collision of particle #1 with its neighbor serves to
restore particle #1 to its original position and displace particle #2 in a forward direction. This back and
forth motion of particles in the direction of energy transport creates regions within the medium where the
particles are pressed together and other regions where the particles are spread apart. Longitudinal waves
can always be quickly identified by the presence of such regions. This process continues along the chain
of particles until the sound wave reaches the ear of the listener. A detailed discussion of sound is
presented in another unit of The Physics Classroom Tutorial.
Waves traveling through a solid medium can be either transverse waves or longitudinal waves. Yet waves
traveling through the bulk of a fluid (such as a liquid or a gas) are always longitudinal waves. Transverse
waves require a relatively rigid medium in order to transmit their energy. As one particle begins to move
it must be able to exert a pull on its nearest neighbor. If the medium is not rigid as is the case with fluids,
the particles will slide past each other. This sliding action that is characteristic of liquids and gases
prevents one particle from displacing its neighbor in a direction perpendicular to the energy transport. It is
for this reason that only longitudinal waves are observed moving through the bulk of liquids such as our
oceans. Earthquakes are capable of producing both transverse and longitudinal waves that travel through
the solid structures of the Earth. When seismologists began to study earthquake waves they noticed that
only longitudinal waves were capable of traveling through the core of the Earth. For this reason,
geologists believe that the Earth's core consists of a liquid - most likely molten iron.
While waves that travel within the depths of the ocean are longitudinal waves, the waves that travel along
the surface of the oceans are referred to as surface waves. A surface wave is a wave in which particles of
the medium undergo a circular motion. Surface waves are neither longitudinal nor transverse. In
longitudinal and transverse waves, all the particles in the entire bulk of the medium move in a parallel and
a perpendicular direction (respectively) relative to the direction of energy transport. In a surface wave, it
is only the particles at the surface of the medium that undergo the circular motion. The motion of particles
tends to decrease as one proceeds further from the surface.
Any wave moving through a medium has a source. Somewhere along the medium, there was an initial
displacement of one of the particles. For a slinky wave, it is usually the first coil that becomes displaced
by the hand of a person. For a sound wave, it is usually the vibration of the vocal chords or a guitar string
that sets the first particle of air in vibrational motion. At the location where the wave is introduced into
the medium, the particles that are displaced from their equilibrium position always moves in the same
direction as the source of the vibration. So if you wish to create a transverse wave in a slinky, then the
first coil of the slinky must be displaced in a direction perpendicular to the entire slinky. Similarly, if you
wish to create a longitudinal wave in a slinky, then the first coil of the slinky must be displaced in a
direction parallel to the entire slinky.
Electromagnetic versus Mechanical Waves
Another way to categorize waves is on the basis of their ability or inability to transmit energy through a
vacuum (i.e., empty space). Categorizing waves on this basis leads to two notable categories:
electromagnetic waves and mechanical waves.
An electromagnetic wave is a wave that is capable of transmitting its energy through a vacuum (i.e.,
empty space). Electromagnetic waves are produced by the vibration of charged particles. Electromagnetic
waves that are produced on the sun subsequently travel to Earth through the vacuum of outer space. Were
it not for the ability of electromagnetic waves to travel to through a vacuum, there would undoubtedly be
no life on Earth. All light waves are examples of electromagnetic waves. Light waves are the topic of
another unit at The Physics Classroom Tutorial. While the basic properties and behaviors of light will be
discussed, the detailed nature of an electromagnetic wave is quite complicated and beyond the scope of
The Physics Classroom Tutorial.
A mechanical wave is a wave that is not capable of transmitting its energy through a vacuum.
Mechanical waves require a medium in order to transport their energy from one location to another. A
sound wave is an example of a mechanical wave. Sound waves are incapable of traveling through a
vacuum. Slinky waves, water waves, stadium waves, and jump rope waves are other examples of
mechanical waves; each requires some medium in order to exist. A slinky wave requires the coils of the
slinky; a water wave requires water; a stadium wave requires fans in a stadium; and a jump rope wave
requires a jump rope.
The above categories represent just a few of the ways in which physicists categorize waves in order to
compare and contrast their behaviors and characteristic properties. This listing of categories is not
exhaustive; there are other categories as well. The five categories of waves listed here will be used
periodically throughout this unit on waves as well as the units on sound and light.
Check Your Understanding
1. A transverse wave is transporting energy from east to west. The particles of the medium will
move_____.
A)
B)
C)
D)
east to west only
both eastward and westward
north to south only
both northward and southward
2. A wave is transporting energy from left to right. The particles of the medium are moving back and
forth in a leftward and rightward direction. This type of wave is known as a ____.
a. mechanical
c. transverse
b. electromagnetic
d. longitudinal
3. Describe how the fans in a stadium must move in order to produce a longitudinal stadium wave.
5. A sound wave is a mechanical wave, not an electromagnetic wave. This means that
1. particles of the medium move perpendicular to the direction of energy transport.
2.
3.
4.
a sound wave transports its energy through a vacuum.
particles of the medium regularly and repeatedly oscillate about their rest position.
a medium is required in order for sound waves to transport energy.
5. A science fiction film depicts inhabitants of one spaceship (in outer space) hearing the sound of a
nearby spaceship as it zooms past at high speeds. Critique the physics of this film.
6. If you strike a horizontal rod vertically from above, what can be said about the waves created in the
rod?
a. The particles vibrate horizontally along the direction of the rod.
b. The particles vibrate vertically, perpendicular to the direction of the rod.
c. The particles vibrate in circles, perpendicular to the direction of the rod.
d. The particles travel along the rod from the point of impact to its end.
7. Which of the following is not a characteristic of mechanical waves?
a. They consist of disturbances or oscillations of a medium.
b. They transport energy.
c. They travel in a direction that is at right angles to the direction of the particles of the medium.
d. They are created by a vibrating source.
8. The sonar device on a fishing boat uses underwater sound to locate fish. Would you expect sonar to be
a longitudinal or a transverse wave?
The Anatomy of a Wave
A transverse wave is a wave in which the particles of the medium are displaced in a direction
perpendicular to the direction of energy transport. A transverse wave can be created in a rope if the rope is
stretched out horizontally and the end is vibrated back-and-forth in a vertical direction. If a snapshot of
such a transverse wave could be taken so as to freeze the shape of the rope in time, then it would look like
the following diagram.
The dashed line drawn through the center of the diagram represents the equilibrium or rest position of the
string. This is the position that the string would assume if there were no disturbance moving through it.
Once a disturbance is introduced into the string, the particles of the string begin to vibrate upwards and
downwards. At any given moment in time, a particle on the medium could be above or below the rest
position. Points A, E and H on the diagram represent the crests of this wave. The crest of a wave is the
point on the medium that exhibits the maximum amount of positive or upward displacement from the rest
position. Points C and J on the diagram represent the troughs of this wave. The trough of a wave is the
point on the medium that exhibits the maximum amount of negative or downward displacement from the
rest position.
The wave shown above can be described by a variety of properties. One such property is amplitude. The
amplitude of a wave refers to the maximum amount of displacement of a particle on the medium from its
rest position. In a sense, the amplitude is the distance from rest to crest. Similarly, the amplitude can be
measured from the rest position to the trough position. In the diagram above, the amplitude could be
measured as the distance of a line segment that is perpendicular to the rest position and extends vertically
upward from the rest position to point A.
The wavelength is another property of a wave that is portrayed in the diagram above. The wavelength of
a wave is simply the length of one complete wave cycle. If you were to trace your finger across the wave
in the diagram above, you would notice that your finger repeats its path. A wave is a repeating pattern. It
repeats itself in a periodic and regular fashion over both time and space. And the length of one such
spatial repetition (known as a wave cycle) is the wavelength. The wavelength can be measured as the
distance from crest to crest or from trough to trough. In fact, the wavelength of a wave can be measured
as the distance from a point on a wave to the corresponding point on the next cycle of the wave. In the
diagram above, the wavelength is the horizontal distance from A to E, or the horizontal distance from B to
F, or the horizontal distance from D to G, or the horizontal distance from E to H. Any one of these
distance measurements would suffice in determining the wavelength of this wave.
A longitudinal wave is a wave in which the particles of the medium are displaced in a direction parallel to
the direction of energy transport. A longitudinal wave can be created in a slinky if the slinky is stretched
out horizontally and the end coil is vibrated back-and-forth in a horizontal direction. If a snapshot of such
a longitudinal wave could be taken so as to freeze the shape of the slinky in time, then it would look like
the following diagram.
Because the coils of the slinky are vibrating longitudinally, there are regions where they become pressed
together and other regions where they are spread apart. A region where the coils are pressed together in a
small amount of space is known as a compression. A compression is a point on a medium through which
a longitudinal wave is traveling that has the maximum density. A region where the coils are spread apart,
thus maximizing the distance between coils, is known as a rarefaction. A rarefaction is a point on a
medium through which a longitudinal wave is traveling that has the minimum density. Points A, C and E
on the diagram above represent compressions and points B, D, and F represent rarefactions. While a
transverse wave has an alternating pattern of crests and troughs, a longitudinal wave has an alternating
pattern of compressions and rarefactions.
As discussed above, the wavelength of a wave is the length of one complete cycle of a wave. For a
transverse wave, the wavelength is determined by measuring from crest to crest. A longitudinal wave
does not have crest; so how can its wavelength be determined? The wavelength can always be determined
by measuring the distance between any two corresponding points on adjacent waves. In the case of a
longitudinal wave, a wavelength measurement is made by measuring the distance from a compression to
the next compression or from a rarefaction to the next rarefaction. On the diagram above, the distance
from point A to point C or from point B to point D would be representative of the wavelength.
Check Your Understanding
Consider the diagram below in order to answer questions #1-2.
1. The wavelength of the wave in the diagram above is given by letter ______.
2. The amplitude of the wave in the diagram above is given by letter _____.
3. Indicate the interval that represents one full wavelength.
a. A to C
b. B to D
c. A to G
d. C to G
Frequency and Period of a Wave
The nature of a wave was discussed in Lesson 1 of this unit. In that lesson, it was mentioned that a wave
is created in a slinky by the periodic and repeating vibration of the first coil of the slinky. This vibration
creates a disturbance that moves through the slinky and transports energy from the first coil to the last
coil. A single back-and-forth vibration of the first coil of a slinky introduces a pulse into the slinky. But
the act of continually vibrating the first coil with a back-and-forth motion in periodic fashion introduces a
wave into the slinky.
Suppose that a hand holding the first coil of a slinky is moved back-and-forth two complete cycles in one
second. The rate of the hand's motion would be 2 cycles/second. The first coil, being attached to the hand,
in turn would vibrate at a rate of 2 cycles/second. The second coil, being
attached to the first coil, would vibrate at a rate of 2 cycles/second. The
third coil, being attached to the second coil, would vibrate at a rate of 2
cycles/second. In fact, every coil of the slinky would vibrate at this rate of 2
cycles/second. This rate of 2 cycles/second is referred to as the frequency of
the wave. The frequency of a wave refers to how often the particles of the
medium vibrate when a wave passes through the medium. Frequency is a
part of our common, everyday language. For example, it is not uncommon to hear a question like "How
frequently do you mow the lawn during the summer months?" Of course the question is an inquiry about
how often the lawn is mowed and the answer is usually given in the form of "1 time per week." In
mathematical terms, the frequency is the number of complete vibrational cycles of a medium per a given
amount of time. Given this definition, it is reasonable that the quantity frequency would have units of
cycles/second, waves/second, vibrations/second, or something/second. Another unit for frequency is the
Hertz (abbreviated Hz) where 1 Hz is equivalent to 1 cycle/second. If a coil of slinky makes 2 vibrational
cycles in one second, then the frequency is 2 Hz. If a coil of slinky makes 3 vibrational cycles in one
second, then the frequency is 3 Hz. And if a coil makes 8 vibrational cycles in 4 seconds, then the
frequency is 2 Hz (8 cycles/4 s = 2 cycles/s).
The quantity frequency is often confused with the quantity period. Period refers to the time that it takes to
do something. When an event occurs repeatedly, then we say that the event is periodic and refer to the
time for the event to repeat itself as the period. The period of a wave is the time for a particle on a
medium to make one complete vibrational cycle. Period, being a time, is measured in units of time such
as seconds, hours, days or years. The period of orbit for the Earth around the Sun is approximately 365
days; it takes 365 days for the Earth to complete a cycle. The period of a typical class at a high school
might be 55 minutes; every 55 minutes a class cycle begins (50 minutes for class and 5 minutes for
passing time means that a class begins every 55 minutes). The period for the minute hand on a clock is
3600 seconds (60 minutes); it takes the minute hand 3600 seconds to complete one cycle around the
clock.
Frequency and period are distinctly different, yet related, quantities. Frequency refers to how often
something happens. Period refers to the time it takes something to happen. Frequency is a rate quantity.
Period is a time quantity. Frequency is the cycles/second. Period is the seconds/cycle. As an example of
the distinction and the relatedness of frequency and period, consider a woodpecker that drums upon a tree
at a periodic rate. If the woodpecker drums upon a tree 2 times in one second, then the frequency is 2 Hz.
Each drum must endure for one-half a second, so the period is 0.5 s. If the woodpecker drums upon a tree
4 times in one second, then the frequency is 4 Hz; each drum must endure for one-fourth a second, so the
period is 0.25 s. If the woodpecker drums upon a tree 5 times in one second, then the frequency is 5 Hz;
each drum must endure for one-fifth a second, so the period is 0.2 s. Do you observe the relationship?
Mathematically, the period is the reciprocal of the frequency and vice versa. In equation form, this is
expressed as follows.
Since the symbol f is used for frequency and the symbol T is used for period, these equations are also
expressed as:
The quantity frequency is also confused with the quantity speed. The speed of
an object refers to how fast an object is moving and is usually expressed as the
distance traveled per time of travel. For a wave, the speed is the distance
traveled by a given point on the wave (such as a crest) in a given period of
time. So while wave frequency refers to the number of cycles occurring per
second, wave speed refers to the meters traveled per second. A wave can
vibrate back and forth very frequently, yet have a small speed; and a wave can vibrate back and forth with
a low frequency, yet have a high speed. Frequency and speed are distinctly different quantities. Wave
speed will be discussed in more detail later in this lesson.
Check Your Understanding
Throughout this unit, internalize the meaning of terms such as period, frequency, and wavelength. Utilize
the meaning of these terms to answer conceptual questions; avoid a formula fixation.
1. A wave is introduced into a thin wire held tight at each end. It has an amplitude of 3.8 cm, a frequency
of 51.2 Hz and a distance from a crest to the neighboring trough of 12.8 cm. Determine the period of such
a wave.
2. Frieda the fly flaps its wings back and forth 121 times each second. The period of the wing flapping is
____ sec.
3. A tennis coach paces back and forth along the sideline 10 times in 2 minutes. The frequency of her
pacing is ________ Hz.
a. 5.0
b. 0.20
c. 0.12
d. 0.083
4. Non-digital clocks (which are becoming more rare) have a second hand that rotates around in a regular
and repeating fashion. The frequency of rotation of a second hand on a clock is _______ Hz.
a. 1/60
d. 1
b. 1/12
e. 60
c. 1/2
5. Olive Udadi accompanies her father to the park for an afternoon of fun. While there, she hops on the
swing and begins a motion characterized by a complete back-and-forth cycle every 2 seconds. The
frequency of swing is _________.
a. 0.5 Hz
b. 1 Hz
c. 2 Hz
6. In problem #5, the period of swing is __________.
a. 0.5 second
b. 1 second
c. 2 second
7. A period of 5.0 seconds corresponds to a frequency of ________ Hertz.
a. 0.2
d. 0.05
b. 0.5
e. 0.002
c. 0.02
8. A common physics lab involves the study of the oscillations of a pendulum. If a pendulum makes 33
complete back-and-forth cycles of vibration in 11 seconds, then its period is ______.
9. A child in a swing makes one complete back and forth motion in 3.2 seconds. This statement provides
information about the child's
a. speed
b. frequency
c. period
10. The period of the sound wave produced by a 440 Hertz tuning fork is ___________.
11. As the frequency of a wave increases, the period of the wave ___________.
a. decreases
b. increases
c. remains the same
Properties of Waves
The Anatomy of a Wave | Frequency and Period of a Wave
Energy Transport and the Amplitude of a Wave | The Speed of a Wave | The Wave Equation
Energy Transport and the Amplitude of a Wave
As mentioned earlier, a wave is an energy transport phenomenon that transports energy along a medium
without transporting matter. A pulse or a wave is introduced into a slinky when a person holds the first
coil and gives it a back-and-forth motion. This creates a disturbance within the medium; this disturbance
subsequently travels from coil to coil, transporting energy as it moves. The energy is imparted to the
medium by the person as he/she does work upon the first coil to give it kinetic energy. This energy is
transferred from coil to coil until it arrives at the end of the slinky. If you were holding the opposite end
of the slinky, then you would feel the energy as it reaches your end. In fact, a high energy pulse would
likely do some rather noticeable work upon your hand upon reaching the end of the medium; the last coil
of the medium would displace you hand in the same direction of motion of the coil. For the same reasons,
a high energy ocean wave can do considerable damage to the rocks and piers along the shoreline when it
crashes upon it.
The amount of energy carried by a wave is related to the amplitude of the wave. A high energy wave is
characterized by a high amplitude; a low energy wave is characterized by a low amplitude. As discussed
earlier in Lesson 2, the amplitude of a wave refers to the maximum amount of displacement of a particle
on the medium from its rest position. The logic underlying the energy-amplitude relationship is as
follows: If a slinky is stretched out in a horizontal direction and a transverse pulse is introduced into the
slinky, the first coil is given an initial amount of displacement. The displacement is due to the force
applied by the person upon the coil to displace it a given amount from rest. The more energy that the
person puts into the pulse, the more work that he/she will do upon the first coil. The more work that is
done upon the first coil, the more displacement that is given to it. The more displacement that is given to
the first coil, the more amplitude that it will have. So in the end, the amplitude of a transverse pulse is
related to the energy which that pulse transports through the medium. Putting a lot of energy into a
transverse pulse will not affect the wavelength, the frequency or the speed of the pulse. The energy
imparted to a pulse will only affect the amplitude of that pulse.
Consider two identical slinkies into which a pulse is introduced. If the same amount of energy is
introduced into each slinky, then each pulse will have the same amplitude. But what if the slinkies are
different? What if one is made of zinc and the other is made of copper? Will the amplitudes now be the
same or different? If a pulse is introduced into two different slinkies by imparting the same amount of
energy, then the amplitudes of the pulses will not necessarily be the same. In a situation such as this, the
actual amplitude assumed by the pulse is dependent upon two types of factors: an inertial factor and an
elastic factor. Two different materials have different mass densities. The imparting of energy to the first
coil of a slinky is done by the application of a force to this coil. More massive slinkies have a greater
inertia and thus tend to resist the force; this increased resistance by the greater mass tends to cause a
reduction in the amplitude of the pulse. Different materials also have differing degrees of springiness or
elasticity. A more elastic medium will tend to offer less resistance to the force and allow a greater
amplitude pulse to travel through it; being less rigid (and therefore more elastic), the same force causes a
greater amplitude.
The energy transported by a wave is directly proportional to the square of the amplitude of the wave. This
energy-amplitude relationship is sometimes expressed in the following manner.
This means that a doubling of the amplitude of a wave is indicative of a
quadrupling of the energy transported by the wave. A tripling of the amplitude of
a wave is indicative of a nine-fold increase in the amount of energy transported
by the wave. And a quadrupling of the amplitude of a wave is indicative of a 16fold increase in the amount of energy transported by the wave. The table at the
right further expresses this energy-amplitude relationship. Observe that whenever
the amplitude increased by a given factor, the energy value is increased by the
same factor squared. For example, changing the amplitude from 1 unit to 2 units
represents a 2-fold increase in the amplitude and is accompanied by a 4-fold (22)
increase in the energy; thus 2 units of energy becomes 4 times bigger - 8 units. As another example,
changing the amplitude from 1 unit to 4 units represents a 4-fold increase in the amplitude and is
accompanied by a 16-fold (42) increase in the energy; thus 2 units of energy becomes 16 times bigger - 32
units.
Check Your Understanding
1. Mac and Tosh stand 8 meters apart and demonstrate the motion of a transverse wave on a snakey. The
wave e can be described as having a vertical distance of 32 cm from a trough to a crest, a frequency of 2.4
Hz, and a horizontal distance of 48 cm from a crest to the nearest trough. Determine the amplitude,
period, and wavelength of such a wave.
2. An ocean wave has an amplitude of 2.5 m. Weather conditions suddenly change such that the wave has
an amplitude of 5.0 m. The amount of energy transported by the wave is __________.
a. halved
b. doubled
c. quadrupled
d. remains the same
3. Two waves are traveling through a container of an inert gas. Wave A has an amplitude of .1 cm. Wave
B has an amplitude of .2 cm. The energy transported by wave B must be __________ the energy
transported by wave A.
a. one-fourth
b. one-half
c. two times larger than
d. four times larger than
The Speed of a Wave
A wave is a disturbance that moves along a medium from one end to the other. If one watches an ocean
wave moving along the medium (the ocean water), one can observe that the crest of the wave is moving
from one location to another over a given interval of time. The crest is observed to cover distance. The
speed of an object refers to how fast an object is moving and is usually expressed as the distance traveled
per time of travel. In the case of a wave, the speed is the distance traveled by a given point on the wave
(such as a crest) in a given interval of time. In equation form,
If the crest of an ocean wave moves a distance of 20 meters in 10 seconds, then the speed of the ocean
wave is 2 m/s. On the other hand, if the crest of an ocean wave moves a distance of 25 meters in 10
seconds (the same amount of time), then the speed of this ocean wave is 2.5 m/s. The faster wave travels a
greater distance in the same amount of time.
Sometimes a wave encounters the end of a medium and the presence of a different medium. For example,
a wave introduced by a person into one end of a slinky will travel through the slinky and eventually reach
the end of the slinky and the presence of the hand of a second person. One behavior that waves undergo at
the end of a medium is reflection. The wave will reflect or bounce off the person's hand. When a wave
undergoes reflection, it remains within the medium and merely reverses its direction of travel. In the case
of a slinky wave, the disturbance can be seen traveling back to the original end. A slinky wave that travels
to the end of a slinky and back has doubled its distance. That is, by reflecting back to the original
location, the wave has traveled a distance that is equal to twice the length
of the slinky.
Reflection phenomena are commonly observed with sound waves. When
you let out a holler within a canyon, you often hear the echo of the holler.
The sound wave travels through the medium (air in this case), reflects off
the canyon wall and returns to its origin (you). The result is that you hear
the echo (the reflected sound wave) of your holler. A classic physics
problem goes like this:
Noah stands 170 meters away from a steep canyon wall. He shouts and
hears the echo of his voice one second later. What is the speed of the
wave?
In this instance, the sound wave travels 340 meters in 1 second, so the speed of the wave is 340 m/s.
Remember, when there is a reflection, the wave doubles its distance. In other words, the distance traveled
by the sound wave in 1 second is equivalent to the 170 meters down to the canyon wall plus the 170
meters back from the canyon wall.
Variables Affecting Wave Speed
What variables affect the speed at which a wave travels through a medium? Does
the frequency or wavelength of the wave affect its speed? Does the amplitude of
the wave affect its speed? Or are other variables such as the mass density of the
medium or the elasticity of the medium responsible for affecting the speed of the
wave? These questions are often investigated in the form of a lab in a physics
classroom.
Suppose a wave generator is used to produce several waves within a rope of a
measurable tension. The wavelength, frequency and speed are determined. Then the frequency of
vibration of the generator is changed to investigate the affect of frequency upon wave speed. Finally, the
tension of the rope is altered to investigate the affect of tension upon wave speed. Sample data for the
experiment are shown below.
Speed of a Wave Lab - Sample Data
Tension
Frequency
Wavelength
Speed
Trial
(N)
(Hz)
(m)
(m/s)
1
2.0
4.05
4.00
16.2
2
2.0
8.03
2.00
16.1
3
2.0
12.30
1.33
16.4
4
2.0
16.2
1.00
16.2
5
2.0
20.2
0.800
16.2
6
5.0
12.8
2.00
25.6
7
5.0
19.3
1.33
25.7
8
5.0
25.5
1.00
25.5
In the first five trials, the tension of the rope was held constant and the frequency was systematically
changed. The data in rows 1-5 of the table above demonstrate that a change in the frequency of a wave
does not affect the speed of the wave. The speed remained a near constant value of approximately 16.2
m/s. The small variations in the values for the speed were the result of experimental error, rather than a
demonstration of some physical law. The data convincingly show that wave frequency does not affect
wave speed. An increase in wave frequency caused a decrease in wavelength while the wave speed
remained constant.
The last three trials involved the same procedure with a different rope tension. Observe that the speed of
the waves in rows 6-8 is distinctly different than the speed of the wave in rows 1-5. The obvious cause of
this difference is the alteration of the tension of the rope. The speed of the waves was significantly higher
at higher tensions. Waves travel through tighter ropes at higher speeds. So while the frequency did not
affect the speed of the wave, the tension in the medium (the rope) did. In fact, the speed of a wave is not
dependent upon (causally affected by) properties of the wave itself. Rather, the speed of the wave is
dependent upon the properties of the medium such as the tension of the rope.
One theme of this unit has been that "a wave is a disturbance moving through a medium." There are two
distinct objects in this phrase - the "wave" and the "medium." The medium could be water, air, or a
slinky. These media are distinguished by their properties - the material they are made of and the physical
properties of that material such as the density, the temperature, the elasticity, etc. Such physical properties
describe the material itself, not the wave. On the other hand, waves are distinguished from each other by
their properties - amplitude, wavelength, frequency, etc. These properties describe the wave, not the
material through which the wave is moving. The lesson of the lab activity described above is that wave
speed depends upon the medium through which the wave is moving. Only an alteration in the properties
of the medium will cause a change in the speed.
Check Your Understanding
1. A teacher attaches a slinky to the wall and begins introducing pulses with different amplitudes. Which
of the two pulses (A or B) below will travel from the hand to the wall in the least amount of time? Justify
your answer.
2. The teacher then begins introducing pulses with a different wavelength. Which of the two pulses (C or
D) will travel from the hand to the wall in the least amount of time ? Justify your answer.
3. The time required for the sound waves (v = 340 m/s) to travel from the tuning fork to point A is ____ .
a. 0.020 second
c. 0.59 second
b. 0.059 second
d. 2.9 second
4. Two waves are traveling through the same container of nitrogen gas. Wave A has a wavelength of 1.5
m. Wave B has a wavelength of 4.5 m. The speed of wave B must be ________ the speed of wave A.
a. one-ninth
c. the same as
b. one-third
d. three times larger than
5. An automatic focus camera is able to focus on objects by use of an ultrasonic sound wave. The camera
sends out sound waves that reflect off distant objects and return to the camera. A sensor detects the time it
takes for the waves to return and then determines the distance an object is from the camera. The camera
lens then focuses at that distance. Now that's a smart camera! In a subsequent life, you might have to be a
camera; so try this problem for practice:
If a sound wave (speed = 340 m/s) returns to the camera 0.150 seconds after leaving the camera, then
how far away is the object?
6. TRUE or FALSE:
Doubling the frequency of a wave source doubles the speed of the waves
7. While hiking through a canyon, Noah Formula lets out a scream. An echo (reflection of the scream off
a nearby canyon wall) is heard 0.82 seconds after the scream. The speed of the sound wave in air is 342
m/s. Calculate the distance from Noah to the nearby canyon wall.
8. Mac and Tosh are resting on top of the water near the end of the pool when Mac creates a surface
wave. The wave travels the length of the pool and back in 25 seconds. The pool is 25 meters long.
Determine the speed of the wave.
9. The water waves below are traveling along the surface of the ocean at a speed of 2.5 m/s and splashing
periodically against Wilbert's perch. Each adjacent crest is 5 meters apart. The crests splash Wilbert's feet
upon reaching his perch. How much time passes between each successive drenching? Answer and explain
using complete sentences.
The Wave Equation
As was discussed in Lesson 1, a wave is produced when a vibrating source periodically disturbs the first
particle of a medium. This creates a wave pattern that begins to travel along the medium from particle to
particle. The frequency at which each individual particle vibrates is equal to the frequency at which the
source vibrates. Similarly, the period of vibration of each individual particle in the medium is equal to the
period of vibration of the source. In one period, the source is able to displace the first particle upwards
from rest, back to rest, downwards from rest, and finally back to rest. This complete back-and-forth
movement constitutes one complete wave cycle.
The diagrams at the right show several "snapshots" of the production of a wave
within a rope. The motion of the disturbance along the medium after every onefourth of a period is depicted. Observe that in the time it takes from the first to
the last snapshot, the hand has made one complete back-and-forth motion. A
period has elapsed. Observe that during this same amount of time, the leading
edge of the disturbance has moved a distance equal to one complete wavelength.
So in a time of one period, the wave has moved a distance of one wavelength.
Combining this information with the equation for speed (speed = distance/time),
it can be said that the speed of a wave is also the wavelength/period.
Since the period is the reciprocal of the frequency, the expression 1/f can be
substituted into the above equation for period. Rearranging the equation yields a
new equation of the form:
Speed = Wavelength • Frequency
The above equation is known as the wave equation. It states the mathematical relationship between the
speed (v) of a wave and its wavelength ( ) and frequency (f). Using the symbols v, , and f, the equation
can be rewritten as
v=f•
As a test of your understanding of the wave equation and its mathematical use in analyzing wave motion,
consider the following three-part question:
Stan and Anna are conducting a slinky experiment. They are studying the possible affect of several
variables upon the speed of a wave in a slinky. Their data table is shown below. Fill in the blanks in the
table, analyze the data, and answer the following questions.
Medium
Zinc,
Wavelength
Frequency
Speed
1.75 m
2.0 Hz
______
0.90 m
3.9 Hz
______
1.19 m
2.1 Hz
______
1-in. dia. coils
Zinc,
1-in. dia. coils
Copper,
1-in. dia. coils
Copper,
0.60 m
4.2 Hz
______
0.95 m
2.2 Hz
______
1.82 m
1.2 Hz
______
1-in. dia. coils
Zinc,
3-in. dia. coils
Zinc,
3-in. dia. Coils
a. As the wavelength of a wave in a uniform medium increases, its speed will _____.
a. decrease
b. increase
c. remain the same
b. As the wavelength of a wave in a uniform medium increases, its frequency will _____.
a. decrease
b. increase
c. remain the same
c. The speed of a wave depends upon (i.e., is causally affected by) ...
a. the properties of the medium through which the wave travels
b. the wavelength of the wave.
c. the frequency of the wave.
d. both the wavelength and the frequency of the wave.
The above example illustrates how to use the wave equation to solve mathematical problems. It also
illustrates the principle that wave speed is dependent upon medium properties and independent of wave
properties. Even though the wave speed is calculated by multiplying wavelength by frequency, an
alteration in wavelength does not affect wave speed. Rather, an alteration in wavelength affects the
frequency in an inverse manner. A doubling of the wavelength results in a halving of the frequency; yet
the wave speed is not changed.
Check Your Understanding
1. Two waves on identical strings have frequencies in a ratio of 2 to 1. If their wave speeds are the same,
then how do their wavelengths compare?
a. 2:1
b. 1:2
c. 4:1
d. 1:4
2. Mac and Tosh stand 8 meters apart and demonstrate the motion of a transverse wave on a snakey. The
wave e can be described as having a vertical distance of 32 cm from a trough to a crest, a frequency of 2.4
Hz, and a horizontal distance of 48 cm from a crest to the nearest trough. Determine the amplitude,
period, and wavelength and speed of such a wave.
3. Dawn and Aram have stretched a slinky between them and begin experimenting with waves. As the
frequency of the waves is doubled,
a. the wavelength is halved and the speed remains constant
b. the wavelength remains constant and the speed is doubled
c. both the wavelength and the speed are halved.
d. both the wavelength and the speed remain constant.
4. A ruby-throated hummingbird beats its wings at a rate of about 70 wing beats per second.
a. What is the frequency in Hertz of the sound wave?
b. Assuming the sound wave moves with a velocity of 350 m/s, what is the wavelength of the wave?
5. Ocean waves are observed to travel along the water surface during a developing storm. A Coast Guard
weather station observes that there is a vertical distance from high point to low point of 4.6 meters and a
horizontal distance of 8.6 meters between adjacent crests. The waves splash into the station once every
6.2 seconds. Determine the frequency and the speed of these waves.
6. Two boats are anchored 4 meters apart. They bob up and
down, returning to the same up position every 3 seconds.
When one is up the other is down. There are never any wave
crests between the boats. Calculate the speed of the waves.
Behavior of Waves
Boundary Behavior | Reflection, Refraction, and Diffraction | Interference of Waves
The Doppler Effect
Boundary Behavior
As a wave travels through a medium, it will often reach the end of the medium and encounter an obstacle
or perhaps another medium through which it could travel. One example of this has already been
mentioned in Lesson 2. A sound wave is known to reflect off canyon walls and other obstacles to produce
an echo. A sound wave traveling through air within a canyon reflects off the canyon wall and returns to its
original source. What affect does reflection have upon a wave? Does reflection of a wave affect the speed
of the wave? Does reflection of a wave affect the wavelength and frequency of the wave? Does reflection
of a wave affect the amplitude of the wave? Or does reflection affect other properties and characteristics
of a wave's motion? The behavior of a wave (or pulse) upon reaching the end of a medium is referred to
as boundary behavior. When one medium ends, another medium begins; the interface of the two media
is referred to as the boundary and the behavior of a wave at that boundary is described as its boundary
behavior. The questions that are listed above are the types of questions we seek to answer when we
investigate the boundary behavior of waves.
Fixed End Reflection
First consider an elastic rope stretched from end to end. One end
will be securely attached to a pole on a lab bench while the other
end will be held in the hand in order to introduce pulses into the
medium. Because the right end of the rope is attached to a pole
(which is attached to a lab bench) (which is attached to the floor
that is attached to the building that is attached to the Earth), the
last particle of the rope will be unable to move when a
disturbance reaches it. This end of the rope is referred to as a fixed end.
If a pulse is introduced at the left end of the rope, it will travel through the rope towards the right end of
the medium. This pulse is called the incident pulse since it is incident towards (i.e., approaching) the
boundary with the pole. When the incident pulse reaches the boundary, two things occur:


A portion of the energy carried by the pulse is reflected and returns towards the left end of the
rope. The disturbance that returns to the left after bouncing off the pole is known as the
reflected pulse.
A portion of the energy carried by the pulse is transmitted to the pole, causing the pole to
vibrate.
Because the vibrations of the pole are not visibly obvious, the energy transmitted to it is not typically
discussed. The focus of the discussion will be on the reflected pulse. What characteristics and properties
could describe its motion?
When one observes the reflected pulse off the fixed end, there are several notable observations. First the
reflected pulse is inverted. That is, if an upward displaced pulse is incident towards a fixed end boundary,
it will reflect and return as a downward displaced pulse. Similarly, if a downward displaced pulse is
incident towards a fixed end boundary, it will reflect and return as an upward displaced pulse.
The inversion of the reflected pulse can be explained by returning to our conceptions of the nature of a
mechanical wave. When a crest reaches the end of a medium ("medium A"), the last particle of the
medium A receives an upward displacement. This particle is attached to the first particle of the other
medium ("medium B") on the other side of the boundary. As the last particle of medium A pulls upwards
on the first particle of medium B, the first particle of medium B pulls downwards on the last particle of
medium A. This is merely Newton's third law of action-reaction. For every action, there is an equal and
opposite reaction. The upward pull on the first particle of medium B has little affect upon this particle due
to the large mass of the pole and the lab bench to which it is attached. The affect of the downward pull on
the last particle of medium A (a pull that is in turn transmitted to the other particles) results in causing the
upward displacement to become a downward displacement. The upward displaced incident pulse thus
returns as a downward displaced reflected pulse. It is important to note that it is the heaviness of the pole
and the lab bench relative to the rope that causes the rope to become inverted upon interacting with the
wall. When two media interact by exerting pushes and pulls upon each other, the most massive medium
wins the interaction. Just like in arm wrestling, the medium that loses receives a change in its state of
motion.
Other notable characteristics of the reflected pulse include:



The speed of the reflected pulse is the same as the speed of the incident pulse.
The wavelength of the reflected pulse is the same as the wavelength of the incident pulse.
The amplitude of the reflected pulse is less than the amplitude of the incident pulse.
Of course, it is not surprising that the speed of the incident and reflected pulse are identical since the two
pulses are traveling in the same medium. Since the speed of a wave (or pulse) is dependent upon the
medium through which it travels, two pulses in the same medium will have the same speed. A similar line
of reasoning explains why the incident and reflected pulses have the same wavelength. Every particle
within the rope will have the same frequency. Being connected to one another, they must vibrate at the
same frequency. Since the wavelength of a wave depends upon the frequency and the speed, two waves
having the same frequency and the same speed must also have the same wavelength. Finally, the
amplitude of the reflected pulse is less than the amplitude of the incident pulse since some of the energy
of the pulse was transmitted into the pole at the boundary. The reflected pulse is carrying less energy
away from the boundary compared to the energy that the incident pulse carried towards the boundary.
Since the amplitude of a pulse is indicative of the energy carried by the pulse, the reflected pulse has a
smaller amplitude than the incident pulse.
Free End Reflection
Now consider what would happen if the end of the rope were
free to move. Instead of being securely attached to a lab pole,
suppose it is attached to a ring that is loosely fit around the
pole. Because the right end of the rope is no longer secured to
the pole, the last particle of the rope will be able to move
when a disturbance reaches it. This end of the rope is referred
to as a free end.
Once more if a pulse is introduced at the left end of the rope, it will travel through the rope towards the
right end of the medium. When the incident pulse reaches the end of the medium, the last particle of the
rope can no longer interact with the first particle of the pole. Since the rope and pole are no longer
attached and interconnected, they will slide past each other. So when a crest reaches the end of the rope,
the last particle of the rope receives the same upward displacement; only now there is no adjoining
particle to pull downward upon the last particle of the rope to cause it to be inverted. The result is that the
reflected pulse is not inverted. When an upward displaced pulse is incident upon a free end, it returns as
an upward displaced pulse after reflection. And when a downward displaced pulse is incident upon a free
end, it returns as a downward displaced pulse after reflection. Inversion is not observed in free end
reflection.
The above discussion of free end and fixed end reflection focuses upon the reflected pulse. As was
mentioned, the transmitted portion of the pulse is difficult to observe when it is transmitted into a pole.
But what if the original medium were attached to another rope with different properties? How could the
reflected pulse and transmitted pulse be described in situations in which an incident pulse reflects off and
transmits into a second medium.
Transmission of a Pulse Across a Boundary from Less to More Dense
Let's consider a thin rope attached to a thick rope, with each rope held at opposite ends by people. And
suppose that a pulse is introduced by the person holding the end of the thin rope. If this is the case, there
will be an incident pulse traveling in the less dense medium (the thin rope) towards the boundary with a
more dense medium (the thick rope).
Upon reaching the boundary, the usual two behaviors will occur.

A portion of the energy carried by the incident pulse is reflected and returns towards the left
end of the thin rope. The disturbance that returns to the left after bouncing off the boundary is
known as the reflected pulse.

A portion of the energy carried by the incident pulse is transmitted into the thick rope. The
disturbance that continues moving to the right is known as the transmitted pulse.
The reflected pulse will be found to be inverted in situations such as this. During the interaction between
the two media at the boundary, the first particle of the more dense medium overpowers the smaller mass
of the last particle of the less dense medium. This causes an upward displaced pulse to become a
downward displaced pulse. The more dense medium on the other hand was at rest prior to the interaction.
The first particle of this medium receives an upward pull when the incident pulse reaches the boundary.
Since the more dense medium was originally at rest, an upward pull can do nothing but cause an upward
displacement. For this reason, the transmitted pulse is not inverted. In fact, transmitted pulses can never
be inverted. Since the particles in this medium are originally at rest, any change in their state of motion
would be in the same direction as the displacement of the particles of the incident pulse.
The Before and After snapshots of the two media are shown in the diagram below.
Comparisons can also be made between the characteristics of the transmitted pulse and those of the
reflected pulse. Once more there are several noteworthy characteristics.



The transmitted pulse (in the more dense medium) is traveling slower than the reflected pulse
(in the less dense medium).
The transmitted pulse (in the more dense medium) has a smaller wavelength than the reflected
pulse (in the less dense medium).
The speed and the wavelength of the reflected pulse are the same as the speed and the
wavelength of the incident pulse.
One goal of physics is to use physical models and ideas to explain the observations made of the physical
world. So how can these three characteristics be explained? First recall from Lesson 2 that the speed of a
wave is dependent upon the properties of the medium. In this case, the transmitted and reflected pulses
are traveling in two distinctly different media. Waves always travel fastest in the least dense medium.
Thus, the reflected pulse will be traveling faster than the transmitted pulse. Second, particles in the more
dense medium will be vibrating with the same frequency as particles in the less dense medium. Since the
transmitted pulse was introduced into the more dense medium by the vibrations of particles in the less
dense medium, they must be vibrating at the same frequency. So the reflected and transmitted pulses have
the different speeds but the same frequency. Since the wavelength of a wave depends upon the frequency
and the speed, the wave with the greatest speed must also have the greatest wavelength. Finally, the
incident and the reflected pulse share the same medium. Since the two pulses are in the same medium,
they will have the same speed. Since the reflected pulse was created by the vibrations of the incident
pulse, they will have the same frequency. And two waves with the same speed and the same frequency
must also have the same wavelength
Transmission of a Pulse Across a Boundary from More to Less Dense
Finally, let's consider a thick rope attached to a thin rope, with the incident pulse originating in the thick
rope. If this is the case, there will be an incident pulse traveling in the more dense medium (thick rope)
towards the boundary with a less dense medium (thin rope). Once again there will be partial reflection and
partial transmission at the boundary. The reflected pulse in this situation will not be inverted. Similarly,
the transmitted pulse is not inverted (as is always the case). Since the incident pulse is in a heavier
medium, when it reaches the boundary, the first particle of the less dense medium does not have sufficient
mass to overpower the last particle of the more dense medium. The result is that an upward displaced
pulse incident towards the boundary will reflect as an upward displaced pulse. For the same reasons, a
downward displaced pulse incident towards the boundary will reflect as a downward displaced pulse.
The Before and After snapshots of the two media are shown in the diagram below.
Comparisons between the characteristics of the transmitted pulse and the reflected pulse lead to the
following observations.



The transmitted pulse (in the less dense medium) is traveling faster than the reflected pulse (in
the more dense medium).
The transmitted pulse (in the less dense medium) has a larger wavelength than the reflected
pulse (in the more dense medium).
The speed and the wavelength of the reflected pulse are the same as the speed and the
wavelength of the incident pulse.
These three observations are explained using the same logic as used above.
The boundary behavior of waves in ropes can be summarized by the following principles:





The wave speed is always greatest in the least dense rope.
The wavelength is always greatest in the least dense rope.
The frequency of a wave is not altered by crossing a boundary.
The reflected pulse becomes inverted when a wave in a less dense rope is heading towards a
boundary with a more dense rope.
The amplitude of the incident pulse is always greater than the amplitude of the reflected pulse.
All the observations discussed here can be explained by the simple application of these principles. Take a
few moments to use these principles to answer the following questions.
Check Your Understanding
Case 1: A pulse in a more dense medium is traveling towards the boundary with a less dense medium.
1. The reflected pulse in medium 1 ________ (will, will not) be inverted because _______.
2. The speed of the transmitted pulse will be ___________ (greater than, less than, the same as) the speed
of the incident pulse.
3. The speed of the reflected pulse will be ______________ (greater than, less than, the same as) the
speed of the incident pulse.
4. The wavelength of the transmitted pulse will be ___________ (greater than, less than, the same as) the
wavelength of the incident pulse.
5. The frequency of the transmitted pulse will be ___________ (greater than, less than, the same as) the
frequency of the incident pulse.
Case 2: A pulse in a less dense medium is traveling towards the boundary with a more dense medium.
6. The reflected pulse in medium 1 ________ (will, will not) be inverted because _____________.
7. The speed of the transmitted pulse will be ___________ (greater than, less than, the same as) the speed
of the incident pulse.
8. The speed of the reflected pulse will be ______________ (greater than, less than, the same as) the
speed of the incident pulse.
9. The wavelength of the transmitted pulse will be ___________ (greater than, less than, the same as) the
wavelength of the incident pulse.
10. The frequency of the transmitted pulse will be ___________ (greater than, less than, the same as) the
frequency of the incident pulse.
Reflection, Refraction, and Diffraction
Previously in Lesson 3, the behavior of waves traveling along a rope from a more dense medium to a less
dense medium (and vice versa) was discussed. The wave doesn't just stop when it reaches the end of the
medium. Rather, a wave will undergo certain behaviors when it encounters the end of the medium.
Specifically, there will be some reflection off the boundary and some transmission into the new medium.
But what if the wave is traveling in a two-dimensional medium such as a water wave traveling through
ocean water? Or what if the wave is traveling in a three-dimensional medium such as a sound wave or a
light wave traveling through air? What types of behaviors can be expected of such two- and threedimensional waves?
The study of waves in two dimensions is often done using a ripple tank. A
ripple tank is a large glass-bottomed tank of water that is used to study the
behavior of water waves. A light typically shines upon the water from
above and illuminates a white sheet of paper placed directly below the
tank. A portion of light is absorbed by the water as it passes through the
tank. A crest of water will absorb more light than a trough. So the bright
spots represent wave troughs and the dark spots represent wave crests. As
the water waves move through the ripple tank, the dark and bright spots
move as well. As the waves encounter obstacles in their path, their
behavior can be observed by watching the movement of the dark and
bright spots on the sheet of paper. Ripple tank demonstrations are
commonly done in a Physics class in order to discuss the principles
underlying the reflection, refraction, and diffraction of waves.
If a linear object attached to an oscillator bobs back and forth within the water, it becomes a source of
straight waves. These straight waves have alternating crests and troughs. As viewed on the sheet of paper
below the tank, the crests are the dark lines stretching across the paper and the
troughs are the bright lines. These waves will travel through the water until they
encounter an obstacle - such as the wall of the tank or an object placed within the
water. The diagram at the right depicts a series of straight waves approaching a long
barrier extending at an angle across the tank of water. The direction that these
wavefronts (straight-line crests) are traveling through the water is represented by the
blue arrow. The blue arrow is called a ray and is drawn perpendicular to the
wavefronts. Upon reaching the barrier placed within the water, these waves bounce
off the water and head in a different direction. The diagram below shows the reflected wavefronts and the
reflected ray. Regardless of the angle at which the wavefronts approach the barrier, one general law of
reflection holds true: the waves will always reflect in such a way that the angle at which they approach
the barrier equals the angle at which they reflect off the barrier. This is known as the law of reflection.
This law will be discussed in more detail in Unit 13 of The Physics Classroom.
The discussion above pertains to the reflection of waves off of straight surfaces. But
what if the surface is curved, perhaps in the shape of a parabola? What
generalizations can be made for the reflection of water waves off parabolic surfaces?
Suppose that a rubber tube having the shape of a parabola is placed within the water.
The diagram at the right depicts such a parabolic barrier in the ripple tank. Several
wavefronts are approaching the barrier; the ray is drawn for these wavefronts. Upon
reflection off the parabolic barrier, the water waves will change direction and head
towards a point. This is depicted in the diagram below. It is as though all the energy being carried by the
water waves is converged at a single point - the point is known as the focal point. After passing through
the focal point, the waves spread out through the water. Reflection of waves off of curved surfaces will be
discussed in more detail in Unit 13 of The Physics Classroom.
Reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves
involves a change in the direction of waves as they pass from one medium to another. Refraction, or the
bending of the path of the waves, is accompanied by a change in speed and wavelength of the waves. In
Lesson 2, it was mentioned that the speed of a wave is dependent upon the properties of the medium
through which the waves travel. So if the medium (and its properties) is changed, the speed of the waves
is changed. The most significant property of water that would affect the speed of waves traveling on its
surface is the depth of the water. Water waves travel fastest when the medium is the deepest. Thus, if
water waves are passing from deep water into shallow water, they will slow down. And as mentioned in
the previous section of Lesson 3, this decrease in speed will also be accompanied by a decrease in
wavelength. So as water waves are transmitted from deep water into shallow water,
the speed decreases, the wavelength decreases, and the direction changes.
This boundary behavior of water waves can be observed in a ripple tank if the tank is
partitioned into a deep and a shallow section. If a pane of glass is placed in the bottom
of the tank, one part of the tank will be deep and the other part of the tank will be
shallow. Waves traveling from the deep end to the shallow end can be seen to refract
(i.e., bend), decrease wavelength (the wavefronts get closer together), and slow down
(they take a longer time to travel the same distance). When traveling from deep water to shallow water,
the waves are seen to bend in such a manner that they seem to be traveling more perpendicular to the
surface. If traveling from shallow water to deep water, the waves bend in the opposite direction. The
refraction of light waves will be discussed in more detail in a later unit of The Physics Classroom.
Reflection involves a change in direction of waves when they bounce off a barrier; refraction of waves
involves a change in the direction of waves as they pass from one medium to
another; and diffraction involves a change in direction of waves as they pass
through an opening or around a barrier in their path. Water waves have the ability to
travel around corners, around obstacles and through openings. This ability is most
obvious for water waves with longer wavelengths. Diffraction can be demonstrated
by placing small barriers and obstacles in a ripple tank and observing the path of the
water waves as they encounter the obstacles. The waves are seen to pass around the
barrier into the regions behind it; subsequently the water behind the barrier is
disturbed. The amount of diffraction (the sharpness of the bending) increases with increasing wavelength
and decreases with decreasing wavelength. In fact, when the wavelength of the waves is smaller than the
obstacle, no noticeable diffraction occurs.
Diffraction of water waves is observed in a harbor as waves bend around small boats and are found to
disturb the water behind them. The same waves however are unable to diffract around larger boats since
their wavelength is smaller than the boat. Diffraction of sound waves is commonly observed; we notice
sound diffracting around corners, allowing us to hear others who are speaking to us from adjacent rooms.
Many forest-dwelling birds take advantage of the diffractive ability of long-wavelength sound waves.
Owls for instance are able to communicate across long distances due to the fact that their long-wavelength
hoots are able to diffract around forest trees and carry farther than the short-wavelength tweets of
songbirds. Diffraction is observed of light waves but only when the waves encounter obstacles with
extremely small wavelengths (such as particles suspended in our atmosphere). Diffraction of sound waves
and of light waves will be discussed in a later unit of The Physics Classroom Tutorial.
Reflection, refraction and diffraction are all boundary behaviors of waves associated with the bending of
the path of a wave. The bending of the path is an observable behavior when the medium is a two- or
three-dimensional medium. Reflection occurs when there is a bouncing off of a barrier. Reflection of
waves off straight barriers follows the law of reflection. Reflection of waves off parabolic barriers results
in the convergence of the waves at a focal point. Refraction is the change in direction of waves that occurs
when waves travel from one medium to another. Refraction is always accompanied by a wavelength and
speed change. Diffraction is the bending of waves around obstacles and openings. The amount of
diffraction increases with increasing wavelength.
Sound is a Mechanical Wave
Sound and music are parts of our everyday sensory experience. Just as humans have eyes for the detection
of light and color, so we are equipped with ears for the detection of sound. We seldom take the time to
ponder the characteristics and behaviors of sound and the mechanisms by which sounds are produced,
propagated, and detected. The basis for an understanding of sound, music and hearing is the physics of
waves. Sound is a wave that is created by vibrating objects and propagated through a medium from one
location to another. In this unit, we will investigate the nature, properties and behaviors of sound waves
and apply basic wave principles towards an understanding of music.
As discussed in the previous unit of The Physics Classroom Tutorial, a wave can be described as a
disturbance that travels through a medium, transporting energy from one location to another location. The
medium is simply the material through which the disturbance is moving; it can be thought of as a series of
interacting particles. The example of a slinky wave is often used to illustrate the nature of a wave. A
disturbance is typically created within the slinky by the back and forth movement of the first coil of the
slinky. The first coil becomes disturbed and begins to push or pull on the second coil. This push or pull on
the second coil will displace the second coil from its equilibrium position. As the second coil becomes
displaced, it begins to push or pull on the third coil; the push or pull on the third coil displaces it from its
equilibrium position. As the third coil becomes displaced, it begins to push or pull on the fourth coil. This
process continues in consecutive fashion, with each individual particle acting to displace the adjacent
particle. Subsequently the disturbance travels through the slinky. As the disturbance moves from coil to
coil, the energy that was originally introduced into the first coil is transported along the medium from one
location to another.
A sound wave is similar in nature to a slinky wave for a variety of reasons. First, there is a medium that
carries the disturbance from one location to another. Typically, this medium is air, though it could be any
material such as water or steel. The medium is simply a series of interconnected and interacting particles.
Second, there is an original source of the wave, some vibrating object capable of disturbing the first
particle of the medium. The disturbance could be created by the vibrating vocal cords of a person, the
vibrating string and soundboard of a guitar or violin, the vibrating tines of a tuning fork, or the vibrating
diaphragm of a radio speaker. Third, the sound wave is transported from one location to another by means
of particle-to-particle interaction. If the sound wave is moving through air, then as one air particle is
displaced from its equilibrium position, it exerts a push or pull on its nearest neighbors, causing them to
be displaced from their equilibrium position. This particle interaction continues throughout the entire
medium, with each particle interacting and causing a disturbance of its nearest neighbors. Since a sound
wave is a disturbance that is transported through a medium via the mechanism of particle-to-particle
interaction, a sound wave is characterized as a mechanical wave.
The creation and propagation of sound waves are often demonstrated in class through the use of a tuning
fork. A tuning fork is a metal object consisting of two tines capable of vibrating if struck by a rubber
hammer or mallet. As the tines of the tuning forks vibrate back and forth, they begin to disturb
surrounding air molecules. These disturbances are passed on to adjacent air molecules by the mechanism
of particle interaction. The motion of the disturbance, originating at the tines of the tuning fork and
traveling through the medium (in this case, air) is what is referred to as a sound wave. The generation and
propagation of a sound wave is demonstrated in the animation below.
Many Physics demonstration tuning forks are mounted on a sound box.
In such instances, the vibrating tuning fork, being connected to the
sound box, sets the sound box into vibrational motion. In turn, the sound
box, being connected to the air inside of it, sets the air inside of the
sound box into vibrational motion. As the tines of the tuning fork, the
structure of the sound box, and the air inside of the sound box begin
vibrating at the same frequency, a louder sound is produced. In fact, the
more particles that can be made to vibrate, the louder or more amplified
the sound. This concept is often demonstrated by the placement of a
vibrating tuning fork against the glass panel of an overhead projector or
on the wooden door of a cabinet. The vibrating tuning fork sets the glass
panel or wood door into vibrational motion and results in an amplified sound.
We know that a tuning fork is vibrating because we hear the sound that is produced by its vibration.
Nonetheless, we do not actually visibly detect any vibrations of the tines.
This is because the tines are vibrating at a very high frequency. If the tuning
fork that is being used corresponds to middle C on the piano keyboard, then
the tines are vibrating at a frequency of 256 Hertz; that is, 256 vibrations per
second. We are unable to visibly detect vibrations of such high frequency. A
common physics demonstration involves slowing down the vibrations by
through the use of a strobe light. If the strobe light puts out a flash of light at
a frequency of 512 Hz (two times the frequency of the tuning fork), then the
tuning fork can be observed to be moving in a back and forth motion. With
the room darkened, the strobe would allow us to view the position of the tines
two times during their vibrational cycle. Thus we would see the tines when they are displaced far to the
left and again when they are displaced far to the right. This would be convincing proof that the tines of
the tuning fork are indeed vibrating to produce sound.
In a previous unit of The Physics Classroom Tutorial, a distinction was made between two categories of
waves: mechanical waves and electromagnetic waves. Electromagnetic waves are waves that have an
electric and magnetic nature and are capable of traveling through a
vacuum. Electromagnetic waves do not require a medium in order to
transport their energy. Mechanical waves are waves that require a
medium in order to transport their energy from one location to
another. Because mechanical waves rely on particle interaction in
order to transport their energy, they cannot travel through regions of
space that are void of particles. That is, mechanical waves cannot
travel through a vacuum. This feature of mechanical waves is often
demonstrated in a Physics class. A ringing bell is placed in a jar and
air inside the jar is evacuated. Once air is removed from the jar, the
sound of the ringing bell can no longer be heard. The clapper is seen
striking the bell; but the sound that it produces cannot be heard because there are no particles inside of the
jar to transport the disturbance through the vacuum. Sound is a mechanical wave and cannot travel
through a vacuum.
Check Your Understanding
1. A sound wave is different than a light wave in that a sound wave is
a. produced by an oscillating object and a light wave is not.
b. not capable of traveling through a vacuum.
c. not capable of diffracting and a light wave is.
d. capable of existing with a variety of frequencies and a light wave has a single frequency.
Sound as a Longitudinal Wave
In the first part of Lesson 1, it was mentioned that sound is a mechanical wave that is created by a
vibrating object. The vibrations of the object set particles in the surrounding medium in vibrational
motion, thus transporting energy through the medium. For a sound wave traveling through air, the
vibrations of the particles are best described as longitudinal. Longitudinal waves are waves in which the
motion of the individual particles of the medium is in a direction that is parallel to the direction of energy
transport. A longitudinal wave can be created in a slinky if the slinky is stretched out in a horizontal
direction and the first coils of the slinky are vibrated horizontally. In such a case, each individual coil of
the medium is set into vibrational motion in directions parallel to the direction that the energy is
transported.
Sound waves in air (and any fluid medium) are longitudinal waves because particles of the medium
through which the sound is transported vibrate parallel to the direction that the sound wave moves. A
vibrating string can create longitudinal waves as depicted in the animation below. As the vibrating string
moves in the forward direction, it begins to push upon surrounding air molecules, moving them to the
right towards their nearest neighbor. This causes the air molecules to the right of the string to be
compressed into a small region of space. As the vibrating string moves in the reverse direction (leftward),
it lowers the pressure of the air immediately to its right, thus causing air molecules to move back leftward.
The lower pressure to the right of the string causes air molecules in that region immediately to the right of
the string to expand into a large region of space. The back and forth vibration of the string causes
individual air molecules (or a layer of air molecules) in the region immediately to the right of the string to
continually vibrate back and forth horizontally. The molecules move rightward as the string moves
rightward and then leftward as the string moves leftward. These back and forth vibrations are imparted to
adjacent neighbors by particle-to-particle interaction. Other surrounding particles begin to move
rightward and leftward, thus sending a wave to the right. Since air molecules (the particles of the
medium) are moving in a direction that is parallel to the direction that the wave moves, the sound wave is
referred to as a longitudinal wave. The result of such longitudinal vibrations is the creation of
compressions and rarefactions within the air.
Sound is a Pressure Wave
Sound is a mechanical wave that results from the back and forth vibration of the particles of the medium
through which the sound wave is moving. If a sound wave is moving from left to right through air, then
particles of air will be displaced both rightward and leftward as the energy of the sound wave passes
through it. The motion of the particles is parallel (and anti-parallel) to the direction of the energy
transport. This is what characterizes sound waves in air as longitudinal waves.
A vibrating tuning fork is capable of creating such a longitudinal wave. As the tines of the fork vibrate
back and forth, they push on neighboring air particles. The forward motion of a tine pushes air molecules
horizontally to the right and the backward retraction of the tine creates a low-pressure area allowing the
air particles to move back to the left.
Because of the longitudinal motion of the air particles, there are regions in the air where the air particles
are compressed together and other regions where the air particles are spread apart. These regions are
known as compressions and rarefactions respectively. The compressions are regions of high air pressure
while the rarefactions are regions of low air pressure. The diagram below depicts a sound wave created by
a tuning fork and propagated through the air in an open tube. The compressions and rarefactions are
labeled.
The wavelength of a wave is merely the distance that a disturbance travels along the medium in one
complete wave cycle. Since a wave repeats its pattern once every wave cycle, the wavelength is
sometimes referred to as the length of the repeating patterns - the length of one complete wave. For a
transverse wave, this length is commonly measured from one wave crest to the next adjacent wave crest
or from one wave trough to the next adjacent wave trough. Since a longitudinal wave does not contain
crests and troughs, its wavelength must be measured differently. A longitudinal wave consists of a
repeating pattern of compressions and rarefactions. Thus, the wavelength is commonly measured as the
distance from one compression to the next adjacent compression or the distance from one rarefaction to
the next adjacent rarefaction.
Since a sound wave consists of a repeating pattern of high-pressure and low-pressure regions moving
through a medium, it is sometimes referred to as a pressure wave. If a detector, whether it is the human
ear or a man-made instrument, were used to detect a sound wave, it would detect fluctuations in pressure
as the sound wave impinges upon the detecting device. At one instant in time, the detector would detect a
high pressure; this would correspond to the arrival of a compression at the detector site. At the next
instant in time, the detector might detect normal pressure. And then finally a low pressure would be
detected, corresponding to the arrival of a rarefaction at the detector site. The fluctuations in pressure as
detected by the detector occur at periodic and regular time intervals. In fact, a plot of pressure versus time
would appear as a sine curve. The peak points of the sine curve correspond to compressions; the low
points correspond to rarefactions; and the "zero points" correspond to the pressure that the air would have
if there were no disturbance moving through it. The diagram below depicts the correspondence between
the longitudinal nature of a sound wave in air and the pressure-time fluctuations that it creates at a fixed
detector location.
The above diagram can be somewhat misleading if you are not careful. The representation of sound by a
sine wave is merely an attempt to illustrate the sinusoidal nature of the pressure-time fluctuations. Do not
conclude that sound is a transverse wave that has crests and troughs. Sound waves traveling through air
are indeed longitudinal waves with compressions and rarefactions. As sound passes through air (or any
fluid medium), the particles of air do not vibrate in a transverse manner. Do not be misled - sound waves
traveling through air are longitudinal waves.
Check Your Understanding
1. A sound wave is a pressure wave; regions of high (compressions) and low pressure (rarefactions) are
established as the result of the vibrations of the sound source. These compressions and rarefactions result
because sound
a. is more dense than air and thus has more inertia, causing the bunching up of sound.
b. waves have a speed that is dependent only upon the properties of the medium.
c. is like all waves; it is able to bend into the regions of space behind obstacles.
d. is able to reflect off fixed ends and interfere with incident waves
e. vibrates longitudinally; the longitudinal movement of air produces pressure fluctuations
Regardless of the source of the sound wave - whether it is a vibrating string or the vibrating tines of a
tuning fork - sound waves traveling through air are longitudinal waves. And the essential characteristic of
a longitudinal wave that distinguishes it from other types of waves is that the particles of the medium
move in a direction parallel to the direction of energy transport.
Sound Properties and Their Perception
Pitch and Frequency | Intensity and the Decibel Scale | The Speed of Sound | The Human Ear
Pitch and Frequency
A sound wave, like any other wave, is introduced into a medium by a vibrating object. The vibrating
object is the source of the disturbance that moves through the medium. The
vibrating object that creates the disturbance could be the vocal cords of a
person, the vibrating string and sound board of a guitar or violin, the
vibrating tines of a tuning fork, or the vibrating diaphragm of a radio speaker.
Regardless of what vibrating object is creating the sound wave, the particles
of the medium through which the sound moves is vibrating in a back and forth motion at a given
frequency. The frequency of a wave refers to how often the particles of the medium vibrate when a wave
passes through the medium. The frequency of a wave is measured as the number of complete back-andforth vibrations of a particle of the medium per unit of time. If a particle of air undergoes 1000
longitudinal vibrations in 2 seconds, then the frequency of the wave would be 500 vibrations per second.
A commonly used unit for frequency is the Hertz (abbreviated Hz), where
1 Hertz = 1 vibration/second
As a sound wave moves through a medium, each particle of the medium vibrates at the same frequency.
This is sensible since each particle vibrates due to the motion of its nearest neighbor. The first particle of
the medium begins vibrating, at say 500 Hz, and begins to set the second particle into vibrational motion
at the same frequency of 500 Hz. The second particle begins vibrating at 500 Hz and thus sets the third
particle of the medium into vibrational motion at 500 Hz. The process continues throughout the medium;
each particle vibrates at the same frequency. And of course the frequency at which each particle vibrates
is the same as the frequency of the original source of the sound wave. Subsequently, a guitar string
vibrating at 500 Hz will set the air particles in the room vibrating at the same frequency of 500 Hz, which
carries a sound signal to the ear of a listener, which is detected as a 500 Hz sound wave.
The back-and-forth vibrational motion of the particles of the medium would not be the only observable
phenomenon occurring at a given frequency. Since a sound wave is a pressure wave, a detector could be
used to detect oscillations in pressure from a high pressure to a low pressure and back to a high pressure.
As the compressions (high pressure) and rarefactions (low pressure) move through the medium, they
would reach the detector at a given frequency. For example, a compression would reach the detector 500
times per second if the frequency of the wave were 500 Hz. Similarly, a rarefaction would reach the
detector 500 times per second if the frequency of the wave were 500 Hz. The frequency of a sound wave
not only refers to the number of back-and-forth vibrations of the particles per unit of time, but also refers
to the number of compressions or rarefactions that pass a given point per unit of time. A detector could be
used to detect the frequency of these pressure oscillations over a given period of time. The typical output
provided by such a detector is a pressure-time plot as shown below.
Since a pressure-time plot shows the fluctuations in pressure over time, the period of the sound wave can
be found by measuring the time between successive high pressure points (corresponding to the
compressions) or the time between successive low pressure points (corresponding to the rarefactions). As
discussed in an earlier unit, the frequency is simply the reciprocal of the period. For this reason, a sound
wave with a high frequency would correspond to a pressure time plot with a small period - that is, a plot
corresponding to a small amount of time between successive high pressure points. Conversely, a sound
wave with a low frequency would correspond to a pressure time plot with a large period - that is, a plot
corresponding to a large amount of time between successive high pressure points. The diagram below
shows two pressure-time plots, one corresponding to a high frequency and the other to a low frequency.
The ears of a human (and other animals) are sensitive detectors capable of detecting the fluctuations in air
pressure that impinge upon the eardrum. The mechanics of the ear's detection ability will be discussed
later in this lesson. For now, it is sufficient to say that the human ear is capable of detecting sound waves
with a wide range of frequencies, ranging between approximately 20 Hz to 20 000 Hz. Any sound with a
frequency below the audible range of hearing (i.e., less than 20 Hz) is known as an infrasound and any
sound with a frequency above the audible range of hearing (i.e., more than 20 000 Hz) is known as an
ultrasound. Humans are not alone in their ability to detect a wide range of frequencies. Dogs can detect
frequencies as low as approximately 50 Hz and as high as 45 000 Hz. Cats can detect frequencies as low
as approximately 45 Hz and as high as 85 000 Hz. Bats, being nocturnal creature, must rely on sound
echolocation for navigation and hunting. Bats can detect frequencies as high as 120 000 Hz. Dolphins can
detect frequencies as high as 200 000 Hz. While dogs, cats, bats, and dolphins have an unusual ability to
detect ultrasound, an elephant possesses the unusual ability to detect infrasound, having an audible range
from approximately 5 Hz to approximately 10 000 Hz.
The sensation of a frequency is commonly referred to as the pitch of a sound. A high pitch sound
corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound
wave. Amazingly, many people, especially those who have been musically trained, are capable of
detecting a difference in frequency between two separate sounds that is as little as 2 Hz. When two
sounds with a frequency difference of greater than 7 Hz are played simultaneously, most people are
capable of detecting the presence of a complex wave pattern resulting from the interference and
superposition of the two sound waves. Certain sound waves when played (and heard) simultaneously will
produce a particularly pleasant sensation when heard, are said to be consonant. Such sound waves form
the basis of intervals in music. For example, any two sounds whose frequencies make a 2:1 ratio are said
to be separated by an octave and result in a particularly pleasing sensation when heard. That is, two sound
waves sound good when played together if one sound has twice the frequency of the other. Similarly two
sounds with a frequency ratio of 5:4 are said to be separated by an interval of a third; such sound waves
also sound good when played together. Examples of other sound wave intervals and their respective
frequency ratios are listed in the table below.
Interval
Frequency Ratio
Examples
Octave
2:1
512 Hz and 256 Hz
Third
5:4
320 Hz and 256 Hz
Fourth
4:3
342 Hz and 256 Hz
Fifth
3:2
384 Hz and 256 Hz
The ability of humans to perceive pitch is associated with the frequency of the sound wave that impinges
upon the ear. Because sound waves traveling through air are longitudinal waves that produce high- and
low-pressure disturbances of the particles of the air at a given frequency, the ear has an ability to detect
such frequencies and associate them with the pitch of the sound. But pitch is not the only property of a
sound wave detectable by the human ear. In the next part of Lesson 2, we will investigate the ability of
the ear to perceive the intensity of a sound wave.
Check Your Understanding
1. Two musical notes that have a frequency ratio of 2:1 are said to be separated by an octave. A musical
note that is separated by an octave from middle C (256 Hz) has a frequency of _____.
a. 128 Hz
d. 345 Hz
b. 254 Hz
c. 258 Hz
e. none of these
Reflection and Its Importance
The Role of Light to Sight | The Line of Sight | The Law of Reflection
Specular vs. Diffuse Reflection
The Role of Light to Sight
The bottom line is: without light, there would be no sight. The visual ability of humans and other animals
is the result of the complex interaction of light, eyes and brain. We are able to see because light from an
object can move through space and reach our eyes. Once light reaches our eyes, signals are sent to our
brain, and our brain deciphers the information in order to detect the appearance,
location and movement of the objects we are sighting at. The whole process, as
complex as it is, would not be possible if it were not for the presence of light. Without light, there would
be no sight.
If you were to turn off the room lights for a moment and then cover all the windows with black
construction paper to prevent any entry of light into the room, then you would notice that nothing in the
room would be visible. There would be objects present that were capable of being seen. There would be
eyes present that would be capable of detecting light from those objects. There would be a brain present
that would be capable of deciphering the information sent to it. But there would be no light! The room
and everything in it would look black. The appearance of black is merely a sign of the absence of light.
When a room full of objects (or a table, a shirt or a sky) looks black, then the objects are not generating
nor reflecting light to your eyes. And without light, there would be no sight.
The objects that we see can be placed into one of two categories: luminous objects and illuminated
objects. Luminous objects are objects that generate their own light. Illuminated objects are objects that
are capable of reflecting light to our eyes. The sun is an example of a luminous object, while the moon is
an illuminated object. During the day, the sun generates sufficient light to illuminate objects on Earth.
The blue skies, the white clouds, the green grass, the colored leaves of fall, the neighbor's house, and the
car approaching the intersection are all seen as a result of light from the sun (the luminous object)
reflecting off the illuminated objects and traveling to our eyes. Without the light from the luminous
objects, these illuminated objects would not be seen. During the evening when the Earth has rotated to a
position where the light from the sun can no longer reach our part of the Earth (due to its inability to bend
around the spherical shape of the Earth), objects on Earth appear black (or at least so dark that we could
say they are nearly black). In the absence of a porch light or a street light, the neighbor's house can no
longer be seen; the grass is no longer green, but rather black; the leaves on the trees are dark; and were it
not for the headlights of the car, it would not be seen approaching the intersection. Without luminous
objects generating light that propagates through space to illuminate non-luminous objects, those nonluminous objects cannot bee seen. Without light, there would be no sight.
A common Physics demonstration involves the directing of a laser beam across the room. With the room
lights off, the laser is turned on and its beam is directed towards a plane mirror. The presence of the light
beam cannot be detected as it travels towards the mirror. Furthermore, the light beam cannot be detected
after reflecting off the mirror and traveling through the air towards a wall in the room. The only locations
where the presence of the light beam can be detected are at the location where the light beam strikes the
mirror and at the location where the light beam strikes a wall. At these two locations, a portion of the light
in the beam is reflecting off the objects (the mirror and the wall) and traveling towards the students' eyes.
And since the detection of objects is dependent upon light traveling from that object to the eye, these are
the only two locations where one can detect the light beam. But in between the laser and the mirror, the
light beam cannot be detected. There is nothing present in the region between the laser and the mirror that
is capable of reflecting the light of the beam to students' eyes.
But then the phenomenal occurred (as it often does in a Physics class). A mister is used to spray water
into the air in the region where the light beam is moving. Small suspended droplets of water are capable
of reflecting light from the beam to your eye. It is only due to the presence of the suspended water
droplets that the light path from the laser to the mirror could be detected. When light from the laser (a
luminous object) strikes the suspended water droplets (the illuminated object), the light is reflected to
students' eyes. The path of the light beam can now be seen. With light, there can be sight. But without
light, there would be no sight.
None of us generate light in the visible region of the electromagnetic spectrum. We are not brilliant
objects (please take no offense) like the sun; rather, we are illuminated objects like the moon. We make
our presence visibly known by reflecting light to the eyes of those who look our way. It is only by
reflection that we, as well as most of the other objects in our physical world, can be seen. And if reflected
light is so essential to sight, then the very nature of light reflection is a worthy topic of study among
students of physics. And in this lesson and the several that follow, we will undertake a study of the way
light reflects off objects and travels to our eyes in order to allow us to view them.
The Line of Sight
In the first section of Lesson 1, it was stated, "without light, there would be no sight." Everything that can
be seen is seen only when light from that object travels to our eyes. Whether it be a luminous object (that
generates light of its own) or an illuminated object (that reflects the light that is incident upon it), you can
only view the object when light from that object travels to your eye. As you look at Mary in class, you are
able to see Mary because she is illuminated with light and that light reflects off of her and travels to your
eye. In the process of viewing Mary, you are directing your sight along a line in the direction of Mary. If
you wish to view the top of Mary's head, then you direct your sight along a line towards the top of her
head. If you wish to view Mary's feet, then you direct your sight along a line towards Mary's feet. And if
you wish to view the image of Mary in a mirror, then you must direct your sight along a line towards the
location of Mary's image. This directing of our sight in a specific direction is sometimes referred to as the
line of sight.
It is a rather simple principle:
In order to view an object, you must sight along a line at that object; and when you do light will come
from that object to your eye along the line of sight.
A luminous object emits light in a variety of directions; and an illuminated object reflects light in a
variety of directions. Although this light diverges from the object in a variety of directions, your eye only
sees the very small diverging cone of rays that is coming towards it. If your eye were located at a different
location, then you would see a different cone of rays. Regardless of the eye location, you will still need to
sight along a line in a specific direction in order to view the object.
While simple, this concept of the line of sight is also profound! This very principle of the line of sight will
assist us in understanding the formation of images in both this unit (reflection) and the next unit
(refraction).
A common Physics lab involves the determination of the image location of a
pencil (or some object) as formed by a plane mirror. In the process of
determining the image location, the manner in which light from the object
travels to your eye is investigated. First, the method of parallax is used to locate
the image of the object. Two pencils are inserted into rubber stoppers; one
stoppered pencil serves as the object and the other serves to assist the student in
locating the image. The object pencil is placed in front of a plane mirror. Then
the student sights at the image of the object pencil in the mirror. As a student
sights along a line (the line of sight) at the image of the pencil, the second
pencil is placed behind the mirror along the same line of sight; this is called the image pencil. When
placed along the line of sight, the portion of the image pencil that extends above the mirror will be aligned
with the image that is seen in the mirror. Then the eye location is repositioned to the other side of the
object pencil and the process is repeated. The precise image location of the object is the location where all
lines of sight intersect regardless of where the eye is located. Two important ideas are gleaned from such
a lab: one pertains to how light travels from the object to the eye and one pertains to the location of the
image of an object.
As you sight at the image of an object in the mirror (whether it be a stoppered pencil or any object), light
travels along your line of sight towards your eye. The object is being illuminated by light in the room; a
countless number of rays of light are reflecting off the object in a variety of directions. When viewing the
image of the object in a plane mirror, one of these rays of light originates at the object location and first
travels along a line towards the mirror (as represented by the blue ray in the diagram below). This ray of
light is known as the incident ray - the light ray approaching the mirror. The incident ray intersects the
mirror at the same location where your line of sight intersects the mirror. The light ray then reflects off
the mirror and travels to your eye (as represented by the red ray in the diagram below); this ray of light is
known as the reflected ray.
So the manner in which light travels to your eye as you view the image of an object in a mirror can be
summarized as follows.
To view the image of an object in a mirror, you must sight along a line at the image. One of the many
rays of light from the object will approach the mirror and reflect along your line of sight to your eye.
The second important idea that can be gleaned from this stoppered pencil lab pertains to the location of
the image. Observe in the diagram above that the image is positioned directly across the mirror along a
line that runs perpendicular to the mirror. The distance from the mirror to the object (known as the object
distance) is equal to the distance from the mirror to the image (known as the image distance). For all
plane mirrors, this equality holds true:
Object distance = Image distance
In the next part of Lesson 1, we will investigate the reflection of light in more detail. Our focus will be to
identify a law that governs light reflection.
Check Your Understanding
The following diagrams depict some ideas about how light might travel from an object location to an eye
location when the image of the object is viewed in a mirror. Comment on the incorrectness of the
following diagrams. Discuss what makes them incorrect.
1.
The Law of Reflection
Light is known to behave in a very predictable manner. If a ray of light could be observed approaching
and reflecting off of a flat mirror, then the behavior of the light as it reflects would follow a predictable
law known as the law of reflection. The diagram below illustrates the law of reflection.
In the diagram, the ray of light approaching the mirror is known as the incident ray (labeled I in the
diagram). The ray of light that leaves the mirror is known as the reflected ray (labeled R in the diagram).
At the point of incidence where the ray strikes the mirror, a line can be drawn perpendicular to the surface
of the mirror. This line is known as a normal line (labeled N in the diagram). The normal line divides the
angle between the incident ray and the reflected ray into two equal angles. The angle between the incident
ray and the normal is known as the angle of incidence. The angle between the reflected ray and the
normal is known as the angle of reflection. (These two angles are labeled with the Greek letter "theta"
accompanied by a subscript; read as "theta-i" for angle of incidence and "theta-r" for angle of reflection.)
The law of reflection states that when a ray of light reflects off a surface, the angle of incidence is equal to
the angle of reflection.
It is common to observe this law at work in a Physics lab such as the one described in the previous part of
Lesson 1. To view an image of a pencil in a mirror, you must sight along a line at the image location. As
you sight at the image, light travels to your eye along the path shown in the diagram below. The diagram
shows that the light reflects off the mirror in such a manner that the angle of incidence is equal to the
angle of reflection.
It just so happens that the light that travels along the line of sight to your eye follows the law of reflection.
(The reason for this will be discussed later in Lesson 2). If you were to sight along a line at a different
location than the image location, it would be impossible for a ray of light to come from the object, reflect
off the mirror according to the law of reflection, and subsequently travel to your eye. Only when you sight
at the image, does light from the object reflect off the mirror in accordance with the law of reflection and
travel to your eye. This truth is depicted in the diagram below.
For example, in Diagram A above, the eye is sighting along a line at a position above the actual image
location. For light from the object to reflect off the mirror and travel to the eye, the light would have to
reflect in such a way that the angle of incidence is less than the angle of reflection. In Diagram B above,
the eye is sighting along a line at a position below the actual image location. In this case, for light from
the object to reflect off the mirror and travel to the eye, the light would have to reflect in such a way that
the angle of incidence is more than the angle of reflection. Neither of these cases would follow the law of
reflection. In fact, in each case, the image is not seen when sighting along the indicated line of sight. It is
because of the law of reflection that an eye must sight at the image location in order to see the image of an
object in a mirror.
Image Formation in Plane Mirrors
Why is an Image Formed? | Image Characteristics in Plane Mirrors
Ray Diagrams for Plane Mirrors | What Portion of a Mirror is Required to View an Image?
Right Angle Mirrors | Other Multiple Mirror Systems
Why is an Image Formed?
In Lesson 1 of this unit of the Physics Classroom Tutorial, the manner in which light reflected off objects
in order to allow us to see them was discussed. A major principle in that lesson was expressed as follows:
In order to view an object, you must sight along a line at that object; and when you do light will come
from that object to your eye along the line of sight.
This very principle can be extended to the task of viewing the image of an object in a plane (i.e., flat)
mirror:
In order to see the image of an object in a mirror, you must sight at the image; when you sight at the
image, light will come to your eye along that line of sight.
The image location is thus located at that position where observers are sighting when viewing the image
of an object. It is the location behind the mirror where all the light appears to diverge from. In the diagram
below, three individuals are sighting at the image of an object along three different lines of sight. Each
person sees the image due to the reflection of light off the mirror in accordance with the law of reflection.
When each line of sight is extended backwards, each line will intersect at the same point. This point is the
image point of the object.
This principle can be illustrated in a Physics class using a 5-foot plane mirror and a pair of large
cylinders. One cylinder is placed in front of the mirror and students from different locations in the room
are asked to sight at its image. The second cylinder is then aligned along the line of sight and readjusted
until it is in line with each person's line of sight. Regardless of who is viewing the image and from where
they are viewing the image, each sight line must intersect in the same location. It is possible that the
second cylinder is aligned with one student's line of sight but not with another student's. If this is so, then
the cylinder is not placed at the exact location of the image. This is depicted in the diagram below.
In a case such as this, the cylinder position is adjusted until it is located at the position where all students
in the classroom can see it extending above the mirror and in line with the image that each student sees
when looking in the mirror. Only, then can we conclude the cylinder is located at the image position.
Since there is only one image for an object placed in front of a plane mirror, it is reasonable that every
sight line would intersect in a single location. This location of intersection is known as the image location.
The image location is simply the one location in space where it seems to every observer that the light is
diverging from. Regardless of where the observer is located, when the observer sights at the image
location, the observer is sighting along a line towards the same location that all other observers are
sighting. And as mentioned in an earlier lesson, the perpendicular distance from this image location to the
mirror is equal to the perpendicular distance from the object location to the mirror. In fact, the image
location is directly across the mirror from the object location and an equal distance from the mirror.
Of course, it is possible that certain individuals in the room will be unable to view the image of an object
in a plane mirror. Because of the person's position relative to the image position and to the extremities of
the mirror, the person is unable to detect a ray of light reflecting to their eye as they sight at the image
location. This does not mean that there is no image. Indeed, any object positioned in front of a plane
mirror (or even to the side of the plane mirror) has an image regardless of whether there are people
positioned in an appropriate location to view it. In the diagram below, there is an image of an object
located on the other side of the mirror. However, Ray Zuvlite is unable to view the image due to his
position in the room. Ray is certainly able to sight in the direction of the image location. However, the
light from the object is unable to reflect off the mirror in accordance with the law of reflection and travel
to his eye along his line of sight. Since light from the object does not make it to his eye, Ray is unable to
see the image of the object in the mirror.
Of course, this problem could be remedied if the mirror were wider, if the object were moved to the left or
closer to the mirror, and/or if Ray moved to the left. Repositioning the object, the mirror, and/or the
person could result in a ray of light from the object reflecting off the mirror and traveling to Ray's eye.
The diagram below depicts this remedy.
So why is an image formed by a plane mirror? An image is formed because light emanates from an object
in a variety of directions. Some of this light (which we represent by rays) reaches the mirror and reflects
off the mirror according to the law of reflection. Each one of these rays of light can be extended
backwards behind the mirror where they will all intersect at a point (the image point). Any person who is
positioned along the line of one of these reflected rays can sight along the line and view the image - a
representation of the object.
This principle of image formation is often applied in a Physics lab. Suppose that a mirror is placed on a
sheet of paper that is placed on top of a piece of cardboard. A pin is positioned in an upright position (and
held in place by the cardboard) at a location in front of the mirror. A student can sight along a line at the
image of the pin from a variety of locations. With one eye closed, a straightedge is used to assist in
drawing the lines of sight. These lines of sight are drawn from a variety of sighting locations. Each line of
sight can be extended backwards beyond the mirror. If the sight lines are drawn correctly, then each line
will intersect at the same location. The location of intersection of all sight lines is the image location.
Validation of the accuracy of your sighting and ray tracing can be accomplished by measuring angles of
incidence and angles of reflection on the diagram. These should be equal for each individual sight line.
That is, angle A should equal angle B; angle C should equal angle D; and angle E should equal angle F.
Finally, the object distance can be compared to the image distance; these should also be equal.
Image Characteristics
As discussed in the previous section of Lesson 2, an image
location is the location in space where all the reflected light appears to diverge from. Since light from the
object appears to diverge from this location, a person who sights along a line at this location will perceive
a replica or likeness of the actual object.
In the case of plane mirrors, the image is said to be a virtual image. Virtual images are images that are
formed in locations where light does not actually reach. Light does not actually pass through the location
on the other side of the mirror; it only appears to an observer as though the light is coming from this
location. Whenever a mirror (whether a plane mirror or otherwise) creates an image that is virtual, it will
be located behind the mirror where light does not really come from. Later in this unit, we will study
instances in which real images are formed by curved mirrors. Such images are formed on the same side
of the mirror as the object and light passes through the actual image location.
Besides the fact that plane mirror images are virtual, there are several other characteristics that are worth
noting. The second characteristic has to do with the
orientation of the image. If you view an image of yourself in
a plane mirror (perhaps a bathroom mirror), you will quickly
notice that there is an apparent left-right reversal of the
image. That is, if you raise your left hand, you will notice
that the image raises what would seem to be it's right hand. If
you raise your right hand, the image raises what would seem
to be its left hand. This is often termed left-right reversal.
This characteristic becomes even more obvious if you wear a
shirt with lettering. For example, a shirt displaying the word
"NIKE" will read "EKIN" when viewed in the mirror; a shirt
displaying the word "ILLINOIS" will read "SIONILLI;" and
a shirt displaying the word "BOB" will read "BOB." (NOTE: Not only will the order of letters appear
reversed, the actual orientation of the letters themselves will appear reversed as well. Of course, this is a
little difficult to do when typing from a keyboard.) While there is an apparent left-right reversal of the
orientation of the image, there is no top-bottom vertical reversal. If you stand on your feet in front of a
plane mirror, the image does not stand on its head. Similarly, the ceiling does not become the floor. The
image is said to be upright, as opposed to inverted.
Students of Physics are usually quite intrigued by this apparent left-right reversal. Exactly what is
happening to cause ILLINOIS to read as SIONILLI? And why is the reversal observed in the left to right
direction and not in the head to toe direction? These questions urge us to ponder the situation more
deeply. Let's suppose for a moment that we could print the name of your favorite school subject on your
shirt and have you look in the mirror. We all know that when you look in the mirror, you will see the
letters SCISYHP written on the shirt of your image - the reversed form of PHYSICS. But can we really
say that the word appearing on your shirt is the word PHYSICS (with the letters un-reversed)? The
answer is no! (But you don't have to believe it yet. Keep reading ... and pondering.)
To further explore the reason for this appearance of left-right reversal, let's suppose we write the word
PHYSICS on a transparency and hold it in front of us in front of a plane mirror. If we look at the image of
the transparency in the mirror, we would observe the expected - SCISYHP. The letters are written
reversed when viewed in the mirror. But what if we look at the letters on the transparency? How are those
letters oriented? When we face the mirror and look at the letters on the transparency, we observe the
unexpected - SCISYHP. When viewed from the perspective of the person holding the transparency (and
facing the mirror, the letters exhibit the same left-right reversal as the mirror image. The letters appear
reversed on the image because they are actually reversed on the shirt. At least they are reversed when
viewed from the perspective of a person who is facing the mirror. Imagine that! All this time you thought
the mirror was reversing the letters on your shirt. But the fact is that the letters were already reversed on
your shirt; at least they were reversed from the person who stands behind the T-shirt. The people who
view your shirt from the front have a different reference frame and thus do not see the letters as being
reversed. The apparent left-right reversal of an image is simply a frame of reference phenomenon. When
viewing the image of your shirt in a plane mirror (or any part of the world), you are viewing your shirt
from the front. This is a switch of reference frames.
A third characteristic of plane mirror images pertains to the relationship between the object's distance to
the mirror and the image's distance to the mirror. For plane mirrors, the object distance (often represented
by the symbol do) is equal to the image distance (often represented by the symbol di). That is the image is
the same distance behind the mirror as the object is in front of the mirror. If you stand a distance of 2
meters from a plane mirror, you must focus at a location 2 meters behind the mirror in order to view your
image.
A fourth and final characteristic of plane mirror images is that the dimensions of the image are the same
as the dimensions of the object. If a 1.6-meter tall person stands in front of a mirror, he/she will see an
image that is 1.6-meters tall. If a penny with a diameter of 18-mm is placed in front of a plane mirror, the
image of the penny has a diameter of 18 mm. The ratio of the image dimensions to the object dimensions
is termed the magnification. Plane mirrors produce images that have a magnification of 1.
In conclusion, plane mirrors produce images with a number of distinguishable characteristics. Images
formed by plane mirrors are virtual, upright, left-right reversed, the same distance from the mirror as the
object's distance, and the same size as the object.
Check Your Understanding
1. You might have noticed that emergency vehicles such as ambulances are often labeled on the front
hood with reversed lettering (e.g., ECNALUBMA). Explain why this is so.
2. If Suzie stands 3 feet in front of a plane mirror, how far from the person will her image be located?
3. If a toddler crawls towards a mirror at a rate of 0.25 m/s, then at what speed will the toddler and the
toddler's image approach each other?
Ray Diagrams
The line of sight principle suggests that in order to view an image of an object in a mirror, a person must
sight along a line at the image of the object. When sighting along such a line, light from the object reflects
off the mirror according to the law of reflection and travels to the person's eye. This process was
discussed and explained earlier in this lesson. One useful tool that is frequently used to depict this idea is
known as a ray diagram. A ray diagram is a diagram that traces the path that light takes in order for a
person to view a point on the image of an object. On the diagram, rays (lines with arrows) are drawn for
the incident ray and the reflected ray. Complex objects such as people are often represented by stick
figures or arrows. In such cases it is customary to draw rays for the extreme positions of such objects.
This section of Lesson 2 details and illustrates the procedure for drawing
ray diagrams. Let's begin with the task of drawing a ray diagram to show
how Suzie will be able to see the image of the green object arrow in the
diagram below. For simplicity sake, we will suppose that Suzie is viewing
the image with her left eye closed. Thus, we will focus on how light travels
from the two extremities of the object arrow (the left and right side) to the mirror and finally to Suzie's
right eye as she sights at the image. The four steps of the process for drawing a ray diagram are listed,
described and illustrated below.
1. Draw the image of the object.
Use the principle that the object distance is equal to the image distance to
determine the exact location of the object. Pick one extreme on the object
and carefully measure the distance from this extreme point to the mirror.
Mark off the same distance on the opposite side of the mirror and mark the
image of this extreme point. Repeat this process for all extremes on the
object until you have determined the complete location and shape of the
image. Note that all distance measurements should be made by measuring along a segment that is
perpendicular to the mirror.
2. Pick one extreme on the image of the object and draw the reflected ray that will travel to the eye as it
sights at this point.
Use the line of sight principle: the eye must sight along a line at the image
of the object in order to see the image of the object. It is customary to draw
a bold line for the reflected ray (from the mirror to the eye) and a dashed
line as an extension of this reflected ray; the dashed line extends behind the
mirror to the location of the image point. The reflected ray should have an
arrowhead upon it to indicate the direction that the light is traveling. The
arrowhead should be pointing towards the eye since the light is traveling from the mirror to the eye,
thus enabling the eye to see the image.
3. Draw the incident ray for light traveling from the corresponding extreme on the object to the mirror.
The incident ray reflects at the mirror's surface according to the law of
reflection. But rather than measuring angles, you can merely draw the
incident ray from the extreme of the object to the point of incidence on the
mirror's surface. Since you drew the reflected ray in step 2, the point of
incidence has already been determined; the point of incidence is merely
the point where the line of sight intersects the mirror's surface. Thus draw
the incident ray from the extreme point to the point of incidence. Once more, be sure to draw an
arrowhead upon the ray to indicate its direction of travel. The arrowhead should be pointing towards
the mirror since light travels from the object to the mirror.
4. Repeat steps 2 and 3 for all other extremities on the object.
After completing steps 2 and 3, you have only shown how light travels from
a single extreme on the object to the mirror and finally to the eye. You will
also have to show how light travels from the other extremes on the object
to the eye. This is merely a matter of repeating steps 2 and 3 for each
individual extreme. Once repeated for each extreme, your ray diagram is
complete.
The best way to learn to draw ray diagrams involves trying it yourself. It's easy. Merely duplicate the two
setups below onto a blank sheet of paper, grab a ruler/straightedge, and begin. If necessary, refer to the
four-step procedure listed above. When finished, compare your diagram with the completed diagrams at
the bottom of this page.
Uses of Ray Diagrams
Ray diagrams can be particularly useful for determining and
explaining why only a portion of the image of an object can be seen
from a given location. The ray diagram at the right shows the lines
of sight used by the eye in order to see a portion of the image in the
mirror. Since the mirror is not long enough, the eye can only view
the topmost portion of the image. The lowest point on the image that
the eye can see is that point in line with the line of sight that
intersects the very bottom of the mirror. As the eye tries to view
even lower points on the image, there is not sufficient mirror present
to reflect light from the lower points on the object to the eye. The portion of the object that cannot be seen
in the mirror is shaded green in the diagram below.
Similarly, ray diagrams are useful tools for determining and explaining what objects might be viewed
when sighting into a mirror from a given location. For example, suppose that six students - Al, Bo, Cy,
Di, Ed, and Fred sit in front of a plane mirror and attempt to see each other in the mirror. And suppose the
exercise involves answering the following questions: Whom can Al see? Whom can Bo see? Whom can
Cy see? Whom can Di see? Whom can Ed see? And whom can Fred see?
The task begins by locating the images of the given students. Then, Al is isolated from the rest of the
students and lines of sight are drawn to see who Al can see. The leftward-most student whom Al can see
is the student whose image is to the right of the line of sight that intersects the left edge of the mirror. This
would be Ed. The rightward-most student who Al can see is the student whose image is to the left of the
line of sight that intersects the right edge of the mirror. This would be Fred. Al could see any student
positioned between Ed and Fred by looking at any other positions along the mirror. However in this case,
there are no other students between Ed and Fred; thus, Ed and Fred are the only students whom Al can
see? The diagram below illustrates this using lines of sight for Al.
Of course the same process can be repeated for the other students by observing their lines of sight.
Perhaps you will want to try to determine whom Bo, Cy, Di, Ed, and Fred can see? Then check
your answers by clicking the button below.
Check Your Understanding
1. Six students are arranged in front of a mirror. Their positions are shown below. The image of each
student is also drawn on the diagram. Make the appropriate line of sight constructions to determine that
students each individual student can see.
Here are completed diagrams for the two examples given above.
What Portion of a Mirror is Required?
In the previous part of this lesson, the use of ray diagrams were introduced and illustrated. Ray diagrams
can be used to determine where a person must sight along a mirror in order to see an image of
him/herself. As such, ray diagrams can be used to determine what portion of a plane mirror must be used
in order to view an image. The diagram below depicts a 6-foot tall man standing in front of a plane
mirror. To see the image of his feet, he must sight along a line towards his feet; and to see the image of
the top of his head, he must sight along a line towards the top of his head. The ray diagram depicts these
lines of sight and the complete path of light from his extremities to the mirror and to the eye. In order to
view his image, the man must look as low as point Y (to see his feet) and as high as point X (to see the tip
of his head). The man only needs the portion of mirror extending between points X and Y in order to view
his entire image. All other portions of the mirror are useless to the task of this man viewing his own
image.
The diagram depicts some important information about plane mirrors. Using a cm-ruler, measure the
height of the man (the vertical arrow) on the computer screen and measure the distance between points X
and Y. What do you notice? The man is twice as tall as the distance between points X and Y. In other
words, to view an image of yourself in a plane mirror, you will need an amount of mirror equal to onehalf of your height. A 6-foot tall man needs 3-feet of mirror (positioned properly) in order to view his
entire image.
But what if the man stood a different distance from the mirror? Wouldn't that cause the man to need a
different amount of mirror to view his image? Maybe less mirror would be required in such an instance?
These questions can be explored with the help of another ray diagram. The diagram below depicts a man
standing different distances from a plane mirror. Ray diagrams for each situation (standing close and
standing far away) are drawn. To assist in distinguishing between the two ray diagrams, they have been
color-coded. Red and blue light rays have been used for the situation in which the man is standing far
away. Green and purple light rays have been used for the situation in which the man is standing close to
the mirror.
The two ray diagrams above demonstrate that the distance that a person stands from the mirror will not
affect the amount of mirror that the person needs to see their image. Indeed in the diagram, the man's line
of sight crosses the mirror at the same locations. A 6-foot tall man needs 3-feet of mirror to view his
whole image regardless of where he is standing. In fact, the man needs the exact same 3-feet of mirror.
A common Physics lab involves using a tall plane mirror to explore the relationship between object height
and the portion of mirror needed to view an image. A student stands a few meters from a planer mirror
and views her image. With the student standing upright and still and staring at her feet, the lab partner
moves a marker up and down the mirror until the sight location on the mirror is identified. The partner
then marks this location on the mirror with an erasable marker. The process is repeated for the student
staring at the tip of her head. Of course, being a lab, the procedure is subject to a variety of procedural and
measurement error that may yield less than ideal results. The mirrors are occasionally mounted on a wall
that is not perfectly vertical. Or a student will lean forward a slight amount, thus reducing his/her
effective height. Or the mirror warps over the years leading to one that is concave or convex rather than
planar. Despite these potential complications, the 1:2 ratio between the portion of mirror required to view
the image and the height of the object is often observed.
Check Your Understanding
1. Ben Phooled is 6-feet tall. He is the tallest person in his family. It just so happens that Ben learned the
important principle of the 2:1 relationship just prior to his family's decision to purchase a mirror that was
to be used by the entire family. Enthused about the recent physics lesson, Ben decided to put it to good
use. Ben convinced his parents that it would be a waste of money to buy a mirror longer than 3 feet.
"After all," Ben argued, "I'm the tallest person in the family and only three feet of mirror would be
required to view my image." Ben's parents conceded and they purchased a 3-foot tall mirror and mounted
it on the bathroom wall.
Comment on the wisdom behind the Phooled family decision.
Refraction at a Boundary
Boundary Behavior | Refraction and Sight | The Cause of Refraction
Optical Density and Light Speed | The Direction of Bending | If I were an Archer Fish
Boundary Behavior
A wave doesn't just stop when it reaches the end of the medium. Rather, a wave will undergo certain
behaviors when it encounters the end of the medium. Specifically, there will be some reflection off the
boundary and some transmission into the new medium. The transmitted wave undergoes refraction (or
bending) if it approaches the boundary at an angle. If the boundary is merely an obstacle implanted within
the medium, and if the dimensions of the obstacle are smaller than the wavelength of the wave, then there
will be very noticeable diffraction of the wave around the object. The reflection, refraction, and
diffraction of waves were first introduced in Unit 10 of The Physics Classroom tutorial. In Unit 11 of The
Physics Classroom Tutorial, the reflection, refraction, and diffraction of sound waves was discussed.
Since light is a wave, it too will undergo these same behaviors when it reaches a boundary between two
media. The boundary behavior of light waves has already been introduced in Unit 12 of The Physics
Classroom Tutorial. In this unit, we will focus on the refraction of light in great detail. We will explore
the conceptual and mathematical principles governing the bending of waves as they cross the boundary
between two media. To understand light refraction, we will need to back up a few steps and investigate
the behavior of waves when they reach the end of a medium.
Boundary Behavior for Waves on a Rope
Suppose that there is a thin rope attached to a thick rope, with each rope held at opposite ends by people.
And suppose that a pulse is introduced by the person holding the end of the thin rope. If this is the case,
there will be an incident pulse traveling in the less dense medium (thin rope) towards the boundary with a
more dense medium (thick rope).
Upon reaching the boundary, two behaviors will occur.


A portion of the energy carried by the incident pulse is reflected and returns towards the left
end of the thin rope. The disturbance that returns to the left after bouncing off the boundary is
known as the reflected pulse.
A portion of the energy carried by the incident pulse is transmitted into the thick rope. The
disturbance that continues moving to the right is known as the transmitted pulse.
These two behaviors - reflection and transmission - were first introduced in Unit 10 of The Physics
Classroom. In that unit, it was mentioned that the passage of the energy from the incident medium into
the transmitted medium was accompanied by a change in speed and wavelength. In the case of a pulse
crossing the boundary from a less dense medium into a more dense medium, the speed and the
wavelength are both decreased. On the other hand, if a pulse crosses the boundary from a more dense
medium into a less dense medium, the speed and the wavelength are both increased.
Refraction of Light Waves
The above discussion was limited to the behavior of a wave on a rope. But what if the wave is a light
wave traveling in a three-dimensional medium? For example, what would happen if a light wave is
traveling through air and reaches the boundary with a glass surface? How can the reflection and
transmission behavior of a light wave be described? First, the light wave behaves like the wave on the
rope: a portion of the wave is transmitted into the new medium (glass) and a
portion of the wave reflects off the air-glass boundary. Second, the same wave
property changes that were observed for the wave on the rope are also observed for
the light wave passing from air into glass; there is a change in speed and
wavelength of the wave as it crosses the air-glass boundary. When passing from air
into glass, both the speed and the wavelength decrease. Finally, and most
importantly, the light is observed to change directions as it crosses the boundary
separating the air and the glass. This bending of the path of light is known as
refraction. A one-word synonym for refraction is bending. The transmitted wave
experiences this refraction at the boundary. As seen in the diagram at the right, each individual wavefront
is bent only along the boundary. Once the wavefront has passed across the boundary, it travels in a
straight line. For this reason, refraction is called a boundary behavior. A ray is drawn perpendicular to the
wavefronts; this ray represents the direction that the light wave is traveling. Observe that the ray is a
straight line inside of each of the two media, but bends at the boundary. Again, refraction is a boundary
behavior
The Ray Model of Light
In this unit, we will rely heavily on the use of rays to represent the direction in which light is moving.
While we often think of light behaving as a wave, we will still find it useful to represent its movement
through a medium using a line segment with an arrowhead (i.e., a ray) to depict the refraction of light.
The ray is constructed in a direction perpendicular to the wavefronts of the light wave; this accurately
depicts the light wave's direction. In this sense, we are viewing light as behaving as a stream of particles
that head in the direction of the ray. The idea that the path of light can be represented by a ray is known as
the ray model of light. The same ray model was utilized in Unit 13 of The Physics Classroom Tutorial to
discuss the reflection of light waves.
Refraction and Sight
In Unit 13 of The Physics Classroom Tutorial, it was emphasized that we are able to see because light
from an object can travel to our eyes. Every object that can be seen is seen only because light from that
object travels to our eyes. As you look at Mary in class, you are able to see Mary because she is
illuminated with light and that light reflects off of her and travels to your eye. In the process of viewing
Mary, you are directing your sight along a line in the direction of Mary. If you wish to view the top of
Mary's head, then you direct your sight along a line towards the top of her head. If you wish to view
Mary's feet, then you direct your sight along a line towards Mary's feet. And if you wish to view the
image of Mary in a mirror, then you must direct your sight along a line towards the location of Mary's
image. This directing of our sight in a specific direction is sometimes referred to as the line of sight.
As light travels through a given medium, it travels in a straight line. However, when light passes from one
medium into a second medium, the light path bends. Refraction takes place. The refraction occurs only at
the boundary. Once the light has crossed the boundary between the two media, it continues to travel in a
straight line. Only now, the direction of that line is different than it was in the former medium. If when
sighting at an object, light from that object changes media on the way to your eye, a visual distortion is
likely to occur. This visual distortion is witnessed if you look at a pencil submerged in a glass half-filled
with water. As you sight through the side of the glass at the portion of the pencil located above the water's
surface, light travels directly from the pencil to your eye. Since this light does not change medium, it will
not refract. (Actually, there is a change of medium from air to glass and back into air. Because the glass is
so thin and because the light starts and finished in air, the refraction into and out of the glass causes little
deviation in the light's original direction.) As you sight at the portion of the pencil that was submerged in
the water, light travels from water to air (or from water to glass to air). This light ray changes medium and
subsequently undergoes refraction. As a result, the image of the pencil appears to be broken. Furthermore,
the portion of the pencil that is submerged in water appears to be wider than the portion of the pencil that
is not submerged. These visual distortions are explained by the refraction of light.
In this case, the light rays that undergo a deviation from their original
path are those that travel from the submerged portion of the pencil,
through the water, across the boundary, into the air, and ultimately to the
eye. At the boundary, this ray refracts. The eye-brain interaction cannot
account for the refraction of light. As was emphasized in Unit 13, the
brain judges the image location to be the location where light rays
appear to originate from. This image location is the location where
either reflected or refracted rays intersect. The eye and brain assume that
light travels in a straight line and then extends all incoming rays of light
backwards until they intersect. Light rays from the submerged portion of
the pencil will intersect in a different location than light rays from the
portion of the pencil that extends above the surface of the water. For this
reason, the submerged portion of the pencil appears to be in a different
location than the portion of the pencil that extends above the water. The
diagram at the right shows a God's-eye view of the light path from the
submerged portion of the pencil to each of your two eyes. Only the left
and right extremities (edges) of the pencil are considered. The blue lines
depict the path of light to your right eye and the red lines depict the path
of light to your left eye. Observe that the light path has bent at the
boundary. Dashed lines represent the extensions of the lines of sight
backwards into the water. Observe that these extension lines intersect at
a given point; the point represents the image of the left and the right edge of the pencil. Finally, observe
that the image of the pencil is wider than the actual pencil. A ray model of light that considers the
refraction of light at boundaries adequately explains the broken pencil observations.
Flickr Physics Photo
A straw is placed with a diagonal orientation within a half-filled beaker of water. At the surface of the
water, the straw appears to be misaligned or broken; the portion of the straw above the water is shifted
relative to the image viewed under the water. The bending of the path of light as it passes from the
water to air causes the observed distortion of the image of the straw.
The broken pencil phenomenon occurs during your everyday spearfishing outing. Fortunately for the fish,
light refracts as it travels from the fish in the water to the eyes of the hunter. The refraction occurs at the
water-air boundary. Due to this bending of the path of light, a fish appears to be at a location where it
isn't. A visual distortion occurs. Subsequently, the hunter launches the spear at the location where the fish
is thought to be and misses the fish. Of course, the fish are never concerned about such hunters; they
know that light refracts at the boundary and that the location where the hunter is sighting is not the same
location as the actual fish. How did the fish get so smart and learn all this? They live in schools.
Now any fish that has done his/her physics homework knows that the amount of refraction that occurs is
dependent upon the angle at which the light approaches the boundary. We will investigate this aspect of
refraction in great detail in Lesson 2. For now, it is sufficient to say that as the hunter with the spear sights
more perpendicular to the water, the amount of refraction decreases. The most successful hunters are
those who sight perpendicular to the water. And the smartest fish are those who head for the deep when
they spot hunters who sight in this direction.
Since refraction of light occurs when it crosses the boundary, visual distortions often occur. These
distortions occur when light changes medium as it travels from the object to our eyes.
Flickr Physics Photo
That frosty mug full of A&W root beer seems to drain so quickly. What seems like a lot of root beer is
actually a lot of glass. The deception is uncovered when the root beer mug is submerged in water. When
looking at the mug above the water level, there appears to be a lot of root beer. But when looking at the
mug below the water level, one observes the truth; the water surrounding the glass minimizes the effect
of refraction and reveals the large quantity of glass. We've been deceived!
The Cause of Refraction
We have learned that refraction occurs as light passes across the boundary between two media. Refraction
is merely one of several possible boundary behaviors by which a light wave could behave when it
encounters a new medium or an obstacle in its path. The transmission of light across a boundary between
two media is accompanied by a change in both the speed and wavelength of the wave. The light wave not
only changes directions at the boundary, it also speeds up or slows down and transforms into a wave with
a larger or a shorter wavelength. The only time that a wave can be transmitted across a boundary, change
its speed, and still not refract is when the light wave approaches the boundary in a direction that is
perpendicular to it. As long as the light wave changes speed and approaches the boundary at an angle,
refraction is observed.
But why does light refract? What is the cause of such behavior? And why is there this one exception to
the refraction of light? An analogy of marching soldiers is often used to address this question. In fact, it is
not uncommon that the analogy be illustrated in a Physics class with a student demonstration. A group of
students forms a straight line (shoulder to shoulder) and connect themselves to their nearest neighbor
using meter sticks. A strip of masking tape divides the room into two media. In one of the media (on one
side of the tape), students walk at a normal pace. In the other media (or on the other side of the tape),
students walk very slowly using baby steps. The group of students walk forward together in a straight line
towards the diagonal strip of masking tape. The students maintain the line as they approach the masking
tape. When an individual student reaches the tape, that student abruptly changes the pace of her/his walk.
The group of students continues walking until all students in the line have entered into the second
medium. The diagram below represents the line of students approaching the boundary (the masking tape)
between the two media. On the diagram, an arrow is used to show the general direction of travel for the
group of students in both media. Observe that the direction of the students changes at the "boundary."
The fundamental feature of the students' motion that leads to this change in direction is the change in
speed. Upon reaching the masking tape, each individual student abruptly changes speed. Because the
students approach the masking tape at an angle, each individual student reaches the tape at a different
time. The student who reaches the tape first, slows down while the rest of the line of students marches
ahead. This occurs for every student in the line of students. Once a student reaches the boundary, that
student slows down while his/her nearest neighbor marches ahead at the original pace. The result is that
the direction that the line of students is heading is altered at the boundary. The change in speed of the line
of students causes a change in direction.
Conditions of Refraction
Will this refractive behavior always occur? No! There are two conditions that are required in order to
observe the change in direction of the path of the students:

The students must change speed when crossing the boundary.

The students must approach the boundary at an angle; refraction will not occur when they
approach the boundary head-on (i.e., heading perpendicular to it).
These are both reasonable enough conditions if you consider the previous paragraph. If the students do
not change speed, then there is no cause factor. Recall that it was the change in speed of the students that
caused the change in direction. The second condition is also reasonable. If the students approach the
masking tape in a direction that is perpendicular to it, then each student will reach the tape at the exact
same time. Recall that the line of student changed their direction because they had reached the masking
tape at different times. The first student reached the tape, slowed down, and observed the rest of the
students marching ahead at the original speed. The change in direction of the line of students only occurs
at the boundary when the students change speed and approach at an angle.
The Marching Soldiers analogy provides an excellent analogy to understanding the cause of light
refraction. The line of students approaching the masking tape are analogous to a wavefront of light. The
masking tape is analogous to a boundary between two media. The change in speed that occurred for the
line of students would also occur for a wave of light. And like the marching students, a light wave will
not undergo refraction if it approaches the boundary in a direction that is perpendicular to it.
The same two conditions that are necessary for bending the path of the line of students are also necessary
for bending the direction of a light ray. Light refracts at a boundary because of a change in speed. There is
a distinct cause-effect relationship. The change in speed is the cause and the change in direction
(refraction) is the effect.
Flickr Physics Photo
Laser light shown passing through a rectangular block of lucite. It enters the block along the normal line
with an angle of incidence of 0°. There is no bending upon entering, nor upon leaving the block. When
light travels along the normal to the boundary, it changes its speed but not its direction.
Optical Density and Light Speed
Refraction is the bending of the path of a light wave as it passes from one material to another material.
The refraction occurs at the boundary and is caused by a change in the speed of the light wave upon
crossing the boundary. The tendency of a ray of light to bend one direction or another is dependent upon
whether the light wave speeds up or slows down upon crossing the boundary. Because a major focus of
our study will be upon the direction of bending, it will be important to understand the factors that effect
the speed at which a light wave is transported through a medium.
The mechanism by which a light wave is transported through a medium occurs in a manner that is similar
to the way that any other wave is transported - by particle-to-particle interaction. In Unit 10 of The
Physics Classroom Tutorial, the particle-to-particle interaction mechanism by which a mechanical wave
transports energy was discussed in detail. In Unit 12 of The Physics Classroom Tutorial, the mechanism
of energy transport by an electromagnetic wave was briefly discussed. Here we will look at this method in
more detail.
An electromagnetic wave (i.e., a light wave) is produced by a vibrating electric charge. As the wave
moves through the vacuum of empty space, it travels at a speed of c (3 x 108 m/s). This value is the speed
of light in a vacuum. When the wave impinges upon a particle of matter, the energy is absorbed and sets
electrons within the atoms into vibrational motion. If the frequency of the electromagnetic wave does not
match the resonant frequency of vibration of the electron, then the energy is reemitted in the form of an
electromagnetic wave. This new electromagnetic wave has the same frequency as the original wave and it
too will travel at a speed of c through the empty space between atoms. The newly emitted light wave
continues to move through the interatomic space until it impinges upon a neighboring particle. The energy
is absorbed by this new particle and sets the electrons of its atoms into vibration motion. And once more,
if there is no match between the frequency of the electromagnetic wave and the resonant frequency of the
electron, the energy is reemitted in the form of a new electromagnetic wave. The cycle of absorption and
reemission continues as the energy is transported from particle to particle through the bulk of a medium.
Every photon (bundle of electromagnetic energy) travels between the interatomic void at a speed of c; yet
time delay involved in the process of being absorbed and reemitted by the atoms of the material lowers
the net speed of transport from one end of the medium to the other. Subsequently, the net speed of an
electromagnetic wave in any medium is somewhat less than its speed in a vacuum - c (3 x 108 m/s).
Optical Density and the Index of Refraction
Like any wave, the speed of a light wave is dependent upon the properties of the medium. In the case of
an electromagnetic wave, the speed of the wave depends upon the optical density of that material. The
optical density of a medium is not the same as its physical density. The physical density of a material
refers to the mass/volume ratio. The optical density of a material relates to the sluggish tendency of the
atoms of a material to maintain the absorbed energy of an electromagnetic wave in the form of vibrating
electrons before reemitting it as a new electromagnetic disturbance. The more optically dense that a
material is, the slower that a wave will move through the material.
One indicator of the optical density of a material is the index of refraction value of the material. Index of
refraction values (represented by the symbol n) are numerical index values that are expressed relative to
the speed of light in a vacuum. The index of refraction value of a material is a number that indicates the
number of times slower that a light wave would be in that material than it is in a vacuum. A vacuum is
given an n value of 1.0000. The n values of other materials are found from the following equation:
The table below lists index of refraction values for a variety of medium. The materials listed at the top of
the table are those through which light travels fastest; these are the least optically dense materials. So as
the index of refraction value increases, the optical density increases, and the speed of light in that material
decreases.
The index of refraction values provide a measure of the relative speed of a light wave in a particular
medium. Knowledge of such relative speeds allows a student of physics to predict which way a light ray
would bend when passing from one medium to another. In the next part of Lesson 1, the rules for the
direction of bending will be discussed in detail.
The Direction of Bending
Refraction is the bending of the path of a light wave as it passes from one material into another material.
The refraction occurs at the boundary and is caused by a change in the speed of the light wave upon
crossing the boundary. The tendency of a ray of light to bend one direction or another is dependent upon
whether the light wave speeds up or slows down upon crossing the boundary. The speed of a light wave is
dependent upon the optical density of the material through which it moves. For this reason, the direction
that the path of a light wave bends depends on whether the light wave is traveling from a more dense
(slow) medium to a less dense (fast) medium or from a less dense medium to a more dense medium. In
this part of Lesson 1, we will investigate this topic of the direction of bending of a light wave.
Recall the Marching Soldiers analogy discussed earlier in this lesson. The analogy served as a model for
understanding the boundary behavior of light waves. As discussed, the analogy is often illustrated in a
Physics classroom by a student demonstration. In the demonstration, a line of students (representing a
light wave) marches towards a masking tape (representing the boundary) and slows down upon crossing
the boundary (representative of entering a new medium). The direction of the line of students changes
upon crossing the boundary. The diagram below depicts this change in direction for a line of students who
slow down upon crossing the boundary.
On the diagram, the direction of the students is represented by two arrows known as rays. The direction
of the students as they approach the boundary is represented by an incident ray (drawn in blue). And the
direction of the students after they cross the boundary is represented by a refracted ray (drawn in red).
Since the students change direction (i.e., refract), the incident ray and the refracted ray do not point in the
same direction. Also, note that a perpendicular line is drawn to the boundary at the point where the
incident ray strikes the boundary (i.e., masking tape). A line drawn perpendicular to the boundary at the
point of incidence is known as a normal line. Observe that the refracted ray lies closer to the normal line
than the incident ray does. In such an instance as this, we would say that the path of the students has bent
towards the normal. We can extend this analogy to light and conclude that:
Light Traveling from a Fast to a Slow Medium
If a ray of light passes across the boundary from a material in which it travels fast into a
material in which travels slower, then the light ray will bend towards the normal line.
The above principle applies to light passing from a material in which it travels fast across a boundary and
into a material in which it travels slowly. But what if light wave does the opposite? What if a light wave
passes from a material in which it travels slowly across a boundary and into a material in which it travels
fast? The answer to this question can be answered if we reconsider the Marching Soldier analogy. Now
suppose that the each individual student in the train of students speeds up once they cross the masking
tape. The first student to reach the boundary will speed up and pull ahead of the other students. When the
second student reaches the boundary, he/she will also speed up and pull ahead of the other students who
have not yet reached the boundary. This continues for each consecutive student, causing the line of
students to now be traveling in a direction further from the normal. This is depicted in the diagram below.
From this analogy and the diagram above, we see that the refracted ray (in red) is further away from the
normal then the incident ray (in blue). In such an instance as this, we would say that the path of the
students has bent away from the normal. We can once more extend this analogy to light and conclude
that:
Light Traveling from a Slow to a Fast Medium
If a ray of light passes across the boundary from a material in which it travels slowly into a
material in which travels faster, then the light ray will bend away from the normal line.
The Tractor Analogy
Now lets consider another analogy to assist in our understanding of
these two important principles. Suppose that a tractor is moving across
an asphalt surface towards a rectangular plot of grass (as shown in the
diagram at the right). Upon entering the grass, the tractors' wheels will
sink into the surface and slow down. Upon exiting the plot of grass on
the opposite side, the tractor wheels will speed up and achieve their
original speed. In effect, this analogy would be representative of a light
wave crossing two boundaries. At the first boundary (the asphalt to
grass boundary), the light wave (or the tractor) would be slowing down;
and at the second boundary (the grass to asphalt boundary), the light
wave (or the tractor) would be speeding up. We can apply our two
important principles listed above and predict the direction of bending
and the path of the tractor as it travels through the rectangular plot of
grass. As indicated on the diagram, upon entering the grass, the wheels
slow down and the path of the tractor bends towards the normal (perpendicular line drawn to the surface).
Upon exiting the plot of grass, the wheels speed up and the path of the tractor bends away from the
normal. The path of the tractor is closer to the normal in the slower medium and farther away from the
normal in the faster medium.
This analogy can be extended to the path of a light wave as it passes from air into and out of a rectangular
block of glass. Since air is less optically dense than glass, the light wave will slow down upon entering
the glass and speed up when exiting the glass. In other words, the light
wave will be undergoing the same change in speed as the tractor in the
above diagram. For this reason, the direction of bending for the light
wave upon entering and exiting the glass will be the same as in the
diagram above. The light ray refracts towards the normal upon
entering the glass (crossing from a fast to a slow medium) and refracts
away from the normal upon exiting the glass (crossing from a slow to
a fast medium). This is shown in the diagram at the right.
There is an important point to be noted in these diagrams of the
rectangular plot of grass and rectangular block of glass. Notice that the
direction of the original incident ray is the same as the direction of the
final refracted ray. Put another way, the direction at which the light is
traveling when entering the rectangular block of glass is the same as
the direction that the light travels after exiting the rectangular block of
glass. There is no ultimate change in the direction that the light is
traveling. This small detail will only be the case under two conditions:


the two sides of the glass through which the light enters and
exits are parallel to each other
the medium surrounding the glass on the side through which the light enters and exits are the
same
These two conditions are met in the case of a rectangular block of glass surrounded by air.
The diagrams below provide a contrast to the rectangular plot of grass and the rectangular block of glass.
Both diagrams involve the refraction of a tractor or a light wave as it passes into and out of a triangular
plot of grass and a triangular block of glass.
Copy this diagram onto a sheet of paper and apply your understanding of refraction principles to
predict the path of the tractor and the light wave as it travels through the triangle-shaped obstacle.
Draw the path on your separate sheet of paper and then click on the button below to check your answer
Least Time Principle
Another means of approaching the subject of the direction that light bends when crossing a boundary
between two media is through the Least Time Principle. This Least Time Principle is sometimes stated
as follows:
Least Time Principle
Of all the possible paths that light might take to get from one point to another, it always
takes the path that requires the least amount of time.
A useful analogy to understanding the principle involves a lifeguard who has become aware of a
drowning swimmer in the water. In order to save the drowning swimmer, the lifeguard must run through
the sand, cross the boundary between the sand and the water, and then swim through the water to the
drowning swimmer. Of course, the guard must reach the swimmer in as little time as possible. Since the
guard can run faster on sand than she can swim in water, it would make sense that the guard covers more
distance in the sand than she does in the water. In other words, she will not run directly at the drowning
swimmer. The optimal entry point into the water is the point that would allow the lifeguard to reach the
drowning swimmer in the least amount of time. Obviously, this point would be at a location closer to the
swimmer than to the guard. The diagram below depicts such an entry point.
Observe in the diagram, that minimizing the time to reach the swimmer means that the lifeguard will
approach the boundary at a steep angle to the normal and then will bend towards the normal upon
crossing the boundary. This analogy demonstrates that the Least Time Principle would predict the
following direction of bending:
A ray of light will bend towards the normal when crossing the boundary from a medium in which it
travels fast into a medium in which it travels slowly.
This is the very generalization that was made earlier on this page.
Some Useful Mnemonics
Using the above principles and logic to explain and predict the direction that light refracts when crossing
a boundary will be a major objective of this unit. Rather than merely restating the principle, you will be
asked to apply it to a variety of situations (such as those in the Check Your Understanding section below).
Part of accomplishing this task will involve remembering the principles. For this reason, the following
useful mnemonics are offered.
FST = Fast to Slow, Towards Normal
If a ray of light passes across the boundary from a material in which it travels fast into a
material in which travels slower, then the light ray will bend towards the normal line.
SFA = Slow to Fast, Away From Normal
If a ray of light passes across the boundary from a material in which it travels slow into a
material in which travels faster, then the light ray will bend away from the normal line.
A mnemonic is a tool used to help one remember and difficult-to-remember idea. Of course, there is
always the risk that the mnemonic will be forgotten. And since FST and SFA might not be the most easily
remembered mnemonics, perhaps the following oddity will help. You can remember FST (fast to slow;
towards) by simply thinking about those Freaky Science Teachers that you have had through the years.
And you can remember SFA by thinking about the disgusting habit of your friend Sara (or Susan or
Sammy or Samir or Somebody ...) - Sarah Farts Alot.
Flickr Physics Photo
Laser light shown passing through a rectangular block of lucite. It bends towards the normal as it enters
the lucite. As it exits the lucite into air, it bends away from the normal.
Check Your Understanding
Test your ability to apply these principles by answering the following questions.
1. When light passes from a more optically dense medium into a less optically dense medium, it will bend
_______ (towards, away from) the normal.
2. When light passes from a less optically dense medium into a more optically dense medium, it will bend
_______ (towards, away from) the normal.
3. When light passes from a medium with a high index of refraction value into a medium with a low index
of refraction value, it will bend _______ (towards, away from) the normal.
4. When light passes from a medium with a low index of refraction value into a medium with a high index
of refraction value, it will bend _______ (towards, away from) the normal.
5. In each diagram, draw the "missing" ray (either incident or refracted) in order to appropriately show
that the direction of bending is towards or away from the normal.
6. Arthur Podd's method of fishing involves spearing the fish while standing on the shore. The actual
location of a fish is shown in the diagram below. Because of the refraction of light, the observed location
of the fish is different than its actual location. Indicate on the diagram the approximate location where
Arthur observes the fish to be. Must Arthur aim above or below where the fish appears to be in order to
strike the fish?
7. For the following two cases, state whether the ray of light will bend towards or away from the normal
upon crossing the boundary.
The Angle of Refraction
Refraction is the bending of the path of a light wave as it passes across the boundary separating two
media. Refraction is caused by the change in speed experienced by a wave when it changes medium. In
Lesson 1, we learned that if a light wave passes from a medium in which it travels slow (relatively
speaking) into a medium in which it travels fast, then the light wave would refract away from the normal.
In such a case, the refracted ray will be farther from the normal line than the incident ray; this is the SFA
rule of refraction. On the other hand, if a light wave passes from a medium in which it travels fast
(relatively speaking) into a medium in which it travels slow, then the light wave will refract towards the
normal. In such a case, the refracted ray will be closer to the normal line than the incident ray is; this is
the FST rule of refraction. These two rules regarding the refraction of light only indicate the direction that
a light ray bends; they do not indicate how much bending occurs. Lesson 1 focused on the topics of
"What causes refraction?" and "Which direction does light refract?" Lesson 2 will focus on the question
of "By how much does light refract when it crosses a boundary?"
The question is: "By how much does light refract when it crosses a boundary?" Perhaps there are
numerous answers to such a question. (For example, "a lot," "a little,"
"like wow! Quite a bit dude," etc.) The concern of this lesson is to
express the amount of refraction of a light ray in terms of a
measurable quantity that has a mathematical value. The diagram to
the right shows a light ray undergoing refraction as it passes from air
into water. As mentioned in Lesson 1, the incident ray is a ray
(drawn perpendicular to the wavefronts) that shows the direction that
light travels as it approaches the boundary. (The meaning of an
incident ray was first introduced in the discussion of Reflection of
Light in Unit 13 of The Physics Classroom Tutorial.) Similarly, the
refracted ray is a ray (drawn perpendicular to the wavefronts) that
shows the direction that light travels after it has crossed over the
boundary. In the diagram, a normal line is drawn to the surface at the point of incidence. This line is
always drawn perpendicular to the boundary. The angle that the incident ray makes with the normal line is
referred to as the angle of incidence. Similarly, the angle that the refracted ray makes with the normal
line is referred to as the angle of refraction. The angle of incidence and angle of refraction are denoted
by the following symbols:
= angle of incidence
= angle of refraction
The amount of bending that a light ray experiences can be expressed in terms of the angle of refraction
(more accurately, by the difference between the angle of refraction and the angle of incidence). A ray of
light may approach the boundary at an angle of incidence of 45-degrees and bend towards the normal. If
the medium into which it enters causes a small amount of refraction, then the angle of refraction might be
a value of about 42-degrees. On the other hand if the medium into which the light enters causes a large
amount of refraction, the angle of refraction might be 22-degrees. (These values are merely arbitrarily
chosen values to illustrate a point.) The diagram below depicts a ray of light approaching three different
boundaries at an angle of incidence of 45-degrees. The refractive medium is different in each case,
causing different amounts of refraction. The angles of refraction are shown on the diagram.
Of the three boundaries in the diagram, the light ray refracts the most at the air-diamond boundary. This is
evident by the fact that the difference between the angle of incidence and the angle of refraction is
greatest for the air-diamond boundary. But how can this be explained? The cause of refraction is a change
in light speed; and wherever the light speed changes most, the refraction is greatest. We have already
learned that the speed is related to the optical density of a material that is related to the index of refraction
of a material. Of the four materials present in the above diagram, air is the least dense material (lowest
index of refraction value) and diamond is the most dense material (largest index of refraction value).
Thus, it would be reasonable that the most refraction occurs for the transmission of light across an air-
diamond boundary.
In this example, the angle of refraction is the measurable quantity that indicates the amount of refraction
taking place at any boundary. A comparison of the angle of refraction to the angle of incidence provides a
good measure of the refractive ability of any given boundary. For any given angle of incidence, the angle
of refraction is dependent upon the speeds of light in each of the two materials. The speed is in turn
dependent upon the optical density and the index of refraction values of the two materials. There is a
mathematical equation relating the angles that the light rays make with the normal to the indices (plural
for index) of refraction of the two materials on each side of the boundary. This mathematical equation is
known as Snell's Law and is the topic of the next section of Lesson 2.
Total Internal Reflection
Boundary Behavior Revisited | Total Internal Reflection | The Critical Angle
Boundary Behavior Revisited
Earlier in this unit, the boundary behavior of light waves was discussed. It was mentioned that a light
wave doesn't just stop when it reaches the end of the medium. Rather, the light wave undergoes certain
behaviors when it encounters the end of the medium - such behaviors include reflection,
transmission/refraction, and diffraction. In Unit 13 of The Physics Classroom Tutorial, the primary focus
was the reflective behavior of light waves at the boundary. In this unit, our primary interest has been the
refractive behavior of light waves at the boundary. In Lesson 3, we will investigate the connection
between light reflection and light refraction.
A light wave, like any wave, is an energy-transport phenomenon. A light wave transports energy from
one location to another. When a light wave strikes a boundary between two distinct media, a portion of
the energy will be transmitted into the new medium and a portion of the energy will be reflected off the
boundary and stay within the original medium. The actual percentage of energy that is transmitted and
reflected is dependent upon a number of variables; these will be discussed as we proceed through Lesson
3. For now, our concern is to review and internalize the basic concepts and terminology associated with
boundary behavior. Reflection of a light wave involves the bouncing of a light wave off the boundary,
while refraction of a light wave involves the bending of the path of a light wave upon crossing a boundary
and entering a new medium. Both reflection and refraction involve a change in direction of a wave, but
only refraction involves a change in medium.
The diagram at the right shows several wavefronts approaching a
boundary between two media. These wavefronts are referred to as the
incident waves and the ray that points in the direction that they are
traveling is referred to as the incident ray. The incident ray is drawn in
blue on the diagram at the right. Notice on the diagram that the incident
ray leads into two other rays at the point of incidence with the
boundary. The reflected waves are the waves that bounce off the
boundary and head back upwards and the reflected ray is the ray that
points in the direction that the reflected waves are traveling. The
reflected ray is drawn in green on the diagram at the right. The refracted
waves are the waves that are transmitted across the boundary and
continue moving downwards, only at a different angle than before. The refracted ray is the ray that points
in the direction that the refracted waves are traveling. The refracted ray is drawn in red on the diagram at
the right. At the point of incidence (the point where the incident ray strikes the boundary), a normal line
is drawn. The normal line is always drawn perpendicular to the surface at the point of incidence. The
normal line creates a variety of angles with the light rays; these angles are important and are given special
names. The angle between the incident ray and the normal is the angle of incidence. The angle between
the reflected ray and the normal is the angle of reflection. And the angle between the refracted ray and
the normal is the angle of refraction.
The fundamental law that governs the reflection of light is called the law of reflection. Whether the light
is reflecting off a rough surface or a smooth surface, a curved surface or a planar surface, the light ray
follows the law of reflection. The law of reflection states that
When a light ray reflects off a surface, the angle of incidence is equal to the angle of reflection.
The fundamental law that governs the refraction of light is Snell's Law. Snell's Law states that
When a light ray is transmitted into a new medium, the relationship between the angle of incidence and
the angle of refraction is given by the following equation
where the ni and nr values represent the indices of refraction of the incident and the refractive medium
respectively.
As we proceed through this Lesson, we will see that there is a connection between the reflection and the
refraction of light. Each of these two behaviors usually occurs together. But as we will see, there are two
conditions, which when both met, will cause the light waves to undergo reflection without any
accompanying refraction.
Flickr Physics Photo
Laser light shown passing into a hemi-cylindrical dish filled with water. The light enters the water (at the
curved side of the dish) along the normal line; no bending occurs upon entry. The light continues
through the water along a straight line until it reaches the boundary with air (at the flat side of the dish).
The angle of incidence in the water is just short of 48°. Since the angle of incidence is just less than the
air-water critical angle of about 48.6°, there is still a little refraction in addition to the reflection. The
refracted ray can be seen at about 80°.
Total Internal Reflection
A common Physics lab is to sight through the long side of an isosceles triangle at a pin or other object
held behind the opposite face. When done so, an unusual observation - a discrepant event - is observed.
The diagram on the left below depicts the physical situation. A ray of light entered the face of the
triangular block at a right angle to the boundary. This ray of light passes across the boundary without
refraction since it was incident along the normal (recall the If I Were An Archer Fish page). The ray of
light then travels in a straight line through the glass until it reaches the second boundary. Now instead of
transmitting across this boundary, all of the light seems to reflect off the boundary and transmit out the
opposite face of the isosceles triangle. This discrepant event bothers many as they spend several minutes
looking for the light to refract through the second boundary. Then finally, to their amazement, they
looked through the third face of the block and clearly see the ray. What happened? Why did light not
refract through the second face?
The phenomenon observed in this part of the lab is known as total internal reflection. Total internal
reflection, or TIR as it is intimately called, is the reflection of the total amount of incident light at the
boundary between two media. TIR is the topic of focus in Lesson 3.
To understand total internal reflection, we will begin with a thought experiment. Suppose that a laser
beam is submerged in a tank of water (don't do this at home) and pointed upwards towards water-air
boundary. Then suppose that the angle at which the beam is directed upwards is slowly altered, beginning
with small angles of incidence and proceeding towards larger and larger angles of incidence. What would
be observed in such an experiment? If we understand the principles of boundary behavior, we would
expect that we would observe both reflection and refraction. And indeed, that is what is observed
(mostly). But that's not the only observation that we could make. We would also observe that the intensity
of the reflected and refracted rays do not remain constant. At angle of incidence close to 0 degrees, most
of the light energy is transmitted across the boundary and very little of it is reflected. As the angle is
increased to greater and greater angles, we would begin to observe less refraction and more reflection.
That is, as the angle of incidence is increased, the brightness of the refracted ray decreases and the
brightness of the reflected ray increases. Finally, we would observe that the angles of the reflection and
refraction are not equal. Since the light waves would refract away from the normal (a case of the SFA
principle of refraction), the angle of refraction would be greater than the angle of incidence. And if this
were the case, the angle of refraction would also be greater than the angle of reflection (since the angles
of reflection and incidence are the same). As the angle of incidence is increased, the angle of refraction
would eventually reach a 90-degree angle. These principles are depicted in the diagram below.
The maximum possible angle of refraction is 90-degrees. If you think about it (a practice that always
helps), you recognize that if the angle of refraction were greater than 90 degrees, then the refracted ray
would lie on the incident side of the medium - that's just not possible. So in the case of the laser beam in
the water, there is some specific value for the angle of incidence (we'll call it the critical angle) that yields
an angle of refraction of 90-degrees. This particular value for the angle of incidence could be calculated
using Snell's Law (ni = 1.33, nr = 1.000, = 90 degrees, = ???) and would be found to be 48.6 degrees.
Any angle of incidence that is greater than 48.6 degrees would not result in refraction. Instead, when the
angles of incidence is greater than 48.6 degrees (the critical angle), all of the energy (the total energy)
carried by the incident wave to the boundary stays within the water (internal to the original medium) and
undergoes reflection off the boundary. When this happens, total internal reflection occurs
Two Requirements for Total Internal Reflection
Total internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off
the boundary. TIR only takes place when both of the following two conditions are met:


the light is in the more dense medium and approaching the less dense medium.
the angle of incidence is greater than the so-called critical angle.
Total internal reflection will not take place unless the incident light is traveling within the more optically
dense medium towards the less optically dense medium. TIR will happen for light traveling from water
towards air, but it will not happen for light traveling from air towards water. TIR would happen for light
traveling from water towards air, but it will not happen for light traveling from water (n=1.333) towards
crown glass (n=1.52). TIR occurs because the angle of refraction reaches a 90-degree angle before the
angle of incidence reaches a 90-degree angle. The only way for the angle of refraction to be greater than
the angle of incidence is for light to bend away from the normal. Since light only bends away from the
normal when passing from a more dense medium into a less dense medium, then this would be a
necessary condition for total internal reflection.
Total internal reflection only occurs with large angles of incidence. Question: How large is large?
Answer: larger than the critical angle. As mentioned above, the critical angle for the water-air boundary is
48.6 degrees. So for angles of incidence greater than 48.6-degrees, TIR occurs. But 48.6 degrees is the
critical angle only for the water-air boundary. The actual value of the critical angle is dependent upon the
two materials on either side of the boundary. For the crown glass-air boundary, the critical angle is 41.1
degrees. For the diamond-air boundary, the critical angle is 24.4 degrees. For the diamond-water
boundary, the critical angle is 33.4 degrees. The critical angle is different for different media. In the next
part of Lesson 3, we will investigate how to determine the critical angle for any two materials. For now,
let's internalize the idea that TIR can only occur if the angle of incidence is greater than the critical angle
for the particular combination of materials.
Light Piping and Optical Fibers
Total internal reflection is often demonstrated in a Physics class through a variety of
demonstrations. In one such demonstration, a beam of laser light is directed into a
coiled plastic thing-a-ma jig. The plastic served as a light pipe, directing the light
through the coils until it finally exits out the opposite end. Once the light entered the
plastic, it was in the more dense medium. Every time the light approached the
plastic-air boundary, it is approaching at angles greater than the critical angle. The
two conditions necessary for TIR are met, and all of the incident light at the plasticair boundary stays internal to the plastic and undergoes reflection. And with the
room lights off, every student becomes quickly aware of the ancient truth that
Physics is better than drugs.
This demonstration helps to illustrate the principle by which optical fibers work. The use of a long strand
of plastic (or other material such as glass) to pipe light from one end of the medium to the other is the
basis for modern day use of optical fibers. Optical fibers are used in
communication systems and micro-surgeries. Since total internal
reflection takes place within the fibers, no incident energy is ever lost
due to the transmission of light across the boundary. The intensity of
the signal remains constant.
Another common Physics demonstration involves the use of a large jug
filled with water and a laser beam. The jug has a pea-sized hole drilled
in its side such that when the cork is removed from the top of the jug,
water begins to stream out the jug's side. The beam of laser light is
then directed into the jug from the opposite side of the hole, through
the water and into the falling stream. The laser light exits the jug
through the hole but is still in the water. As the stream of water begins to fall as a projectile along a
parabolic path to the ground, the laser light becomes trapped within the water due to total internal
reflection. Being in the more dense medium (water) and heading towards a boundary with a less dense
medium (air), and being at angles of incidence greater than the critical angle, the light never leaves the
stream of water. In fact, the stream of water acts as a light pipe to pipe the laser beam along its trajectory.
Once more, students viewing the demonstration are convinced of the fact that Physics is better than drugs.
Flickr Physics Photo
Laser light shown passing into a hemi-cylindrical dish filled with water. The light enters the water (at the
curved side of the dish) along the normal line; no bending occurs upon entry. The light continues
through the water along a straight line until it reaches the boundary with air (at the flat side of the dish).
The angle of incidence in the water is 50°. Since the angle of incidence is greater the air-water critical
angle of about 48°, it undergoes total internal reflection (TIR). Rather than refract out of the dish of
water at the flat side, the laser light is seen reflecting and exiting along the curved side of the dish.
Check Your Understanding
1. For each combination of media, which light ray (A or B) will undergo total internal reflection if the
incident angle is gradually increased?
The Critical Angle
In the previous part of Lesson 3, the phenomenon of total internal reflection was introduced. Total
internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off the
boundary. TIR only takes place when both of the following two conditions are met:


a light ray is in the more dense medium and approaching the less dense medium.
the angle of incidence for the light ray is greater than the so-called critical angle.
In our introduction to TIR, we used the example of light traveling through water towards the boundary
with a less dense material such as air. When the angle of incidence in water reaches a certain critical
value, the refracted ray lies along the boundary, having an angle of refraction of 90-degrees. This angle of
incidence is known as the critical angle; it is the largest angle of incidence for which refraction can still
occur. For any angle of incidence greater than the critical angle, light will undergo total internal
reflection.
So the critical angle is defined as the angle of incidence that provides an angle of refraction of 90-degrees.
Make particular note that the critical angle is an angle of incidence value. For the water-air boundary, the
critical angle is 48.6-degrees. For the crown glass-water boundary, the critical angle is 61.0-degrees. The
actual value of the critical angle is dependent upon the combination of materials present on each side of
the boundary.
Let's consider two different media - creatively named medium i (incident medium) and medium r
(refractive medium). The critical angle is the that gives a value of 90-degrees. If this information is
substituted into Snell's Law equation, a generic equation for predicting the critical angle can be derived.
The derivation is shown below.
ni *• si e
ni • sine(
) = nr • si e
)
) = nr • sine (90 degrees)
ni • sine(
sine(
) = nr
) = nr/ni
= sine-1 (nr/ni) = invsine (nr/ni)
The critical angle can be calculated by taking the inverse-sine of the ratio of the indices of refraction. The
ratio of nr/ni is a value less than 1.0. In fact, for the equation to even give a correct answer, the ratio of
nr/ni must be less than 1.0. Since TIR only occurs if the refractive medium is less dense than the incident
medium, the value of ni must be greater than the value of nr. If at any time the values for the numerator
and denominator become accidentally switched, the critical angle value cannot be calculated.
Mathematically, this would involve finding the inverse-sine of a number greater than 1.00 - which is not
possible. Physically, this would involve finding the critical angle for a situation in which the light is
traveling from the less dense medium into the more dense medium - which again, is not possible.
This equation for the critical angle can be used to predict the critical angle for any boundary, provided
that the indices of refraction of the two materials on each side of the boundary are known. Examples of its
use are shown below:
Example A
Calculate the critical angle for the crown glass-air boundary. Refer to the table of indices of refraction
if necessary.
The solution to the problem involves the use of the above equation for the critical angle.
= sin-1 (nr/ni) = invsine (nr/ni)
= sin-1 (1.000/1.52) = 41.1 degrees
Example B
Calculate the critical angle for the diamond-air boundary. Refer to the table of indices of refraction if
necessary.
The solution to the problem involves the use of the above equation for the critical angle.
= sin-1 (nr/ni) = invsine (nr/ni)
= sin-1 (1.000/2.42) = 24.4 degrees
TIR and the Sparkle of Diamonds
Relatively speaking, the critical angle for the diamond-air boundary is an extremely small number. Of all
the possible combinations of materials that could interface to form a boundary, the combination of
diamond and air provides one of the largest differences in the index of refraction values. This means that
there will be a very small nr/ni ratio and subsequently a small critical angle. This peculiarity about the
diamond-air boundary plays an important role in the brilliance of a diamond gemstone. Having a small
critical angle, light has the tendency to become "trapped" inside of a diamond once it enters. A light ray
will typically undergo TIR several times before finally refracting out of the diamond. Because the
diamond-air boundary has such a small critical angle (due to diamond's large index of refraction), most
rays approach the diamond at angles of incidence greater than the critical angle. This gives diamond a
tendency to sparkle. The effect can be enhanced by the cutting of a diamond gemstone with a strategically
planned shape. The diagram below depicts the total internal reflection within a diamond gemstone with a
strategic and a non-strategic cut.
Interesting Refraction Phenomena
Dispersion of Light by a Prism | Rainbow Formation | Mirages
Dispersion of Light by Prisms
In the Light and Color unit of The Physics Classroom Tutorial, the visible light spectrum was introduced
and discussed. Visible light, also known as white light, consists of a collection of component colors.
These colors are often observed as light passes through a triangular prism. Upon passage through the
prism, the white light is separated into its component colors - red, orange, yellow, green, blue and violet.
The separation of visible light into its different colors is known as dispersion. It was mentioned in the
Light and Color unit that each color is characteristic of a distinct wave frequency; and different
frequencies of light waves will bend varying amounts upon passage through a prism. In this unit, we will
investigate the dispersion of light in more detail, pondering the reasons why different frequencies of light
bend or refract different amounts when passing through the prism.
Earlier in this unit, the concept of optical density was introduced. Different materials are distinguished
from each other by their different optical densities. The optical density is simply a measure of the
tendency of a material to slow down light as it travels through it. As mentioned earlier, a light wave
traveling through a transparent material interacts with the atoms of that material. When a light wave
impinges upon an atom of the material, it is absorbed by that atom. The absorbed energy causes the
electrons in the atom to vibrate. If the frequency of the light wave does not match the resonance
frequency of the vibrating electrons, then the light will be reemitted by the atom at the same frequency at
which it impinged upon it. The light wave then travels through the interatomic vacuum towards the next
atom of the material. Once it impinges upon the next atom, the process of absorption and re-emission is
repeated.
The optical density of a material is the result of the tendency of the atoms of a material to maintain the
absorbed energy of the light wave in the form of vibrating electrons before reemitting it as a new
electromagnetic disturbance. Thus, while a light wave travels through a vacuum at a speed of c (3.00 x
108 m/s), it travels through a transparent material at speeds less than c. The index of refraction value (n)
provides a quantitative expression of the optical density of a given medium. Materials with higher index
of refraction values have a tendency to hold onto the absorbed light energy for greater lengths of time
before reemitting it to the interatomic void. The more closely that the frequency of the light wave matches
the resonant frequency of the electrons of the atoms of a material, the greater the optical density and the
greater the index of refraction. A light wave would be slowed down to a greater extent when passing
through such a material
What was not mentioned earlier in this unit is that the index of refraction values are dependent upon the
frequency of light. For visible light, the n value does not show a large variation with frequency, but
nonetheless it shows a variation. For instance for some types of glass, the n value for frequencies of violet
light is 1.53; and the n value for frequencies of red light is 1.51. The absorption and re-emission process
causes the higher frequency (lower wavelength) violet light to travel slower through crown glass than the
lower frequency (higher wavelength) red light. It is this difference in n value for the varying frequencies
(and wavelengths) that causes the dispersion of light by a triangular prism. Violet
light, being slowed down to a greater extent by the absorption and re-emission
process, refracts more than red light. Upon entry of white light at the first boundary
of a triangular prism, there will be a slight separation of the white light into the
component colors of the spectrum. Upon exiting the triangular prism at the second
boundary, the separation becomes even greater and ROYGBIV is observed in its
splendor.
The Angle of Deviation
The amount of overall refraction caused by the passage of a light ray through a prism is often expressed in
terms of the angle of deviation ( ). The angle of deviation is the angle made between the incident ray of
light entering the first face of the prism and the refracted ray that emerges from the second face of the
prism. Because of the different indices of refraction for the different wavelengths of visible light, the
angle of deviation varies with wavelength. Colors of the visible light spectrum that have shorter
wavelengths (BIV) will deviated more from their original path than the colors with longer wavelengths
(ROY). The emergence of different colors of light from a triangular prism at different angles leads an
observer to see the component colors of visible light separated from each other.
Of course the discussion of the dispersion of light by triangular prisms begs the following question: Why
doesn't a square or rectangular prism cause the dispersion of a narrow beam of white light? The short
answer is that it does. The long answer is provided in the following discussion and illustrated by the
diagram below.
Suppose that a flashlight could be covered with black paper with a slit across it so as to create a beam of
white light. And suppose that the beam of white light with its component colors unseparated were
directed at an angle towards the surface of a rectangular glass prism. As would be expected, the light
would refract towards the normal upon entering the glass and away from the normal upon exiting the
glass. But since the violet light has a shorter wavelength, it would refract more than the longer
wavelength red light. The refraction of light at the entry location into the rectangular glass prism would
cause a little separation of the white light. However, upon exiting the glass prism, the refraction takes
place in the opposite direction. The light refracts away from the normal, with the violet light bending a bit
more than the red light. Unlike the passage through the triangular prism with non-parallel sides, there is
no overall angle of deviation for the various colors of white light. Both the red and the violet components
of light are traveling in the same direction as they were traveling before entry into the prism. There is
however a thin red fringe present on one end of the beam and thin violet fringe present on the opposite
side of the beam. This fringe is evidence of dispersion. Because there is a different angle of deviation of
the various components of white light after transmission across the first boundary, the violet is separated
ever so slightly from the red. Upon transmission across the second boundary, the direction of refraction is
reversed; yet because the violet light has traveled further downward when passing through the rectangle it
is the primary color present in the lower edge of the beam. The same can be said for red light on the upper
edge of the beam.
Dispersion of light provides evidence for the existence of a spectrum of wavelengths present in visible
light. It is also the basis for understanding the formation of rainbows. Rainbow formation is the next topic
of discussion in Lesson 4.
Rainbow Formation
One of nature's most splendid masterpieces is the rainbow. A rainbow is an excellent demonstration of the
dispersion of light and one more piece of evidence that visible light is composed of a spectrum of
wavelengths, each associated with a distinct color. To view a rainbow, your back must be to the sun as
you look at an approximately 40 degree angle above the ground into a region of the atmosphere with
suspended droplets of water or even a light mist. Each individual droplet of water acts as a tiny prism that
both disperses the light and reflects it back to your eye. As you sight into the sky, wavelengths of light
associated with a specific color arrive at your eye from the collection of droplets. The net effect of the
vast array of droplets is that a circular arc of ROYGBIV is seen across the sky. But just exactly how do
the droplets of water disperse and reflect the light? And why does the pattern always appear as
ROYGBIV from top to bottom? These are the questions that we will seek to understand on this page of
The Physics Classroom Tutorial. To understand these questions, we will need to draw upon our
understanding of refraction, internal reflection and dispersion.
The Path of Light Through a Droplet
A collection of suspended water droplets in the atmosphere serves as a refractor of light. The water
represents a medium with a different optical density than the surrounding air. Light waves refract when
they cross over the boundary from one medium to another. The decrease in speed upon entry of light into
a water droplet causes a bending of the path of light towards the normal. And upon exiting the droplet,
light speeds up and bends away from the normal. The droplet causes a deviation in the path of light as it
enters and exits the drop.
There are countless
paths by which light rays from the sun can
pass through a drop.
Each path is characterized by this bending
towards and away
from the normal. One path of great
significance in the
discussion of rainbows is the path in
which light refracts
into the droplet, internally reflects, and
then refracts out of
the droplet. The diagram at the right
depicts such a path.
A light ray from the sun enters the droplet
with a slight
downward trajectory. Upon refracting
twice and reflecting
once, the light ray is dispersed and bent
downward towards
an observer on earth's surface. Other entry
locations into the
droplet may result in similar paths or even
in light continuing
through the droplet and out the opposite
side without
significant internal reflection. But for the
entry location shown in the diagram at the right, there is an optimal concentration of light exiting the
airborne droplet at an angle towards the ground. As in the case of the refraction of light through prisms
with nonparallel sides, the refraction of light at two boundaries of the droplet results in the dispersion of
light into a spectrum of colors. The shorter wavelength blue and violet light refract a slightly greater
amount than the longer wavelength red light. Since the boundaries are not parallel to each other, the
double refraction results in a distinct separation of the sunlight into its component colors.
The angle of deviation between the incoming light rays from
the sun and the refracted rays directed to the observer's eyes is
approximately 42 degrees for the red light. Because of the
tendency of shorter wavelength blue light to refract more than
red light, its angle of deviation from the original sun rays is
approximately 40 degrees. As shown in the diagram, the red
light refracts out of the droplet at a steeper angle toward an
observer on the ground. There are a multitude of paths by
which the original ray can pass through a droplet and
subsequently angle towards the ground. Some of the paths are
dependent upon which part of the droplet the incident rays
contact. Other paths are dependent upon the location of the sun in the sky and the subsequent trajectory of
the incoming rays towards the droplet. Yet the greatest concentration of outgoing rays is found at these
40-42 degree angles of deviation. At these angles, the dispersed light is bright enough to result in a
rainbow display in the sky. Now that we understand the path of light through an individual droplet, we
can approach the topic of how the rainbow forms.
The Formation of the Rainbow
A rainbow is most often viewed as a circular arc in the sky. An observer on the ground observes a halfcircle of color with red being the color perceived on the outside or top of the bow. Those who are
fortunate enough to have seen a rainbow from an airplane in the sky may know that a rainbow can
actually be a complete circle. Observers on the ground only view the top half of the circle since the
bottom half of the circular arc is prevented by the presence of the ground (and the rather obvious fact that
suspended water droplets aren't present below ground). Yet observers in an airborne plane can often look
both upward and downward to view the complete circular bow.
The circle (or half-circle) results because there are a collection of suspended droplets in the atmosphere
that are capable concentrating the dispersed light at angles of deviation of 40-42 degrees relative to the
original path of light from the sun. These droplets actually form a circular arc, with each droplet within
the arc dispersing light and reflecting it back towards the observer. Every droplet within the arc is
refracting and dispersing the entire visible light spectrum (ROYGBIV). As described above, the red light
is refracted out of a droplet at steeper angles towards the ground than the blue light. Thus, when an
observer sights at a steeper angle with respect to the ground, droplets of water within this line of sight are
refracting the red light to the observer's eye. The blue light from these same droplets is directed at a less
steep angle and is directed along a trajectory that passes over the observer's head. Thus, it is the red light
that is seen when looking at the steeper angles relative to the ground. Similarly, when sighting at less
steep angles, droplets of water within this line of sight are directing blue light to the observer's eye while
the red light is directed downwards at a more steep angle towards the observer's feet. This discussion
explains why it is the red light that is observed at the top and on the outer perimeter of a rainbow and the
blue light that is observed on the bottom and the inner perimeter of the rainbow.
Rainbows are not limited to the dispersion of light by raindrops. The splashing of water at the base of a
waterfall caused a mist of water in the air that often results in the formation of rainbows. A backyard
water sprinkler is another common source of a rainbow. Bright sunlight, suspended droplets of water and
the proper angle of sighting are the three necessary components for viewing one of nature's most splendid
masterpieces.
Mirages
Most of our discussion of refraction in this unit has pertained to the refraction of light at a distinct
boundary. As light is transmitted across the boundary from one material to another, there is a change in
speed, which causes a change in direction of the light wave. The boundaries that we have been focusing
on have been distinct interfaces between two recognizably different materials. The boundary between the
glass of a fish tank and the surrounding air or the boundary between the water in a
pool and the surrounding air are examples of distinct interfaces between two
recognizably different materials.
It has been mentioned in our discussion that the refraction or bending of light
occurs at the boundary between two materials; and once a light wave has crossed
the boundary it travels in a straight line. The discussion has presumed that the
medium is a uniform medium. A uniform medium is a medium whose optical
density is everywhere the same within the medium. A uniform medium is the same
everywhere from its top boundary to its bottom boundary and from its left
boundary to its right boundary. But not every medium is a uniform medium, and the fact that air can
sometimes form a non-uniform medium leads to an interesting refraction phenomenon - the formation of
mirages.
A mirage is an optical phenomenon that creates the illusion of water and results from the refraction of
light through a non-uniform medium. Mirages are most commonly observed on sunny days when driving
down a roadway. As you drive down the roadway, there appears to be a puddle of water on the road
several yards (maybe one-hundred yards) in front of the car. Of course, when you arrive at the perceived
location of the puddle, you recognize that the puddle is not there. Instead, the puddle of water appears to
be another one-hundred yards in front of you. You could carefully match the perceived location of the
water to a roadside object; but when you arrive at that object, the puddle of water is still not on the
roadway. The appearance of the water is simply an illusion.
Mirages occur on sunny days. The role of the sun is to heat the roadway to high temperatures. This heated
roadway in turn heats the surrounding air, keeping the air just above the roadway at higher temperatures
than that day's average air temperature. Hot air tends to be less optically dense than cooler air. As such, a
non-uniform medium has been created by the heating of the roadway and the air just above it. While light
will travel in a straight line through a uniform medium, it will refract when traveling through a nonuniform medium. If a driver looks down at the roadway at a very low angle (that is, at a position nearly
one hundred yards away), light from objects above the roadway will follow a curved path to the driver's
eye as shown in the diagram below.
Light that is traveling downward into this less optically dense air begins to speed up. Though there isn't a
distinct boundary between two media, there is a change in speed of a light wave. As expected, a change in
speed is accompanied by a change in direction. If there were a distinct boundary between two media, then
there would be a bending of this light ray away from the normal. For this light ray to bend away from the
normal (towards the boundary), the ray would begin to bend more parallel to the roadway and then bend
upwards towards the cooler air. As such, a person in a car sighting downward at the roadway will see an
object located above the roadway.
Of course, this is not a usual event. When was the last time that you looked downward at a surface and
saw an object above the surface? While not a usual event, it does happen. For instance, suppose you place
a mirror on the floor and look downward at the floor; you will see objects located above the floor due to
the reflection of light by the mirror. Even a glass window placed on the floor will reflect light from
objects above the floor. If you look downward at the glass window at a low enough angle, then you will
see objects located above the floor. Or suppose that you are standing on the shore of a calm pond and look
downward at the water; you might see objects above the pond due to the reflection of light by the water.
So when you experience this sunny day phenomenon, your mind must quickly make sense of how you
can look downward at the roadway and see an object located above the road. In the process of making
sense of this event, your mind draws upon past experiences. Searching the database of stored experiences,
your mind is interested in an explanation of why the eye can sight downward at a surface and see an
object that is located above the surface. In the process of searching, it comes up with three possible
explanations based upon past experiences. Your mind subtly ponders these three options.



There is a mirror on the road. Someone must have for some reason placed a mirror on the road.
The mirror is reflecting light and that is why I see an image of the oncoming truck when I look
downward at the road.
There is a glass window on the road. My gosh, do you believe it! Someone has left a glass
window on the road. The glass window is reflecting light and that is why I see an image of the
oncoming truck when I look downward at the road.
There is water on the road. It must have rained last night and there is a puddle of water left on
the road. The water is reflecting light and that is why I see an image of the oncoming truck when
I look downward at the road.
Of the three possible explanations of the image of the truck, only one makes a lot of sense to the mind there is water on the road. After all, while both glass windows and mirrors can reflect light, nowhere in
your mind's database of past experiences is there an account of a mirror or glass window being seen on a
roadway. Yet there are plenty of times that a water puddle has been observed to be present on a roadway.
Smart person that you are, you then conclude that there is a puddle of water on the road that is causing
you to see objects located above the road when you sight downward at the road. The illusion is complete.
Image Formation by Lenses
The Anatomy of a Lens | Refraction by Lenses | Image Formation Revisited
Converging Lenses: Ray Diagrams | Object-Image Relations
Diverging Lenses: Ray Diagrams | Object-Image Relations
The Mathematics of Lenses
The Anatomy of a Lens
If a piece of glass or other transparent material takes on the appropriate shape, it is possible that parallel
incident rays would either converge to a point or appear to be diverging from a point. A piece of glass that
has such a shape is referred to as a lens.
A lens is merely a carefully ground or molded piece of transparent material that refracts light rays in such
as way as to form an image. Lenses can be thought of as a series of tiny refracting prisms, each of which
refracts light to produce their own image. When these prisms act together, they produce a bright image
focused at a point.
There are a variety of types of lenses. Lenses differ from one another in terms of their shape and the
materials from which they are made. Our focus will be upon lenses that are symmetrical across their
horizontal axis - known as the principal axis. In this unit, we will categorize lenses as converging lenses
and diverging lenses. A converging lens is a lens that converges rays of light that are traveling parallel to
its principal axis. Converging lenses can be identified by their shape; they are relatively thick across their
middle and thin at their upper and lower edges. A diverging lens is a lens that diverges rays of light that
are traveling parallel to its principal axis. Diverging lenses can also be identified by their shape; they are
relatively thin across their middle and thick at their upper and lower edges.
A double convex lens is symmetrical across both its horizontal and vertical axis. Each of the lens' two
faces can be thought of as originally being part of a sphere. The fact that a double convex lens is thicker
across its middle is an indicator that it will converge rays of light that travel parallel to its principal axis.
A double convex lens is a converging lens. A double concave lens is also symmetrical across both its
horizontal and vertical axis. The two faces of a double concave lens can be thought of as originally being
part of a sphere. The fact that a double concave lens is thinner across its middle is an indicator that it will
diverge rays of light that travel parallel to its principal axis. A double concave lens is a diverging lens.
These two types of lenses - a double convex and a double concave lens will be the only types of lenses
that will be discussed in this unit of The Physics Classroom Tutorial.
As we begin to discuss the refraction of light rays and the formation of images by these two types of
lenses, we will need to use a variety of terms. Many of these terms should be familiar to you because they
have already been discussed during Unit 13. If you are uncertain of the meaning of the terms, spend some
time reviewing them so that their meaning is firmly internalized in your mind. They will be essential as
we proceed through Lesson 5. These terms describe the various parts of a lens and include such words as
Principal axis
Vertical Plane
Focal Point
Focal Length
If a symmetrical lens were thought of as being a slice of a sphere, then there would be a line passing
through the center of the sphere and attaching to the mirror in the exact center of the lens. This imaginary
line is known as the principal axis. A lens also has an imaginary vertical axis that bisects the
symmetrical lens into halves. As mentioned above, light rays incident towards either face of the lens and
traveling parallel to the principal axis will either converge or diverge. If the light rays converge (as in a
converging lens), then they will converge to a point. This point is known as the focal point of the
converging lens. If the light rays diverge (as in a diverging lens), then the diverging rays can be traced
backwards until they intersect at a point. This intersection point is known as the focal point of a diverging
lens. The focal point is denoted by the letter F on the diagrams below. Note that each lens has two focal
points - one on each side of the lens. Unlike mirrors, lenses can allow light to pass through either face,
depending on where the incident rays are coming from. Subsequently, every lens has two possible focal
points. The distance from the mirror to the focal point is known as the focal length (abbreviated by f).
Technically, a lens does not have a center of curvature (at least not one that has any importance to our
discussion). However a lens does have an imaginary point that we refer to as the 2F point. This is the
point on the principal axis that is twice as far from the vertical axis as the focal point is.
As we discuss the characteristics of images produced by converging and diverging lenses, these
vocabulary terms will become increasingly important. Remember that this page is here and refer to it as
often as needed.
Refraction by Lenses
We have already learned that a lens is a carefully ground or molded piece of transparent material that
refracts light rays in such as way as to form an image. Lenses serve to refract light at each boundary. As a
ray of light enters a lens, it is refracted; and as the same ray of light exits the lens, it is refracted again.
The net effect of the refraction of light at these two boundaries is that the light ray has changed directions.
Because of the special geometric shape of a lens, the light rays are refracted such that they form images.
Before we approach the topic of image formation, we will investigate the refractive ability of converging
and diverging lenses.
First lets consider a double convex lens. Suppose that several rays of light approach the lens; and suppose
that these rays of light are traveling parallel to the principal axis. Upon reaching the front face of the lens,
each ray of light will refract towards the normal to the surface. At this boundary, the light ray is passing
from air into a more dense medium (usually plastic or glass). Since the light ray is passing from a medium
in which it travels fast (less optically dense) into a medium in which it travels relatively slow (more
optically dense), it will bend towards the normal line. This is the FST principle of refraction. This is
shown for two incident rays on the diagram below. Once the light ray refracts across the boundary and
enters the lens, it travels in a straight line until it reaches the back face of the lens. At this boundary, each
ray of light will refract away from the normal to the surface. Since the light ray is passing from a medium
in which it travels slow (more optically dense) to a medium in which it travels fast (less optically dense),
it will bend away from the normal line; this is the SFA principle of refraction.
The above diagram shows the behavior of two incident rays approaching parallel to the principal axis.
Note that the two rays converge at a point; this point is known as the focal point of the lens. The first
generalization that can be made for the refraction of light by a double convex lens is as follows:
Refraction Rule for a Converging Lens
Any incident ray traveling parallel to the principal axis of a converging lens will refract through the
lens and travel through the focal point on the opposite side of the lens.
Now suppose that the rays of light are traveling through the focal point on the way to the lens. These rays
of light will refract when they enter the lens and refract when they leave the lens. As the light rays enter
into the more dense lens material, they refract towards the normal; and as they exit into the less dense air,
they refract away from the normal. These specific rays will exit the lens traveling parallel to the principal
axis.
The above diagram shows the behavior of two incident rays traveling through the focal point on the way
to the lens. Note that the two rays refract parallel to the principal axis. A second generalization for the
refraction of light by a double convex lens can be added to the first generalization.
Refraction Rules for a Converging Lens


Any incident ray traveling parallel to the principal axis of a converging lens will refract
through the lens and travel through the focal point on the opposite side of the lens.
Any incident ray traveling through the focal point on the way to the lens will refract through
the lens and travel parallel to the principal axis.
The Thin Lens Approximation
These two "rules" will greatly simplify the task of determining the
image location for objects placed in front of converging lenses.
This topic will be discussed in the next part of Lesson 5. For now,
internalize the meaning of the rules and be prepared to use them.
As the rules are applied in the construction of ray diagrams, do not
forget the fact that Snells' Law of refraction of light holds for each
of these rays. It just so happens that geometrically, when Snell's
Law is applied for rays that strike the lens in the manner described
above, they will refract in close approximation with these two
rules. The tendency of incident light rays to follow these rules is
increased for lenses that are thin. For such thin lenses, the path of
the light through the lens itself contributes very little to the overall
change in the direction of the light rays. We will use this so-called thin-lens approximation in this unit.
Furthermore, to simplify the construction of ray diagrams, we will avoid refracting each light ray twice upon entering and emerging from the lens. Instead, we will continue the incident ray to the vertical axis of
the lens and refract the light at that point. For thin lenses, this simplification will produce the same result
as if we were refracting the light twice.
A Third Rule of Refraction for Lenses
The above discussion focuses on the manner in which converging and diverging lenses refract incident
rays that are traveling parallel to the principal axis or are traveling through (or towards) the focal point.
But these are not the only two possible incident rays. There are a multitude of incident rays that strike the
lens and refract in a variety of ways. Yet, there are three specific rays that behave in a very predictable
manner. The third ray that we will investigate is the ray that passes through the precise center of the lens through the point where the principal axis and the vertical axis intersect. This ray will refract as it enters
and refract as it exits the lens, but the net affect of this dual refraction is that the path of the light ray is not
changed. For a thin lens, the refracted ray is traveling in the same direction as the incident ray and is
approximately in line with it. The behavior of this third incident ray is depicted in the diagram below.
Now we have three incident rays whose refractive behavior is easily predicted. These three rays lead to
our three rules of refraction for converging and diverging lenses. These three rules are summarized
below.
Refraction Rules for a Converging Lens



Any incident ray traveling parallel to the principal axis of a converging lens will refract
through the lens and travel through the focal point on the opposite side of the lens.
Any incident ray traveling through the focal point on the way to the lens will refract through
the lens and travel parallel to the principal axis.
An incident ray that passes through the center of the lens will in affect continue in the same
direction that it had when it entered the lens.
Refraction Rules for a Diverging Lens



Any incident ray traveling parallel to the principal axis of a diverging lens will refract through
the lens and travel in line with the focal point (i.e., in a direction such that its extension will
pass through the focal point).
Any incident ray traveling towards the focal point on the way to the lens will refract through
the lens and travel parallel to the principal axis.
An incident ray that passes through the center of the lens will in affect continue in the same
direction that it had when it entered the lens.
These three rules of refraction for converging and diverging lenses will be applied through the remainder
of this lesson. The rules merely describe the behavior of three specific incident rays. While there is a
multitude of light rays being captured and refracted by a lens, only two rays are needed in order to
determine the image location. So as we proceed with this lesson, pick your favorite two rules (usually, the
ones that are easiest to remember) and apply them to the construction of ray diagrams and the
determination of the image location and characteristics.
Image Formation Revisited
One major principle discussed in both Unit 13 and Unit 14 of The Physics Classroom Tutorial is the line
of sight principle:
In order to view an object, you must sight along a line at that object; and when you do light will come
from that object to your eye along the line of sight.
This very principle is combined with rules of reflection and refraction in order to explain how an image is
formed and what the characteristics of such images will be.
Plane Mirror Image Formation
In the plane mirror Lesson of Unit 13, it was mentioned that an
image is formed by a plane mirror as light emanates from an
object in a variety of directions. Some of this light reaches the
mirror and reflects off the mirror according to the law of
reflection. Each one of these rays of light can be extended
backwards behind the mirror where they will all intersect at a
point (the image point). Any person who is positioned along the
line of one of these reflected rays can sight along the line and
view the image - a representation of the object. Thus, an image
location is a location in space where all the reflected light appears
to come from. Since light from the object appears to diverge from
this location, a person who sights along a line at this location will perceive a replica or likeness of the
actual object. In the case of plane mirrors, the image is said to be a virtual image. Virtual images are
images that are formed in locations where light does not actually reach. Light does not actually pass
through the location on the other side of the plane mirror; it only appears to an observer as though the
light were coming from this position.
For plane mirrors, virtual images are formed. Light does not actually pass through the virtual image
location; it only appears to an observer as though the light was emanating from the virtual image location.
The image formed by this concave mirror is a real image. When a real image is formed, it still appears to
an observer as though light is diverging from the real image location. Only in the case of a real image,
light is actually passing through the image location.
Converging Lens Image Formation
Converging lenses can produce both real and virtual images while diverging images can only produce
virtual images. The process by which images are formed for lenses is the same as the process by which
images are formed for plane and curved mirrors. Images are formed at locations where any observer is
sighting as they view the image of the object through the lens. So if the path of several light rays through
a lens is traced, each of these light rays will intersect at a point upon refraction through the lens. Each
observer must sight in the direction of this point in order to view the image of the object. While different
observers will sight along different lines of sight, each line of sight intersects at the image location. The
diagram below shows several incident rays emanating from an object - a light bulb. Three of these
incident rays correspond to our three strategic and predictable light rays. Each incident ray will refract
through the lens and be detected by a different observer (represented by the eyes). The location where the
refracted rays are intersecting is the image location.
In this case, the image is a real image since the light rays are actually passing through the image location.
To each observer, it appears as though light is coming from this location.
Images of Objects That Do Not Occupy a Single Point
The above discussion relates to the formation of an image by a "point object" - in this case, a small light
bulb. The same principles apply to objects that occupy more than one point in space. For example, a
person occupies a multitude of points in space. As you sight at a person through a lens, light emanates
from each individual point on that person in all directions. Some of this light reaches the lens and refracts.
All the light that originates from one single point on the object will refract and intersect at one single
point on the image. This is true for all points on the object; light from each point intersects to create an
image of this point. The result is that a replica or likeness of the object is created as we sight at the object
through the lens. This replica or likeness is the image of that object. This is depicted in the diagram
below.
Now that we have discussed how an image is formed, we will turn our attention to the use of ray diagrams
to predict the location and characteristics of images formed by converging and diverging lenses.
Converging Lenses - Ray Diagrams
One theme of the Reflection and Refraction units of The Physics Classroom Tutorial has been that we see
an object because light from the object travels to our eyes as we sight along a line at the object. Similarly,
we see an image of an object because light from the object reflects off a mirror or refracts through a
transparent material and travel to our eyes as we sight at the image location of the object. From these two
basic premises, we have defined the image location as the location in space where light appears to diverge
from. Because light emanating from the object converges or appears to diverge from this location, a
replica or likeness of the object is created at this location. For both reflection and refraction scenarios, ray
diagrams have been a valuable tool for determining the path of light from the object to our eyes.
In this section of Lesson 5, we will investigate the method for drawing ray diagrams for objects placed at
various locations in front of a double convex lens. To draw these ray diagrams, we will have to recall the
three rules of refraction for a double convex lens:



Any incident ray traveling parallel to the principal axis of a converging lens will refract through
the lens and travel through the focal point on the opposite side of the lens.
Any incident ray traveling through the focal point on the way to the lens will refract through the
lens and travel parallel to the principal axis.
An incident ray that passes through the center of the lens will in effect continue in the same
direction that it had when it entered the lens.
Earlier in this lesson, the following diagram illustrating the path of light from an object through a lens to
an eye placed at various locations was shown.
In this diagram, five incident rays are drawn along with their corresponding refracted rays. Each ray
intersects at the image location and then travels to the eye of an observer. Every observer would observe
the same image location and every light ray would follow the Snell's Law of refraction. Yet only two of
these rays would be needed to determine the image location since it only requires two rays to find the
intersection point. Of the five incident rays drawn, three of them correspond to the incident rays described
by our three rules of refraction for converging lenses. We will use these three rays through the remainder
of this lesson, merely because they are the easiest rays to draw. Certainly two rays would be all that is
necessary; yet the third ray will provide a check of the accuracy of our process.
Step-by-Step Method for Drawing Ray Diagrams
The method of drawing ray diagrams for double convex lens is described below. The description is
applied to the task of drawing a ray diagram for an object located beyond the 2F point of a double convex
lens.
1. Pick a point on the top
rays traveling towards the
of the object and draw three incident
lens.
Using a straight edge,
passes exactly through
Draw the second ray
the principal axis. Draw
travels directly to the
arrowheads upon the rays to indicate their direction of travel.
accurately draw one ray so that it
the focal point on the way to the lens.
such that it travels exactly parallel to
the third incident ray such that it
exact center of the lens. Place
2. Once these incident rays strike the lens, refract them according
to the three rules of refraction for converging lenses.
The ray that passes through the focal point on the way to the
lens will refract and travel parallel to the principal axis. Use a
straight edge to accurately draw its path. The ray that traveled
parallel to the principal axis on the way to the lens will refract
and travel through the focal point. And the ray that traveled to
the exact center of the lens will continue in the same direction. Place arrowheads upon the rays to
indicate their direction of travel. Extend the rays past their point of intersection.
3. Mark the image of the top of the object.
The image point of the top of the object is the point where the
three refracted rays intersect. All three rays should intersect at
exactly the same point. This point is merely the point where all
light from the top of the object would intersect upon refracting
through the lens. Of course, the rest of the object has an image
as well and it can be found by applying the same three steps to
another chosen point. (See note below.)
4. Repeat the process for the bottom of the object.
One goal of a ray diagram is to determine the location, size,
orientation, and type of image that is formed by the double
convex lens. Typically, this requires determining where the
image of the upper and lower extreme of the object is located
and then tracing the entire image. After completing the first
three steps, only the image location of the top extreme of the
object has been found. Thus, the process must be repeated for the point on the bottom of the object. If
the bottom of the object lies upon the principal axis (as it does in this example), then the image of this
point will also lie upon the principal axis and be the same distance from the mirror as the image of the
top of the object. At this point the entire image can be filled in.
Some students have difficulty understanding how the entire image of an object can be deduced once a
single point on the image has been determined. If the object is merely a vertical object (such as the arrow
object used in the example below), then the process is easy. The image is merely a vertical line. In theory,
it would be necessary to pick each point on the object and draw a separate ray diagram to determine the
location of the image of that point. That would require a lot of ray diagrams as illustrated in the diagram
below.
Fortunately, a shortcut exists. If the object is a vertical line, then the image is also a vertical line. For our
purposes, we will only deal with the simpler situations in which the object is a vertical line that has its
bottom located upon the principal axis. For such simplified situations, the image is a vertical line with the
lower extremity located upon the principal axis.
The ray diagram above illustrates that when the object is located at a position beyond the 2F point, the
image will be located at a position between the 2F point and the focal point on the opposite side of the
lens. Furthermore, the image will be inverted, reduced in size (smaller than the object), and real. This is
the type of information that we wish to obtain from a ray diagram. These characteristics of the image will
be discussed in more detail in the next section of Lesson 5.
Once the method of drawing ray diagrams is practiced a couple of times, it becomes as natural as
breathing. Each diagram yields specific information about the image. The two diagrams below show how
to determine image location, size, orientation and type for situations in which the object is located at the
2F point and when the object is located between the 2F point and the focal point.
It should be noted that the process of constructing a ray diagram is the same regardless of where the
object is located. While the result of the ray diagram (image location, size, orientation, and type) is
different, the same three rays are always drawn. The three rules of refraction are applied in order to
determine the location where all refracted rays appear to diverge from (which for real images, is also the
location where the refracted rays intersect).
Ray Diagram for Object Located in Front of the Focal Point
In the three cases described above - the case of the object being located beyond 2F, the case of the object
being located at 2F, and the case of the object being located between 2F and F - light rays are converging
to a point after refracting through the lens. In such cases, a real image is formed. As discussed
previously, a real image is formed whenever refracted light passes through the image location. While
diverging lenses always produce virtual images, converging lenses are capable of producing both real and
virtual images. As shown above, real images are produced when the object is located a distance greater
than one focal length from the lens. A virtual image is formed if the object is located less than one focal
length from the converging lens. To see why this is so, a ray diagram can be used.
A ray diagram for the case in which the object is located in front
of the focal point is shown in the diagram at the right. Observe
that in this case the light rays diverge after refracting through
the lens. When refracted rays diverge, a virtual image is formed.
The image location can be found by tracing all light rays
backwards until they intersect. For every observer, the refracted
rays would seem to be diverging from this point; thus, the point
of intersection of the extended refracted rays is the image point.
Since light does not actually pass through this point, the image
is referred to as a virtual image. Observe that when the object in
located in front of the focal point of the converging lens, its
image is an upright and enlarged image that is located on the
object's side of the lens. In fact, one generalization that can be
made about all virtual images produced by lenses (both
converging and diverging) is that they are always upright and
always located on the object's side of the lens.
Ray Diagram for Object Located at the Focal Point
Thus far we have seen via ray diagrams that a real image is produced when an object is located more than
one focal length from a converging lens; and a virtual image is formed when an object is located less than
one focal length from a converging lens (i.e., in front of F). But what happens when the object is located
at F? That is, what type of image is formed when the object is located exactly one focal length from a
converging lens? Of course a ray diagram is always one tool to help find the answer to such a question.
However, when a ray diagram is used for this case, an immediate difficulty is encountered. The diagram
below shows two incident rays and their corresponding refracted rays.
For the case of the object located at the focal point (F), the light rays neither converge nor diverge after
refracting through the lens. As shown in the diagram above, the refracted rays are traveling parallel to
each other. Subsequently, the light rays will not converge to form a real image; nor can they be extended
backwards on the opposite side of the lens to intersect to form a virtual image. So how should the results
of the ray diagram be interpreted? The answer: there is no image!! Surprisingly, when the object is
located at the focal point, there is no location in space at which an observer can sight from which all the
refracted rays appear to be coming. An image cannot be found when the object is located at the focal
point of a converging lens.
Converging Lenses - Object-Image Relations
Previously in Lesson 5, ray diagrams were constructed in order to determine the general location, size,
orientation, and type of image formed by double convex lenses. Perhaps you noticed that there is a
definite relationship between the image characteristics and the location where an object placed in front of
a double convex lens. The purpose of this portion of the lesson is to summarize these object-image
relationships. The best means of summarizing this relationship is to divide the possible object locations
into five general areas or points:





Case 1: the object is located beyond the 2F point
Case 2: the object is located at the 2F point
Case 3: the object is located between the 2F point and the focal point (F)
Case 4: the object is located at the focal point (F)
Case 5: the object is located in front of the focal point (F)
Case 1: The object is located beyond 2F
When the object is located at a location
will always be located somewhere in
between the 2F point and the focal point
(F) on the other side of the lens.
Regardless of exactly where the object is
located, the image will be located in this
specified region. In this case, the image will be an inverted image.
That is to say, if the object is right side up, then the image is upside
beyond the 2F point, the image
down. In this case, the image is reduced in size; in other words, the image dimensions are smaller than
the object dimensions. If the object is a six-foot tall person, then the image is less than six feet tall. Earlier
in Unit 13, the term magnification was introduced; the magnification is the ratio of the height of the
object to the height of the image. In this case, the magnification is a number with an absolute value less
than 1. Finally, the image is a real image. Light rays actually converge at the image location. If a sheet of
paper were placed at the image location, the actual replica or likeness of the object would appear
projected upon the sheet of paper.
Case 2: The object is located at 2F
When the object is located at the 2F point, the image will also be located at
the 2F point on the other side of the lens. In this case, the image will be
inverted (i.e., a right side up object results in an upside-down image). The
image dimensions are equal to the object dimensions. A six-foot tall person
would have an image that is six feet tall; the absolute value of the
magnification is exactly 1. Finally, the image is a real image. Light rays
actually converge at the image location. As such, the image of the object could be projected upon a sheet
of paper.
Case 3: The object is located between 2F and F
When the object is located in front of the 2F point, the image will be
located beyond the 2F point on the other side of the lens. Regardless of
exactly where the object is located between 2F and F, the image will be
located in the specified region. In this case, the image will be inverted
(i.e., a right side up object results in an upside-down image). The image
dimensions are larger than the object dimensions. A six-foot tall person
would have an image that is larger than six feet tall. The absolute value of the magnification is greater
than 1. Finally, the image is a real image. Light rays actually converge at the image location. As such, the
image of the object could be projected upon a sheet of paper.
Case 4: The object is located at F
When the object is located at the focal point, no image is formed. As
discussed earlier in Lesson 5, the refracted rays neither converge nor diverge.
After refracting, the light rays are traveling parallel to each other and cannot
produce an image.
Case 5: The object is located in front of F
When the object is located at a location in front of the focal point, the
image will always be located somewhere on the same side of the lens as the object. Regardless of exactly
where in front of F the object is located, the image will always be located on the object's side of the lens
and somewhere further from the lens. The image is located behind the object. In this case, the image will
be an upright image. That is to say, if the object is right side up, then the image will also be right side up.
In this case, the image is enlarged; in other words, the image dimensions are greater than the object
dimensions. A six-foot tall person would have an image that is larger than six feet tall. The magnification
is greater than 1. Finally, the image is a virtual image. Light rays diverge upon refraction; for this reason,
the image location can only be found by extending the refracted rays backwards on the object's side the
lens. The point of their intersection is the virtual image location. It would appear to any observer as
though light from the object were diverging from this location. Any attempt to project such an image
upon a sheet of paper would fail since light does not actually pass through the image location.
It might be noted from the above descriptions that there is a relationship between the object distance and
object size and the image distance and image size. Starting from a large value, as the object distance
decreases (i.e., the object is moved closer to the lens), the image distance increases; meanwhile, the image
height increases. At the 2F point, the object distance equals the image distance and the object height
equals the image height. As the object distance approaches one focal length, the image distance and image
height approaches infinity. Finally, when the object distance is equal to exactly one focal length, there is
no image. Then altering the object distance to values less than one focal length produces images that are
upright, virtual and located on the same side of the lens as the object. Finally, if the object distance
approaches 0, the image distance approaches 0 and the image height ultimately becomes equal to the
object height. These patterns are depicted in the diagram below. Eight different object locations are drawn
in red and labeled with a number; the corresponding image locations are drawn in blue and labeled with
the identical number.
Check Your Understanding
1. Identify the means by which you can use a converging lens to form a real image.
2. Identify the means by which you can use a converging lens to form a virtual image.
3. A converging lens is sometimes used as a magnifying glass. Explain how this works; specifically,
identify the general region where the object must be placed in order to produce the magnified effect.
The Mathematics of Lenses
Ray diagrams can be used to determine the image location, size,
orientation and type of image formed of objects when placed at a given
location in front of a lens. The use of these diagrams was demonstrated
earlier in Lesson 5 for both converging and diverging lenses. Ray
diagrams provide useful information about object-image relationships,
yet fail to provide the information in a quantitative form. While a ray
diagram may help one determine the approximate location and size of the image, it will not provide
numerical information about image distance and image size. To obtain this type of numerical information,
it is necessary to use the Lens Equation and the Magnification Equation. The lens equation expresses
the quantitative relationship between the object distance (do), the image distance (di), and the focal length
(f). The equation is stated as follows:
The magnification equation relates the ratio of the image distance and object distance to the ratio of the
image height (hi) and object height (ho). The magnification equation is stated as follows:
These two equations can be combined to yield information about the image distance and image height if
the object distance, object height, and focal length are known.
As a demonstration of the effectiveness of the lens equation and magnification equation, consider the
following sample problem and its solution.
Sample Problem #1
A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length
of 15.2 cm. Determine the image distance and the image size.
Like all problems in physics, begin by the identification of the unknown information.
ho = 4.00 cm
do = 45.7 cm
f = 15.2 cm
Next identify the unknown quantities that you wish to solve for.
di = ???
hi = ???
To determine the image distance, the lens equation must be used. The following lines represent the
solution to the image distance; substitutions and algebraic steps are shown.
1/f = 1/do + 1/di
1/(15.2 cm) = 1/(45.7 cm) + 1/di
0.0658 cm-1 = 0.0219 cm-1 + 1/di
0.0439 cm-1 = 1/di
di = 22.8 cm
The numerical values in the solution above were rounded when written down, yet unrounded numbers
were used in all calculations. The final answer is rounded to the third significant digit.
To determine the image height, the magnification equation is needed. Since three of the four quantities in
the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown
below.
hi/ho = - di/do
hi /(4.00 cm) = - (22.8 cm)/(45.7 cm)
hi = - (4.00 cm) • (22.8 cm)/(45.7 cm)
hi = -1.99 cm
The negative values for image height indicate that the image is an inverted image. As is often the case in
physics, a negative or positive sign in front of the numerical value for a physical quantity represents
information about direction. In the case of the image height, a negative value always indicates an inverted
image.
From the calculations in this problem it can be concluded that if a 4.00-cm tall object is placed 45.7 cm
from a double convex lens having a focal length of 15.2 cm, then the image will be inverted, 1.99-cm tall
and located 22.8 cm from the lens. The results of this calculation agree with the principles discussed
earlier in this lesson. In this case, the object is located beyond the 2F point (which would be two focal
lengths from the lens) and the image is located between the 2F point and the focal point. This falls into
the category of Case 1: The object is located beyond 2F for a converging lens.
Now lets try a second sample problem:
Sample Problem #2
A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex lens having a focal length
of 15.2 cm. (NOTE: this is the same object and the same lens, only this time the object is placed closer to
the lens.) Determine the image distance and the image size.
Again, begin by the identification of the unknown information.
ho = 4.00 cm
do = 8.3 cm
f = 15.2 cm
Next identify the unknown quantities that you wish to solve for.
di = ???
hi = ???
To determine the image distance, the lens equation will have to be used. The following lines represent the
solution to the image distance; substitutions and algebraic steps are shown.
1/f = 1/do + 1/di
1/(15.2 cm) = 1/(8.30 cm) + 1/di
0.0658 cm-1 = 0.120 cm-1 + 1/di
-0.0547 cm-1 = 1/di
di = -18.3 cm
The numerical values in the solution above were rounded when written down, yet unrounded numbers
were used in all calculations. The final answer is rounded to the third significant digit.
To determine the image height, the magnification equation is needed. Since three of the four quantities in
the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown
below.
hi/ho = - di/do
hi /(4.00 cm) = - (-18.3 cm)/(8.30 cm)
hi = - (4.00 cm) • (-18.3 cm)/(8.30 cm)
hi = 8.81 cm
The negative value for image distance indicates that the image is a virtual image located on the object's
side of the lens. Again, a negative or positive sign in front of the numerical value for a physical quantity
represents information about direction. In the case of the image distance, a negative value always means
the image is located on the object's side of the lens. Note also that the image height is a positive value,
meaning an upright image. Any image that is upright and located on the object's side of the lens is
considered to be a virtual image.
From the calculations in the second sample problem it can be concluded that if a 4.00-cm tall object is
placed 8.30 cm from a double convex lens having a focal length of 15.2 cm, then the image will be
enlarged, upright, 8.81-cm tall and located 18.3 cm from the lens on the object's side. The results of this
calculation agree with the principles discussed earlier in this lesson. In this case, the object is located in
front of the focal point (i.e., the object distance is less than the focal length) and the image is located
behind the lens. This falls into the category of Case 5: The object is located in front of F (for a converging
lens).
The third sample problem will pertain to a diverging lens.
Sample Problem #3
A 4.00-cm tall light bulb is placed a distance of 35.5 cm from a diverging lens having a focal length of 12.2 cm. Determine the image distance and the image size.
Like all problems in physics, begin by the identification of the unknown information.
ho = 4.00 cm
do = 35.5 cm
f = -12.2 cm
Next identify the unknown quantities that you wish to solve for.
di = ???
hi = ???
To determine the image distance, the lens equation will have to be used. The following lines represent the
solution to the image distance; substitutions and algebraic steps are shown.
1/f = 1/do + 1/di
1/(-12.2 cm) = 1/(35.5 cm) + 1/di
-0.0820 cm-1 = 0.0282 cm-1 + 1/di
-0.110 cm-1 = 1/di
di = -9.08 cm
The numerical values in the solution above were rounded when written down, yet unrounded numbers
were used in all calculations. The final answer is rounded to the third significant digit.
To determine the image height, the magnification equation is needed. Since three of the four quantities in
the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown
below.
hi/ho = - di/do
hi /(4.00 cm) = - (-9.08 cm)/(35.5 cm)
hi = - (4.00 cm) * (-9.08 cm)/(35.5 cm)
hi = 1.02 cm
The negative values for image distance indicate that the image is located on the object's side of the lens.
As mentioned, a negative or positive sign in front of the numerical value for a physical quantity represents
information about direction. In the case of the image distance, a negative value always indicates the
existence of a virtual image located on the object's side of the lens. In the case of the image height, a
positive value indicates an upright image.
From the calculations in this problem it can be concluded that if a 4.00-cm tall object is placed 35.5 cm
from a diverging lens having a focal length of 12.2 cm, then the image will be upright, 1.02-cm tall and
located 9.08 cm from the lens on the object's side. The results of this calculation agree with the principles
discussed earlier in this lesson. Diverging lenses always produce images that are upright, virtual, reduced
in size, and located on the object's side of the lens.
Sign Conventions
The sign conventions for the given quantities in the lens equation and magnification equations are as
follows:



f is + if the lens is a double convex lens (converging lens)
f is - if the lens is a double concave lens (diverging lens)
di is + if the image is a real image and located on the opposite side of the lens.



di is - if the image is a virtual image and located on the object's side of the lens.
hi is + if the image is an upright image (and therefore, also virtual)
hi is - if the image an inverted image (and therefore, also real)
Like many mathematical problems in physics, the skill is only acquired through much personal practice.
Perhaps you would like to take some time to try the following problems
Check Your Understanding
1. Determine the image distance and image height for a 5-cm tall object placed 45.0 cm from a double
convex lens having a focal length of 15.0 cm
2. Determine the image distance and image height for a 5-cm tall object placed 30.0 cm from a double
convex lens having a focal length of 15.0 cm.
3. Determine the image distance and image height for a 5-cm tall object placed 20.0 cm from a double
convex lens having a focal length of 15.0 cm.
4. Determine the image distance and image height for a 5-cm tall object placed 10.0 cm from a double
convex lens having a focal length of 15.0 cm.
5. A magnified, inverted image is located a distance of 32.0 cm from a double convex lens with a focal
length of 12.0 cm. Determine the object distance and tell whether the image is real or virtual.
6. ZINGER: An inverted image is magnified by 2 when the object is placed 22 cm in front of a double
convex lens. Determine the image distance and the focal length of the lens.
7. A double concave lens has a focal length of -10.8 cm. An object is placed 32.7 cm from the lens's
surface. Determine the image distance.
8. Determine the focal length of a double concave lens that produces an image that is 16.0 cm behind the
lens when the object is 28.5 cm from the lens
9.A 2.8-cm diameter coin is placed a distance of 25.0 cm from a double concave lens that has a focal
length of -12.0 cm. Determine the image distance and the diameter of the image.
10. The focal point is located 20.0 cm from a double concave lens. An object is placed 12 cm from the
lens. Determine the image distance.
The Electromagnetic and Visible Spectra
As discussed in Unit 10 of The Physics Classroom Tutorial, electromagnetic waves are waves
that are capable of traveling through a vacuum. Unlike mechanical waves that require a medium
in order to transport their energy, electromagnetic waves are capable of transporting energy
through the vacuum of outer space. Electromagnetic waves are produced by a vibrating electric
charge and as such, they consist of both an electric and a magnetic component. The precise
nature of such electromagnetic waves is not discussed in The Physics Classroom Tutorial.
Nonetheless, there are a variety of statements that can be made about such waves.
Electromagnetic waves exist with an enormous range of frequencies. This continuous range of
frequencies is known as the electromagnetic spectrum. The entire range of the spectrum is
often broken into specific regions. The subdividing of the entire spectrum into smaller spectra is
done mostly on the basis of how each region of electromagnetic waves interacts with matter. The
diagram below depicts the electromagnetic spectrum and its various regions. The longer
wavelength, lower frequency regions are located on the far left of the spectrum and the shorter
wavelength, higher frequency regions are on the far right. Two very narrow regions within the
spectrum are the visible light region and the X-ray region. You are undoubtedly familiar with
some of the other regions of the electromagnetic spectrum.
Visible Light Spectrum
The focus of Lesson 2 will be upon the visible light region - the very narrow band of
wavelengths located to the right of the infrared region and to the left of the ultraviolet region.
Though electromagnetic waves exist in a vast range of wavelengths, our eyes are sensitive to
only a very narrow band. Since this narrow band of wavelengths is the means by which humans
see, we refer to it as the visible light spectrum. Normally when we use the term "light," we are
referring to a type of electromagnetic wave that stimulates the retina of our eyes. In this sense,
we are referring to visible light, a small spectrum from the enormous range of frequencies of
electromagnetic radiation. This visible light region consists of a spectrum of wavelengths that
range from approximately 700 nanometers (abbreviated nm) to approximately 400 nm.
Expressed in more familiar units, the range of wavelengths extends from 7 x 10-7 meter to 4 x 107
meter. This narrow band of visible light is affectionately known as ROYGBIV.
Each individual wavelength within the spectrum of visible light wavelengths is representative of
a particular color. That is, when light of that particular wavelength strikes the retina of our eye,
we perceive that specific color sensation. Isaac Newton showed that light
shining through a prism will be separated into its different wavelengths and
will thus show the various colors that visible light is comprised of. The
separation of visible light into its different colors is known as dispersion.
Each color is characteristic of a distinct wavelength; and different
wavelengths of light waves will bend varying amounts upon passage through a prism. For these
reasons, visible light is dispersed upon passage through a prism. Dispersion of visible light
produces the colors red (R), orange (O), yellow (Y), green (G), blue (B), and violet (V). It is
because of this that visible light is sometimes referred to as ROY G. BIV. (Incidentally, the
indigo is not actually observed in the spectrum but is traditionally added to the list so that there is
a vowel in Roy's last name.) The red wavelengths of light are the longer wavelengths and the
violet wavelengths of light are the shorter wavelengths. Between red and violet, there is a
continuous range or spectrum of wavelengths. The visible light spectrum is shown in the diagram
below.
When all the wavelengths of the visible light spectrum strike your eye at the same time, white is
perceived. The sensation of white is not the result of a single color of light. Rather, the sensation
of white is the result of a mixture of two or more colors of light. Thus, visible light - the mix of
ROYGBIV - is sometimes referred to as white light. Technically speaking, white is not a color
at all - at least not in the sense that there is a light wave with a wavelength that is characteristic
of white. Rather, white is the combination of all the colors of the visible light spectrum. If all the
wavelengths of the visible light spectrum give the appearance of white, then none of the
wavelengths would lead to the appearance of black. Once more, black is not actually a color.
Technically speaking, black is merely the absence of the wavelengths of the visible light
spectrum. So when you are in a room with no lights and everything around you appears black, it
means that there are no wavelengths of visible light striking your eye as you sight at the
surroundings.
SEEE MORE ON WAVESE AT
http://science.hq.nasa.gov/kids/imagers/ems/
MISSING (TRY THEM FROM IGC TEMP)
MEASUREMENTS including density
HOOKE’S LAW
TURNING EFFECTS OF FORCES
Other forms of energy and energy conversion and conservation
THERMAL PHYSICS
END WITH ELECTROMAGNETIC SPECTRUM
Electricity And magnetism
Static Electricity - Lesson 1
Charge and Charge Interactions
The Structure of Matter
There is a large overlap of the world of static electricity and the everyday world that you
experience. Clothes tumble in the dryer and cling together. You walk across the carpeting to exit
a room and receive a door knob shock. You pull a wool sweater off at the end of the day and see
sparks of electricity. During the dryness of winter, you step out of your car and receive a car door
shock as you try to close the door. Sparks of electricity are seen as you pull a wool blanket off
the sheets of your bed. You stroke your cat's fur and observe the fur standing up on its end. Bolts
of lightning dash across the evening sky during a spring thunderstorm. And most tragic of all,
you have a bad hair day. These are all static electricity events - events that can only be explained
by an understanding of the physics of electrostatics.
Not only do electrostatic occurrences permeate the events of everyday life, without the forces
associated with static electricity, life as we know it would be impossible. Electrostatic forces -
both attractive and repulsive in nature - hold the world of atoms and molecules together in
perfect balance. Without this electric force, material things would not exist. Atoms as the
building blocks of matter depend upon these forces. And material objects, including us
Earthlings, are made of atoms and the acts of standing and walking, touching and feeling,
smelling and tasting, and even thinking is the result of electrical phenomenon. Electrostatic
forces are foundational to our existence.
One of the primary questions to be asked in this unit of The Physics Classroom is: How can an
object be charged and what affect does that charge have upon other objects in its vicinity? The
answer to this question begins with an understanding of the structure of matter. Understanding
charge as a fundamental quantity demands that we have an understanding of the structure of an
atom. So we begin this unit with what might seem to many students to be a short review of a unit
from a Chemistry course.
Application of Atomic Structure to Static Electricity
This brief excursion into the history of atomic theory leads to some important conclusions about
the structure of matter that will be of utmost importance to our study of static electricity. Those
conclusions are summarized here:



All material objects are composed of atoms. There are different kinds of atoms known as
elements; these elements can combine to form compounds. Different compounds have
distinctly different properties. Material objects are composed of atoms and molecules of these
elements and compounds, thus providing different materials with different electrical properties.
An atom consists of a nucleus and a vast region of space outside the nucleus. Electrons are
present in the region of space outside the nucleus. They are negatively charged and weakly
bound to the atom. Electrons are often removed from and added to an atom by normal
everyday occurrences. These occurrences are the focus of this Static Electricity unit of The
Physics Classroom.
The nucleus of the atom contains positively charged protons and neutral neutrons. These
protons and neutrons are not removable or perturbable by usual everyday methods. It would
require some form of high-energy nuclear occurrence to disturb the nucleus and subsequently
dislodge its positively charged protons. These high-energy occurrences are fortunately not an
everyday event and they are certainly not the subject of this unit of The Physics Classroom. One
sure truth of this unit is that the protons and neutrons will remain within the nucleus of the
atom. Electrostatic phenomenon can never be explained by the movement of protons.
Summary of Subatomic Particles
Proton
In nucleus
Neutron
In nucleus
Electron
Outside
nucleus
Tightly Bound Tightly Bound
Weakly Bound
Positive
Charge
No Charge
Massive
Negative
Charge
Massive
Not very
massive
A variety of phenomena will be pondered, investigated and explained through the course of this
Static Electricity unit. Each phenomenon will be explained using a model of matter described by
the above three statements. The phenomena will range from a rubber balloon sticking to a
wooden door to the clinging together of clothes that have tumbled in the dryer to the bolt of
lightning seen in the evening sky. Each of these phenomena will be explained in terms of
electron movement - both within the atoms and molecules of a material and from the atoms and
molecules of one material to those of another. In the next section of Lesson 1 we will explore
how electron movement can be used to explain how and why objects acquire an electrostatic
charge.
Neutral vs. Charged Objects
As discussed in a previous section of Lesson 1, atoms are the building blocks of matter. There
are different types of atoms, known as elements. Atoms of each element are distinguished from
each other by the number of protons that are present in their nucleus. An atom containing one
proton is a hydrogen atom (H). An atom containing 6 protons is a carbon atom. And an atom
containing 8 protons is an oxygen atom.
The number of electrons that surround the nucleus will determine whether or not an atom is
electrically charged or electrically neutral. The amount of charge on a single proton is equal to
the amount of charge possessed by a single electron. A proton and an electron have an equal
amount but an opposite type of charge. Thus, if an atom contains equal numbers of protons and
electrons, the atom is described as being electrically neutral. On the other hand, if an atom has
an unequal number of protons and electrons, then the atom is electrically charged (and in fact, is
then referred to as an ion rather than an atom). Any particle, whether an atom, molecule or ion,
that contains less electrons than protons is said to be positively charged. Conversely, any
particle that contains more electrons than protons is said to be negatively charged.
Charged versus Uncharged Particles
Positively
Charged
Possesses more
protons than
electrons
Negatively
Uncharged
Charged
Possesses more Equal numbers of
electrons than
protons and
protons
electrons
Charged Objects as an Imbalance of Protons and Electrons
In the previous section of Lesson 1, an atom was described as being a small and dense core of
positively charged protons and neutral neutrons surrounded by shells of negatively charged
electrons. The protons are tightly bound within the nucleus and not removable by ordinary
measures. While the electrons are attracted to the protons of the nucleus, the addition of energy
to an atom can persuade the electrons to leave an atom. Similarly, electrons within atoms of
other materials can be persuaded to leave their own electron shells and become members of the
electrons shells of other atoms of different materials. In short, electrons are migrants - constantly
on the move and always ready to try out a new atomic environment.
All objects are composed of these atoms. The electrons contained within
the objects are prone to move or migrate to other objects. The process of
an electron leaving one material object to reside (perhaps only
temporarily) in another object is a common everyday occurrence. Even as
you read the words of this web page, some electrons are likely moving
through the monitor and adhering to your clothing (assuming that you are
using this resource online) (and wearing clothes). If you were to walk
across the carpeting towards the door of the room, electrons would likely
be scuffed off the atoms of your shoes and moved onto the atoms of the
carpet. And as clothes tumble in the dryer, it is highly likely that electrons
on one piece of clothing will move from the atoms of the clothing onto the atoms of another
piece of clothing. In general, for electrons to make a move from the atoms of one material to the
atoms of another material, there must be an energy source, a motive, and a low-resistance
pathway.
The cause and mechanisms by which this movement of electrons occurs will be the subject of
Lesson 2. For now, it is sufficient to say that objects that are charged contain unequal numbers of
protons and electrons. Charged objects have an imbalance of charge - either more negative
electrons than positive protons or vice versa. And neutral objects have a balance of charge equal numbers of protons and electrons. The principle stated earlier for atoms can be applied to
objects. Objects with more electrons than protons are charged negatively; objects with fewer
electrons than protons are charged positively.
In this discussion of electrically charged versus electrically neutral objects, the neutron has been
neglected. Neutrons, being electrically neutral play no role in this unit. Their presence (or
absence) will have no direct bearing upon whether an object is charged or uncharged. Their role
in the atom is merely to provide stability to the nucleus, a subject not discussed in The Physics
Classroom. When it comes to the drama of static electricity, electrons and protons become the
main characters.
Charge as a Quantity
Like mass, the charge of an object is a measurable quantity. The charge possessed by an object is
often expressed using the scientific unit known as the Coulomb. Just as mass is measured in
grams or kilograms, charge is measured in units of Coulombs (abbreviated C). Because one
Coulomb of charge is an abnormally large quantity of charge, the units of microCoulombs (µC)
or nanoCoulombs (nC) are more commonly used as the unit of measurement of charge. To
illustrate the magnitude of 1 Coulomb, an object would need an excess of 6.25 x 1018 electrons to
have a total charge of -1 C. And of course an object with a shortage of 6.25 x 1018 electrons
would have a total charge of +1 C.
The charge on a single electron is -1.6 x 10 -19 Coulomb. The charge on a single proton is +1.6 x
10 -19 Coulomb. The quantity of charge on an object reflects the amount of imbalance between
electrons and protons on that object. Thus, to determine the total charge of a positively charged
object (an object with an excess of protons), one must subtract the total number of electrons from
the total number of protons. This operation yields the number of excess protons. Since a single
proton contributes a charge of +1.6 x 10 -19 Coulomb to the overall charge of an atom, the total
charge can be computed by multiplying the number of excess protons by +1.6 x 10 -19 Coulomb.
A similar process is used to determine the total charge of a negatively charged object (an object
with an excess of electrons), except that the number of protons is first subtracted from the
number of electrons.
This principle is illustrated in the following table.
Object
# of Excess
Protons/Electrons
Quantity and Kind of Charge (Q) on
Object in Coulombs (C)
A
1 x 106 excess electrons
-1.6 x 10-13 C
B
1 x 106 excess protons
+1.6 x 10-13 C
C
2 x 1010 excess electrons
-3.2 x 10-9 C
D
3.5 x 108 excess protons
+5.6 x 10-11 C
E
4.67 x 1010 excess
electrons
-7.5 x 10-9 C
In conclusion, an electrically neutral object is an object that has a balance of protons and
electrons. In contrast, a charged object has an imbalance of protons and electrons. Determining
the quantity of charge on such an object involves a counting process; the total number of
electrons and protons are compared to determine the difference between the number of protons
and electrons. This difference is multiplied by 1.6 x 10 -19 Coulombs to determine the overall
quantity of charge on the object. The type of charge (positive or negative) is determined by
whether the protons or the electrons are in excess.
Charge Interactions
Suppose that you rubbed a balloon with a sample of animal fur such as a wool sweater or even
your own hair. The balloon would likely become charged and its charge would exert a strange
influence upon other objects in its vicinity. If some small bits of paper were placed upon a table
and the balloon were brought near and held above the paper bits, then the presence of the
charged balloon might create a sufficient attraction for the paper bits to raise them off the table.
This influence - known as an electric force - occurs even when the charged balloon is held some
distance away from the paper bits. The electric force is a non-contact force. Any charged object
can exert this force upon other objects - both charged and uncharged objects. One goal of this
unit of The Physics Classroom is to understand the nature of the electric force. In this part of
Lesson 1, two simple and fundamental statements will be made and explained about the nature of
the electric force.
Perhaps you have heard it said so many times that it sounds like a cliché.
Opposites attract. And likes repel.
These two fundamental principles of charge interactions will be used throughout the unit to
explain the vast array of static electricity phenomena. As mentioned in the previous section of
Lesson 1, there are two types of electrically charged objects - those that contain more protons
than electrons and are said to be positively charged and those that contain less protons than
electrons and are said to be negatively charged. These two types of electrical charges - positive
and negative - are said to be opposite types of charge. And consistent with our fundamental
principle of charge interaction, a positively charged object will attract a negatively charged
object. Oppositely charged objects will exert an attractive influence upon each other. In contrast
to the attractive force between two objects with opposite charges, two objects that are of like
charge will repel each other. That is, a positively charged object will exert a repulsive force upon
a second positively charged object. This repulsive force will push the two objects apart.
Similarly, a negatively charged object will exert a repulsive force upon a second negatively
charged object. Objects with like charge repel each other.
The Electric Force and Newton's Third Law
This electric force exerted between two oppositely charged objects or two like charged objects is
a force in the same sense that friction, tension, gravity and air resistance are forces. And being a
force, the same laws and principles that describe any force describe the electrical force. One of
those laws was Newton's law of action-reaction (discussed in Unit 2 of The Physics Classroom).
According to Newton's third law, a force is simply a mutual interaction between two objects
that results in an equal and opposite push or pull upon
those objects. Let's apply Newton's third law to describe
the interaction between Object A and Object B, both
having positive charge.
Object A exerts a rightward push upon Object B. Object B
exerts a leftward push upon Object A. See diagram at
right. These two pushing forces have equal magnitudes and
are exerted in opposite directions of each other. Each
object does its own pushing upon the other. The push upon Object B (by Object A) is directed
away from Object A; and the push upon Object A (by Object B) is directed away from Object B.
Because of the away from nature of the mutual interaction, the force is said to be repulsive.
Now let's apply the same action-reaction principle to two
oppositely charged objects - Object C (positive) and Object
D (negative). See diagram at right. Object C exerts a
leftward pull upon object D. Object D exerts a rightward
pull upon Object C. Again, each object does its own pulling
of the other. Just as before, these two forces have equal
magnitudes and are exerted in opposite directions of each
other. However in this instance, the direction of the force on
Object D is towards Object C and the direction of the force on Object C is towards object D.
Because of the towards each other nature of the mutual interaction, the force is described as
being attractive.
Interaction Between Charged and Neutral Objects
The interaction between two like-charged objects is repulsive. The interaction between two
oppositely charged objects is attractive. What type of interaction is observed between a charged
object and a neutral object? The answer is quite surprising to many students of physics. Any
charged object - whether positively charged or negatively charged - will have an attractive
interaction with a neutral object. Positively charged objects and neutral objects attract each other;
and negatively charged objects and neutral objects attract each other.
This third interaction between charged and neutral
objects is often demonstrated by physics teachers or
experienced by students in physics lab activities. For
instance, if a charged balloon is held above neutral
bits of paper, the force of attraction for the paper bits
will be strong enough to overwhelm the downward
force of gravity and raise the bits of paper off the
table. If a charged plastic tube is held above some bits of paper, the tube will exert an attractive
influence upon the paper to raise it off the table. And to the bewilderment of many, a charged
rubber balloon can be attracted to a wooden cabinet with enough force that it sticks to the
cabinet. Any charged object - plastic, rubber, or aluminum - will exert an attractive force upon a
neutral object. And in accordance with Newton's law of action-reaction, the neutral object
attracts the charged object.
Conductors and Insulators
The behavior of an object that has been charged is dependent upon whether the object is made of
a conductive or a nonconductive material. Conductors are materials that permit electrons to flow
freely from atom to atom and molecule to molecule. An object made of a conducting material
will permit charge to be transferred across the entire surface of the object. If charge is transferred
to the object at a given location, that charge is quickly distributed across the entire surface of the
object. The distribution of charge is the result of electron movement. Since conductors allow for
electrons to be transported from particle to particle, a charged object will always distribute its
charge until the overall repulsive forces between excess electrons is minimized. If a charged
conductor is touched to another object, the conductor can even transfer its charge to that object.
The transfer of charge between objects occurs more readily if the second object is made of a
conducting material. Conductors allow for charge transfer through the free movement of
electrons.
In contrast to conductors, insulators are materials that impede the free flow of electrons from
atom to atom and molecule to molecule. If charge is transferred to an insulator at a given
location, the excess charge will remain at the initial location of charging. The particles of the
insulator do not permit the free flow of electrons; subsequently charge is seldom distributed
evenly across the surface of an insulator.
While insulators are not useful for transferring charge, they do serve a critical role in electrostatic
experiments and demonstrations. Conductive objects are often mounted upon insulating objects.
This arrangement of a conductor on top of an insulator prevents charge from being transferred
from the conductive object to its surroundings. This arrangement also allows for a student (or
teacher) to manipulate a conducting object without touching it. The insulator serves as a handle
for moving the conductor around on top of a lab table. If charging experiments are performed
with aluminum pop cans, then the cans should be mounted on top of Styrofoam cups. The cups
serve as insulators, preventing the pop cans from discharging their charge. The cups also serve as
handles when it becomes necessary to move the cans around on the table.
Examples of conductors include metals, aqueous solutions of salts (i.e., ionic compounds
dissolved in water), graphite, water and the human body. Examples of insulators include plastics,
Styrofoam, paper, rubber, glass and dry air. The division of materials into the categories of
conductors and insulators is a somewhat artificial division. It is more appropriate to think of
materials as being placed somewhere along a continuum. Those materials that are super
conductive (known as superconductors) would be placed at on end and the least conductive
materials (best insulators) would be placed at the other end. Metals would be placed near the
most conductive end and glass would be placed on the opposite end of the continuum. The
conductivity of a metal might be as much as a million trillion times greater than that of glass.
Along the continuum of conductors and insulators, one might find the human body somewhere
towards the conducting side of the middle. When the body acquires a static
charge it has a tendency to distribute that charge throughout the surface of the
body. Given the size of the human body, relative to the size of typical objects
used in electrostatic experiments, it would require an abnormally large
quantity of excess charge before its affect is noticeable. The affects of excess
charge on the body are often demonstrated using a Van de Graaff generator.
When a student places their hand upon the static ball, excess charge from the ball is shared with
the human body. Being a conductor, the excess charge could flow to the human body and spread
throughout the surface of the body, even onto strands of hair. As the individual strands of hair
become charged, they begin to repel each other. Looking to distance themselves from their likecharged neighbors, the strands of hair begin to rise upward and outward - a truly hair-raising
experience.
Many are familiar with the impact that humidity can have upon static charge buildups. You have
likely noticed that bad hair days, doorknob shocks and static clothing are most common during
winter months. Winter months tend to be the driest months of the year with humidity levels in
the air dropping to lower values. Water, being a conductor, has a tendency to gradually remove
excess charge from objects. When the humidity is high, a person acquiring an excess charge will
tend to lose that charge to water molecules in the surrounding air. On the other hand, dry air
conditions are more conducive to the buildup of static charge and more frequent electric shocks.
Since humidity levels tend to vary from day to day and season to season, it is expected that
electrical affects (and even the success of electrostatic demonstrations) can vary from day to day.
Distribution of Charge via Electron Movement
Predicting the direction that electrons would move within a conducting material is a simple
application of the two fundamental rules of charge interaction. Opposites attract and likes repel.
Suppose that some method is used to impart a negative charge to an object at a given location. At
the location where the charge is imparted, there is an excess of electrons. That is, the multitude
of atoms in that region possess more electrons than protons. Of course, there are a number of
electrons that could be thought of as being quite contented since there is an accompanying
positively charged proton to satisfy their attraction for an opposite. However, the so-called
excess electrons have a repulsive response to each other and would prefer more space. Electrons,
like human beings, wish to manipulate their surroundings in an effort to reduce repulsive affects.
Since these excess electrons are present in a conductor, there is little hindrance to their ability to
migrate to other parts of the object. And that is exactly what they do. In an effort to reduce the
overall repulsive affects within the object, there is a mass migration of excess electrons
throughout the entire surface of the object. Excess electrons migrate to distance themselves from
their repulsive neighbors. In this sense, it is said that excess negative charge distributes itself
throughout the surface of the conductor.
But what happens if the conductor acquires an excess of positive charge? What if electrons are
removed from a conductor at a given location, giving the object an overall positive charge? If
protons cannot move, then how can the excess of positive charge distribute itself across the
surface of the material? While the answers to these questions are not as obvious, it still involves
a rather simple explanation that once again relies on the two fundamental rules of charge
interaction. Opposites attract and likes repel. Suppose that a conducting metal sphere is charged
on its left side and imparted an excess of positive charge. (Of course, this requires that electrons
be removed from the object at the location of charging.) A multitude of atoms in the region
where the charging occurs have lost one or more electrons and have an excess of protons. The
imbalance of charge within these atoms creates affects that can be thought of as disturbing the
balance of charge within the entire object. The presence of these excess protons in a given
location draws electrons from other atoms. Electrons in other parts of the object can be thought
of as being quite contented with the balance of charge that they are experiencing. Yet there will
always be some electrons that will feel the attraction for the excess protons some distance away.
In human terms, we might say these electrons are drawn by curiosity or by the belief that the
grass is greener on the other side of the fence. In the language of electrostatics, we simply assert
that opposites attract - the excess protons and both the neighboring and distant electrons attract
each other. The protons cannot do anything about this attraction since they are bound within the
nucleus of their own atoms. Yet, electrons are loosely bound within atoms; and being present in a
conductor, they are free to move. These electrons make the move for the excess protons, leaving
their own atoms with their own excess of positive charge. This electron migration happens across
the entire surface of the object, until the overall sum of repulsive affects between electrons across
the whole surface of the object are minimized.
Check Your Understanding
Use your understanding of charge to answer the following questions. When finished, click the
button to view the answers.
1. One of these isolated charged spheres is copper and the other is rubber. The diagram below
depicts the distribution of excess negative charge over the surface of two spheres. Label which is
which and support your answer with an explanation.
2. Which of the following materials are likely to exhibit more conductive properties than
insulating properties? _____ Explain your answers.
a. rubber
b.
c. silver
aluminum
d. plastic
e. wet
skin
3. A conductor differs from an insulator in that a conductor ________.
a. has an excess of protons
b. has an excess of electrons
c. can become charged and an insulator cannot
d. has faster moving molecules
e. does not have any neutrons to get in the way of electron flow
f. none of these
4. Suppose that a conducting sphere is charged positively by some method. The charge is
initially deposited on the left side of the sphere. Yet because the object is conductive, the charge
spreads uniformly throughout the surface of the sphere. The uniform distribution of charge is
explained by the fact that ____.
a. the charged atoms at the location of charge move throughout the surface of the sphere
b. the excess protons move from the location of charge to the rest of the sphere
c. excess electrons from the rest of the sphere are attracted towards the excess protons
5. When an oil tanker car has arrived at its destination, it prepares to empty its fuel into a
reservoir or tank. Part of the preparation involves connecting the body of the tanker car with a
metal wire to the ground. Suggest a reason for why is this done.
Lightning
Perhaps the most known and powerful display of electrostatics in nature is a lightning storm.
Lightning storms are inescapable from humankind's attention. They are never invited, never
planned and never gone unnoticed. The rage of a lightning strike will wake a person in the
middle of the night. They send children rushing into parent's bedrooms, crying for assurance that
everything will be safe. The fury of a lightning strike is capable of interrupting midday
conversations and activities. They're the frequent cause of canceled ball games and golf outings.
Children and adults alike crowd around windows to watch the lightning displays in the sky,
standing in awe of the power of static discharges. Indeed, a lightning storm is the most powerful
display of electrostatics in nature.
In this part of Lesson 4, we will ponder two questions:


What is the cause and mechanism associated with lightning strikes?
How do lightning rods serve to protect buildings from the devastating affects of a lightning
strike?
Static Charge Buildup in the Clouds
The scientific community has long pondered the cause of lightning strikes. Even today, it is the
subject of a good deal of scientific research and theorizing. The details of how a cloud becomes
statically charged are not completely understood (as of this writing). Nonetheless there are
several theories that make a good deal of sense and that demonstrate many concepts previously
discussed in this unit of The Physics Classroom.
The precursor of any lightning strike is the polarization of positive and negative charges within a
storm cloud. The tops of the storm clouds are known to acquire an excess of positive charge and
the bottoms of the storm clouds acquire an excess of negative charge. Two mechanisms seem
important to the polarization process. One mechanism involves a separation of charge by a
process that bears resemblance to frictional charging. Clouds are known to contain countless
millions of suspended water droplets and ice particles moving and whirling about in turbulent
fashion. Additional water from the ground evaporates, rises upward and forms clusters of
droplets as it approaches a cloud. This upwardly rising moisture collides with water droplets
within the clouds. In the collisions, electrons are ripped off the rising droplets, causing a
separation of negative electrons from a positively charged water droplet or a cluster of droplets.
The second mechanism that contributes to the polarization of a storm cloud involves a freezing
process. Rising moisture encounters cooler temperatures at higher altitudes. These cooler
temperatures cause the cluster of water droplets to undergo freezing. The frozen particles tend to
cluster more tightly together and form the central regions of the cluster of droplets. The frozen
portion of the cluster of rising moisture becomes negatively charged and the outer droplets
acquire a positive charge. Air currents within the clouds can rip the outer portions off the clusters
and carry them upward toward the top of the clouds. The frozen portion of the droplets with their
negative charge tends to gravitate towards the bottom of the storm clouds. Thus, the clouds
become further polarized.
These two mechanisms are believed to be the primary causes of
the polarization of storm clouds. In the end, a storm cloud
becomes polarized with positive charges carried to the upper
portions of the clouds and negative portions gravitating towards
the bottom of the clouds. The polarization of the clouds has an
equally important affect on the surface of the Earth. The cloud's
electric field stretches through the space surrounding it and
induces movement of electrons upon Earth. Electrons on Earth's
outer surface are repelled by the negatively charged cloud's
bottom surface. This creates an opposite charge on the Earth's
surface. Buildings, trees and even people can experience a
buildup of static charge as electrons are repelled by the cloud's
bottom. With the cloud polarized into opposites and with a positive charge induced upon Earth's
surface, the stage is set for Act 2 in the drama of a lightning strike.
The Mechanics of a Lightning Strike
As the static charge buildup in a storm cloud increases, the electric field surrounding the cloud
becomes stronger. Normally, the air surrounding a cloud would be a good enough insulator to
prevent a discharge of electrons to Earth. Yet, the strong electric fields surrounding a cloud are
capable of ionizing the surrounding air and making it more conductive. The ionization involves
the shredding of electrons from the outer shells of gas molecules. The gas molecules that
compose air are thus turned into a soup of positive ions and free electrons. The insulating air is
transformed into a conductive plasma. The ability of a storm cloud's electric fields to transform
air into a conductor makes charge transfer (in the form of a lightning bolt) from the cloud to the
ground (or even to other clouds) possible.
A lightning bolt begins with the development of a step leader. Excess electrons on the bottom of
the cloud begin a journey through the conducting air to the ground at speeds up to 60 miles per
second. These electrons follow zigzag paths towards the ground, branching at various locations.
The variables that affect the details of the actual pathway are not well known. It is believed that
the presence of impurities or dust particles in various parts of the air might create regions
between clouds and earth that are more conductive than other regions. As the step leader grows,
it might be illuminated by the purplish glow that is characteristic of ionized air molecules.
Nonetheless, the step leader is not the actual lightning strike; it merely provides the roadway
between cloud and Earth along which the lightning bolt will eventually travel.
As the electrons of the step leader approach the Earth, there is an additional repulsion of
electrons downward from Earth's surface. The quantity of positive
charge residing on the Earth's surface becomes even greater. This
charge begins to migrate upward through buildings, trees and
people into the air. This upward rising positive charge - known as
a streamer - approaches the step leader in the air above the surface
of the Earth. The streamer might meet the leader at an altitude
equivalent to the length of a football field. Once contact is made
between the streamer and the leader, a complete conducting
pathway is mapped out and the lightning begins. The contact point
between ground charge and cloud charge rapidly ascends upward
at speeds as high as 50 000 miles per second. As many as a billion
trillion electrons can transverse this path in less than a millisecond.
This initial strike is followed by several secondary strikes or charge surges in rapid succession.
These secondary surges are spaced apart so closely in time that may appear as a single strike.
The enormous and rapid flow of charge along this pathway between the cloud and Earth heats
the surrounding air, causing it to expand violently. The expansion of the air creates a shockwave
that we observe as thunder.
Lightning Rods and Other Protective Measures
Tall buildings, farmhouses and other structures susceptible to
lightning strikes are often equipped with lightning rods. The
attachment of a grounded lightning rod to a building is a
protective measure that is taken to protect the building in the
event of a lightning strike. The concept of a lightning rod was
originally developed by Ben Franklin. Franklin proposed that
lightning rods should consist of a pointed metal pole that extends
upward above the building that it is intended to protect. Franklin
suggested that a lightning rod protects a building by one of two
methods. First, the rod serves to prevent a charged cloud from
releasing a bolt of lightning. And second, the lightning rod serves
to safely divert the lightning to the ground in event that the cloud does discharge its lightning via
a bolt. Franklin's theories on the operation of lightning rods have endured for a couple of
centuries. And not until the most recent decades have scientific studies provided evidence to
confirm the manner in which they operate to protect buildings from lightning damage.
The first of Franklin's two proposed theories is often referred to as the lightning dissipation
theory. According to the theory, the use of a lightning rod on a building protects the building by
preventing the lightning strike. The idea is based upon the principle that the electric field strength
is great around a pointed object. The intense electric fields surrounding a pointed object serve to
ionize the surrounding air, thus enhancing its conductive ability. The dissipative theory states
that as a storm cloud approaches, there is a conductive pathway established between the
statically charged cloud and the lightning rod. According to the theory, static charges gradually
migrate along this pathway to the ground, thus reducing the likelihood of a sudden and explosive
discharge. Proponents of the lightning dissipation theory argue that the primary role of a
lightning rod is to discharge the cloud over a longer length of time, thus preventing the excessive
charge buildup that is characteristic of a lightning strike.
The second of Franklin's proposed theories on the operation of the lightning rod is the basis of
the lightning diversion theory. The lightning diversion theory states that a lighting rod protects
a building by providing a conductive pathway of the charge to the Earth. A lightning rod is
typically attached by a thick copper cable to a grounding rod that is buried in the Earth below.
The sudden discharge from the cloud would be drawn towards the elevated lightning rod but
safely directed to the Earth, thus preventing damage from occurring to the building. The
lightning rod and the attached cable and ground pole provide a low resistance pathway from the
region above the building to the ground below. By diverting the charge through the lightning
protection system, the building is spared of the damage associated with a large quantity of
electric charge passing through it.
Lightning researchers are now generally convinced that the lightning dissipation theory provides
an inaccurate model of how lightning rods work. It is indeed true that the tip of a lightning rod is
capable of ionizing the surrounding air and making it more conductive. However, this affect only
extends for a few meters above the tip of the lightning rod. A few meters of enhanced
conductivity above the tip of the rod is not capable of discharging a large cloud that stretches
over several kilometers of distance. Unfortunately, there are currently no scientifically verified
methods of lightning prevention. Furthermore, recent field studies have further shown that the tip
of the lightning rod does not need to be sharply pointed as Ben Franklin suggested. Blunt-tipped
lightning rods have been found to be more receptive to lightning strikes and thus provide a more
likely path of discharge of a charged cloud. When installing a lightning rod on a building as a
lightning protection measure, it is imperative that the rod be elevated above the building and
connected by a low resistance wire to the ground.
Current Electricity - Lesson 4
Circuit Connections
Circuit Symbols and Circuit Diagrams | Two Types of Connections
Series Circuits | Parallel Circuits | Combination Circuits
Circuit Symbols and Circuit Diagrams
Thus far, this unit of The Physics Classroom tutorial has focused on the key ingredients of an
electric circuit and upon the concepts of electric potential difference, current and resistance.
Conceptual meaning of terms have been introduced and applied to simple circuits. Mathematical
relationships between electrical quantities have been discussed and their use in solving problems
has been modeled. Lesson 4 will focus on the means by which two or more electrical devices can
be connected to form an electric circuit. Our discussion will progress from simple circuits to
mildly complex circuits. Former principles of electric potential difference, current and resistance
will be applied to these complex circuits and the same mathematical formulas will be used to
analyze them.
Electric circuits, whether simple or complex, can be described in a variety of ways. An electric
circuit is commonly described with mere words. Saying something like "A light bulb is
connected to a D-cell" is a sufficient amount of words to describe a simple circuit. On many
occasions in Lessons 1 through 3, words have been used to describe simple circuits. Upon
hearing (or reading) the words, a person grows accustomed to quickly picturing the circuit in
their mind. But another means of describing a circuit is to simply draw it. Such drawings provide
a quicker mental picture of the actual circuit. Circuit drawings like the one below have been used
many times in Lessons 1 through 3.
Describing Circuits with Words
Describing Circuits with
Drawings
"A circuit contains a light bulb
and a 1.5-Volt D-cell."
A final means of describing an electric circuit is by use of conventional circuit symbols to
provide a schematic diagram of the circuit and its components. Some circuit symbols used in
schematic diagrams are shown below.
A single cell or other power source is represented by a long and a short parallel line. A collection
of cells or battery is represented by a collection of long and short parallel lines. In
both cases, the long line is representative of the positive terminal of the energy
source and the short line represents the negative terminal. A straight line is used to
represent a connecting wire between any two components of the circuit. An
electrical device that offers resistance to the flow of charge is generically referred to as a resistor
and is represented by a zigzag line. An open switch is generally represented by providing a break
in a straight line by lifting a portion of the line upward at a diagonal. These circuit symbols will
be frequently used throughout the remainder of Lesson 4 as electric circuits are represented by
schematic diagrams. It will be important to either memorize these symbols or to refer to this
short listing frequently until you become accustomed to their use.
As an illustration of the use of electrical symbols in schematic diagrams, consider the following
two examples.
Example 1:
Description with Words: Three D-cells are placed in a battery pack to power a circuit containing three
light bulbs.
Using the verbal description, one can acquire a mental picture of the circuit being described. This verbal
description can then be represented by a drawing of three cells and three light bulbs connected by
wires. Finally, the circuit symbols presented above can be used to represent the same circuit. Note that
three sets of long and short parallel lines have been used to represent the battery pack with its three Dcells. And note that each light bulb is represented by its own individual resistor symbol. Straight lines
have been used to connect the two terminals of the battery to the resistors and the resistors to each
other.
The above circuits presumed that the three light bulbs were connected in such a way that the
charge flowing through the circuit would pass through each one of the three light bulbs in
consecutive fashion. The path of a positive test charge leaving the positive terminal of the battery
and traversing the external circuit would involve a passage through each one of the three
connected light bulbs before returning to the negative terminal of the battery. But is this the only
way that three light bulbs can be connected? Do they have to be connected in consecutive
fashion as shown above? Absolutely not! In fact, example 2 below contains the same verbal
description with the drawing and the schematic diagrams being drawn differently.
Example 2:
Description with Words: Three D-cells are placed in a battery pack to power a circuit containing three
light bulbs.
Using the verbal description, one can acquire a mental picture of the circuit being described. But this
time, the connections of light bulbs is done in a manner such that there is a point on the circuit where
the wires branch off from each other. The branching location is referred to as a node. Each light bulb is
placed in its own separate branch. These branch wires eventually connect to each other to form a
second node. A single wire is used to connect this second node to the negative terminal of the battery.
These two examples illustrate the two common types of connections made in electric circuits.
When two or more resistors are present in a circuit, they can be connected in series or in parallel.
The remainder of Lesson 4 will be devoted to a study of these two types of connections and the
affect that they have upon electrical quantities such as current, resistance and electric potential.
The next part of Lesson 4 will introduce the distinction between series and parallel connections.
Check Your Understanding
1. Use circuit symbols to construct schematic diagrams for the following circuits:
a. A single cell, light bulb and switch are placed together in a circuit such that the switch can be
opened and closed to turn the light bulb on.
b. A three-pack of D-cells is placed in a circuit to power a flashlight bulb.
c.
d.
2. Use the concept of conventional current to draw an unbroken line on the
schematic diagram at the right that indicates the direction of the
conventional current. Place an arrowhead on your unbroken line.
Two Types of Connections
When there are two or more electrical devices present in a circuit with an energy source, there
are a couple of basic means by which to connect them. They can be connected in series or
connected in parallel. Suppose that there are three light bulbs connected together in the same
circuit. If connected in series, then they are connected in such a way that an individual charge
would pass through each one of the light bulbs in consecutive fashion. When in series, charge
passes through every light bulb. If connected in parallel, a single charge passing through the
external circuit would only pass through one of the light bulbs. The light bulbs are placed within
a separate branch line, and a charge traversing the external circuit will pass through only one of
the branches during its path back to the low potential terminal. The means by which the resistors
are connected will have a major affect upon the overall resistance of the circuit, the total current
in the circuit, and the current in each resistor. In Lesson 4, we will explore the affect of the type
of connection upon the overall current and resistance of the circuit.
A common physics lab activity involves constructing both types of circuits with bulbs connected
in series and bulbs connected in parallel. A comparison and contrast is made between the two
circuits.
The main questions of concern in a lab activity such as this are typically the following:



As the number of resistors (light bulbs) increases, what happens to the overall current
within the circuit?
As the number of resistors (light bulbs) increases, what happens to the overall resistance
within the circuit?
If one of the resistors is turned off (i.e., a light bulb goes out), what happens to the other
resistors (light bulbs) in the circuit? Do they remain on (i.e., lit)?
In conducting the lab activity, distinctly different observations are made for the two types of
circuits. A series circuit can be constructed by connecting light bulbs in such a manner that there
is a single pathway for charge flow; the bulbs are added to the same line with no branching point.
As more and more light bulbs are added, the brightness of each bulb gradually decreases. This
observation is an indicator that the current within the circuit is decreasing.
So for series circuits, as more resistors are added the overall current
within the circuit decreases. This decrease in current is consistent with
the conclusion that the overall resistance increases.
A final observation that is unique to series circuits is the affect of
removing a bulb from a socket. If one of three bulbs in a series circuit is
unscrewed from its socket, then it is observed that the other bulbs
immediately go out. In order for the devices in a series circuit to work,
each device must work. If one goes out, they all go out. Suppose that all
the appliances in a household kitchen were all connected in series. In order for the refrigerator to
work in that kitchen, the toaster oven, dishwasher, garbage disposal and overhead light would all
have to be on. In order for one device in series to work, they all must work. If current is cut from
any one of them, it is cut from all of them. Quite obviously, the appliances in the kitchen are not
connected in series.
Using the same collection of wires, D-cells and bulbs, parallel circuits can be explored in the
same manner. The affect of the number of resistors upon the overall current and the overall
resistance can be investigated. The diagrams below depict the usual means of constructing the
circuit with parallel connections of light bulbs. One will note that a study of the overall current
for parallel connections requires the addition of an indicator bulb. The indicator bulb is placed
outside of the branches and allows one to observe the affect of additional resistors upon the
overall current. The bulbs that are placed in the parallel branches only provide an indicator of the
current through that particular branch. So if investigating the affect of the number of resistors
upon the overall current and resistance, one must make careful observations of the indicator bulb,
not the bulbs that are placed in the branches. The diagram below depicts the typical observations.
It is clear from observing the indicator bulbs in the above diagrams that the addition of more
resistors causes the indicator bulb to get brighter. For parallel circuits, as the number of resistors
increases, the overall current also increases. This increase in current is consistent with a decrease
in overall resistance. Adding more resistors in a separate branch has the unexpected result of
decreasing the overall resistance!
If an individual bulb in a parallel branch is unscrewed from its
socket, then there is still current in the overall circuit and current in
the other branches. Removing the third bulb from its socket has the
affect of transforming the circuit from a three-bulb parallel circuit
to a two-bulb parallel circuit. If the appliances in a household
kitchen were connected in parallel, then the refrigerator could
function without having to have the dishwasher, toaster, garbage
disposal and overhead lights on. One appliance can work without
the other appliances having to be on. Since each appliance is in its own separate branch, turning
that appliance off merely cuts off the flow of charge to that branch. There will still be charge
flowing through the other branches to the other appliances. Quite obviously, the appliances in a
home are wired with parallel connections.
The affect of adding resistors is quite different if added in parallel compared to adding them in
series. Adding more resistors in series means that there is more overall resistance; yet adding
more resistors in parallel means that there is less overall resistance. The fact that one can add
more resistors in parallel and produce less resistance is quite bothersome to many. An analogy
may help to clarify the reason behind this initially bothersome truth.
The flow of charge through the wires of a circuit can be compared to the flow of cars along a
tollway system in a very crowded metropolitan area. The main sources of resistance on a tollway
system are the tollbooths. Stopping cars and forcing them to pay a toll at a tollbooth not only
slows the cars down, but in a highly trafficked area, will also cause a bottleneck with a backup
for miles. The rate at which cars flow past a point on that tollway system is reduced significantly
by the presence of a tollbooth. Clearly, tollbooths are the main resistor to car flow.
Now suppose that in an effort to increase the flow rate the Tollway Authority decides to add two
more tollbooths at a particular toll station where the bottleneck is troublesome to travelers. They
consider two possible means of connecting their tollbooths - in series versus in parallel. If adding
the tollbooths (i.e., resistors) in series, they would add them in a manner that every car flowing
along the highway would have to stop at each tollbooth in consecutive fashion. With only one
pathway through the tollbooths, each car would have to stop and pay a toll at each booth. Instead
of paying 60 cents one time at one booth, they would now have to pay 20 cents three times at
each of the three tollbooths. Quite obviously, adding tollbooths in series would have the overall
affect of increasing the total amount of resistance and decreasing the overall car flow rate (i.e.,
current).
The other means of adding the two additional tollbooths at this particular toll station would be to
add the tollbooths in parallel fashion. Each tollbooth could be placed in a separate branch. Cars
flowing along the tollway would stop at only one of the three booths. There would be three
possible pathways for cars to flow through the toll station and each car would chose only one of
the pathways. Quite obviously, adding tollbooths in parallel would have the overall affect of
decreasing the total amount of resistance and increasing the overall car flow rate (i.e., current)
along the tollway. Just as is the case for adding more electrical resistors in parallel, adding more
tollbooths in parallel branches creates less overall resistance. By allowing for more pathways
(i.e., branches) by which charge and cars can flow through the bottleneck areas, the flow rate can
be increased.
Check Your Understanding
1. Observe the electrical wiring below. Indicate whether the connections are series or parallel
connections. Explain each choice.
2. Two electric circuits are diagrammed below. For each circuit, indicate which two devices are
connected in series and which two devices are connected in parallel.
In series? ___________________
In series? ___________________
In parallel? _________________
Circuit Connections
Series Circuits
As mentioned in the previous section of Lesson 4, two or more electrical devices in a circuit can
be connected by series connections or by parallel connections. When all the devices are
connected using series connections, the circuit is referred to as a
series circuit. In a series circuit, each device is connected in a
manner such that there is only one pathway by which charge can
traverse the external circuit. Each charge passing through the loop of
the external circuit will pass through each resistor in consecutive
fashion.
A short comparison and contrast between series and parallel circuits was made in the previous
section of Lesson 4. In that section, it was emphasized that the act of adding more resistors to a
series circuit results in the rather expected result of having more overall resistance. Since there is
only one pathway through the circuit, every charge encounters the resistance of every device; so
adding more devices results in more overall resistance. This increased resistance serves to reduce
the rate at which charge flows (also known as the current).
Equivalent Resistance and Current
Charge flows together through the external circuit at a rate that is everywhere the same. The
current is no greater at one location as it is at another location. The actual amount of current
varies inversely with the amount of overall resistance. There is a clear relationship between the
resistance of the individual resistors and the overall resistance of the collection of resistors. As
far as the battery that is pumping the charge is concerned, the presence of two 6- resistors in
series would be equivalent to having one 12- resistor in the circuit. The presence of three 6resistors in series would be equivalent to having one 18- resistor in the circuit. And the
presence of four 6- resistors in series would be equivalent to having one 24- resistor in the
circuit.
This is the concept of equivalent resistance. The equivalent resistance of a circuit is the amount
of resistance that a single resistor would need in order to equal the overall affect of the collection
of resistors that are present in the circuit. For series circuits, the mathematical formula for
computing the equivalent resistance (Req) is
Req = R1 + R2 + R3 + ...
where R1, R2, and R3 are the resistance values of the individual resistors that are connected in
series.
The current in a series circuit is everywhere the same. Charge does NOT pile up and begin to
accumulate at any given location such that the current at one location is more than at other
locations. Charge does NOT become used up by resistors such that there is less of it at one
location compared to another. The charges can be thought of as marching together through the
wires of an electric circuit, everywhere marching at the same rate. Current - the rate at which
charge flows - is everywhere the same. It is the same at the first resistor as it is at the last resistor
as it is in the battery. Mathematically, one might write
Ibattery = I1 = I2 = I3 = ...
where I1, I2, and I3 are the current values at the individual resistor locations.
These current values are easily calculated if the battery voltage is known and the individual
resistance values are known. Using the individual resistor values and the equation above, the
equivalent resistance can be calculated. And using Ohm's law ( V = I • R), the current in the
battery and thus through every resistor can be determined by finding the ratio of the battery
voltage and the equivalent resistance.
Ibattery = I1 = I2 = I3 = Vbattery / Req
Electric Potential Difference and Voltage Drops
As discussed in Lesson 1, the electrochemical cell of a circuit supplies energy to the charge to
move it through the cell and to establish an electric potential difference across the two ends of
the external circuit. A 1.5-volt cell will establish an electric potential difference across the
external circuit of 1.5 volts. This is to say that the electric potential at the positive terminal is 1.5
volts greater than at the negative terminal. As charge moves through the external circuit, it
encounters a loss of 1.5 volts of electric potential. This loss in electric potential is referred to as a
voltage drop. It occurs as the electrical energy of the charge is transformed to other forms of
energy (thermal, light, mechanical, etc.) within the resistors or loads. If an electric circuit
powered by a 1.5-volt cell is equipped with more than one resistor, then the cumulative loss of
electric potential is 1.5 volts. There is a voltage drop for each resistor, but the sum of these
voltage drops is 1.5 volts - the same as the voltage rating of the power supply. This concept can
be expressed mathematically by the following equation:
Vbattery = V1 + V2 + V3 + ...
To illustrate this mathematical principle in action, consider the two circuits shown below
in Diagrams A and B. Suppose that you were to asked to determine the two unknown
values of the electric potential difference across the light bulbs in each circuit. To determine their
values, you would have to use the equation above. The battery is depicted by its customary
schematic symbol and its voltage is listed next to it. Determine the voltage drop for the two light
bulbs and then click the Check Answers button to see if you are correct.
Earlier in Lesson 1, the use of an electric potential diagram was discussed. An electric potential
diagram is a conceptual tool for representing the electric potential difference between several
points on an electric circuit. Consider the circuit diagram below and its corresponding electric
potential diagram.
The circuit shown in the diagram above is powered by a 12-volt energy source. There are three
resistors in the circuit connected in series, each having its own voltage drop. The negative sign
for the electric potential difference simply denotes that there is a loss in electric potential when
passing through the resistor. Conventional current is directed through the external circuit from
the positive terminal to the negative terminal. Since the schematic symbol for a voltage source
uses a long bar to represent the positive terminal, location A in the diagram is at the positive
terminal or the high potential terminal. Location A is at 12 volts of electric potential and location
H (the negative terminal) is at 0 volts. In passing through the battery, the charge gains 12 volts of
electric potential. And in passing through the external circuit, the charge loses 12 volts of electric
potential as depicted by the electric potential diagram shown
to the right of the schematic diagram. This 12 volts of electric
potential is lost in three steps with each step corresponding to
the flow through a resistor. In passing through the connecting wires between resistors, there is
little loss in electric potential due to the fact that a wire offers relatively little resistance to the
flow of charge. Since locations A and B are separated by a wire, they are at virtually the same
electric potential of 12 V. When a charge passes through its first resistor, it loses 3 V of electric
potential and drops down to 9 V at location C. Since location D is separated from location C by a
mere wire, it is at virtually the same 9 V electric potential as C. When a charge passes through its
second resistor, it loses 7 V of electric potential and drops down to 2 V at location E. Since
location F is separated from location E by a mere wire, it is at virtually the same 2 V electric
potential as E. Finally, as a charge passes through its last resistor, it loses 2 V of electric potential
and drops down to 0 V at G. At locations G and H, the charge is out of energy and needs an
energy boost in order to traverse the external circuit again. The energy boost is provided by the
battery as the charge is moved from H to A.
In Lesson 3, Ohm's law ( V = I • R) was introduced as an equation that relates the voltage drop
across a resistor to the resistance of the resistor and the current at the resistor. The Ohm's law
equation can be used for any individual resistor in a series circuit. When combining Ohm's law
with some of the principles already discussed on this page, a big idea
emerges.
In series circuits, the resistor with the greatest resistance has the greatest
voltage drop.
Since the current is everywhere the same within a series circuit, the I value of V = I • R is the
same in each of the resistors of a series circuit. So the voltage drop ( V) will vary with varying
resistance. Wherever the resistance is greatest, the voltage drop will be greatest about that
resistor. The Ohm's law equation can be used to not only predict that resistor in a series circuit
will have the greatest voltage drop, it can also be used to calculate the actual voltage drop values.
V1 = I • R1
V2 = I • R2
V3 = I • R3
Mathematical Analysis of Series Circuits
The above principles and formulae can be used to analyze a series circuit and determine the
values of the current at and electric potential difference across each of the resistors in a series
circuit. Their use will be demonstrated by the mathematical analysis of the circuit shown below.
The goal is to use the formulae to determine the equivalent resistance of the circuit (Req), the
current at the battery (Itot), and the voltage drops and current for each of the three resistors.
The analysis begins by using the resistance values for the individual resistors in order to
determine the equivalent resistance of the circuit.
Req = R1 + R2 + R3 = 17 + 12 + 11 = 40
Now that the equivalent resistance is known, the current at the battery can be determined using
the Ohm's law equation. In using the Ohm's law equation ( V = I • R) to determine the current in
the circuit, it is important to use the battery voltage for V and the equivalent resistance for R.
The calculation is shown here:
Itot = Vbattery / Req = (60 V) / (40 ) = 1.5 amp
The 1.5 amp value for current is the current at the battery location. For a series circuit with no
branching locations, the current is everywhere the same. The current at the battery location is the
same as the current at each resistor location. Subsequently, the 1.5 amp is the value of I1, I2, and
I3.
Ibattery = I1 = I2 = I3 = 1.5 amp
There are three values left to be determined - the voltage drops across each of the individual
resistors. Ohm's law is used once more to determine the voltage drops for each resistor - it is
simply the product of the current at each resistor (calculated above as 1.5 amp) and the resistance
of each resistor (given in the problem statement). The calculations are shown below.
V1 = I1 • R1
V2 = I2 • R2
V3 = I3 • R3
V1 = (1.5 A)
• (17 )
V2 = (1.5 A)
• (12 )
V3 = (1.5 A)
• (11 )
V1 = 25.5 V
V2 = 18 V
V3 = 16.5 V
As a check of the accuracy of the mathematics performed, it is wise to see if the calculated
values satisfy the principle that the sum of the voltage drops for each individual resistor is equal
to the voltage rating of the battery. In other words, is Vbattery = V1 + V2 + V3 ?
Is Vbattery = V1 + V2 + V3 ?
Is 60 V = 25.5 V + 18 V + 16.5 V ?
Is 60 V = 60 V?
Yes!!
The mathematical analysis of this series circuit involved a blend of concepts and equations. As is
often the case in physics, the divorcing of concepts from equations when embarking on the
solution to a physics problem is a dangerous act. Here, one must consider the concepts that the
current is everywhere the same and that the battery voltage is equivalent to the sum of the
voltage drops across each resistor in order to complete the mathematical analysis. In the next part
of Lesson 4, parallel circuits will be analyzed using Ohm's law and parallel circuit concepts. We
will see that the approach of blending the concepts with the equations will be equally important
to that analysis.
Check Your Understanding
1. Use your understanding of equivalent resistance to complete the following statements:
a. Two 3- resistors placed in series would provide a resistance that is equivalent to one _____resistor.
b. Three 3- resistors placed in series would provide a resistance that is equivalent to one _____resistor.
c. Three 5- resistors placed in series would provide a resistance that is equivalent to one _____resistor.
d. Three resistors with resistance values of 2- , 4- , and 6- are placed in series. These would
provide a resistance that is equivalent to one _____- resistor.
e. Three resistors with resistance values of 5- , 6- , and 7- are placed in series. These would
provide a resistance that is equivalent to one _____- resistor.
f. Three resistors with resistance values of 12- , 3- , and 21- are placed in series. These
would provide a resistance that is equivalent to one _____- resistor.
2. As the number of resistors in a series circuit increases, the overall resistance __________
(increases, decreases, remains the same) and the current in the circuit __________ (increases,
decreases, remains the same).
3. Consider the following two diagrams of series circuits. For each diagram, use arrows to
indicate the direction of the conventional current. Then, make comparisons of the voltage and the
current at the designated points for each diagram.
4. Three identical light bulbs are connected to a D-cell as shown at the
right. Which one of the following statements is true?
a. All three bulbs will have the same brightness.
b. The bulb between X and Y will be the brightest.
c. The bulb between Y and Z will be the brightest.
d. The bulb between Z and the battery will be the brightest.
5. Three identical light bulbs are connected to a battery as shown at the
right. Which adjustments could be made to the circuit that would increase
the current being measured at X? List all that apply.
a. Increase the resistance of one of the bulbs.
b. Increase the resistance of two of the bulbs.
c. Decrease the resistance of two of the bulbs.
d. Increase the voltage of the battery.
e. Decrease the voltage of the battery.
f. Remove one of the bulbs.
6. Three identical light bulbs are connected to a battery as shown at the
right. W, X,Y and Z represent locations along the circuit. Which one of
the following statements is true?
a. The potential difference between X and Y is greater than that between Y
and Z.
b. The potential difference between X and Y is greater than that between Y and W.
c. The potential difference between Y and Z is greater than that between Y and W.
d. The potential difference between X and Z is greater than that between Z and W.
e. The potential difference between X and W is greater than that across the battery.
f. The potential difference between X and Y is greater than that between Z and W.
7. Compare circuit X and Y below. Each is powered by a 12volt battery. The voltage drop across the 12-ohm resistor in
circuit Y is ____ the voltage drop across the single resistor
in X.
a. smaller than
b. larger than
c. the same as
8. A 12-V battery, a 12-ohm resistor and a light bulb are
connected as shown in circuit X below. A 6-ohm resistor
is added to the 12-ohm resistor and bulb to create circuit
Y as shown. The bulb will appear ____.
a. dimmer in circuit X
b. dimmer in circuit Y
c. the same brightness in both circuits
9. Three resistors are connected in series. If placed in a circuit with a 12-volt power supply.
Determine the equivalent resistance, the total circuit current, and the voltage drop across and
current at each resistor.
Parallel Circuits
As mentioned in a previous section of Lesson 4, two or more electrical devices in a circuit can be
connected by series connections or by parallel connections. When all the devices are connected
using parallel connections, the circuit is referred to as a parallel
circuit. In a parallel circuit, each device is placed in its own separate
branch. The presence of branch lines means that there are multiple
pathways by which charge can traverse the external circuit. Each
charge passing through the loop of the external circuit will pass
through a single resistor present in a single branch. When arriving at the branching location or
node, a charge makes a choice as to which branch to travel through on its journey back to the low
potential terminal.
A short comparison and contrast between series and parallel circuits was made in an earlier
section of Lesson 4. In that section, it was emphasized that the act of adding more resistors to a
parallel circuit results in the rather unexpected result of having less overall resistance. Since
there are multiple pathways by which charge can flow, adding another resistor in a separate
branch provides another pathway by which to direct charge through the main area of resistance
within the circuit. This decreased resistance resulting from increasing the number of branches
will have the effect of increasing the rate at which charge flows (also known as the current). In
an effort to make this rather unexpected result more reasonable, a tollway analogy was
introduced. A tollbooth is the main location of resistance to car flow on a tollway. Adding
additional tollbooths within their own branch on a tollway will provide more pathways for cars
to flow through the toll station. These additional tollbooths will decrease the overall resistance to
car flow and increase the rate at which they flow.
Current
The rate at which charge flows through a circuit is known as the current. Charge does NOT pile
up and begin to accumulate at any given location such that the current at one location is more
than at other locations. Charge does NOT become used up by resistors in such a manner that
there is less current at one location compared to another. In a parallel circuit, charge divides up
into separate branches such that there can be more current in one branch than there is in another.
Nonetheless, when taken as a whole, the total amount of current in all the branches when added
together is the same as the amount of current at locations outside the branches. The rule that
current is everywhere the same still works, only with a twist. The current outside the branches is
the same as the sum of the current in the individual branches. It is still the same amount of
current, only split up into more than one pathway.
In equation form, this principle can be written as
Itotal = I1 + I2 + I3 + ...
where Itotal is the total amount of current outside the branches (and in the battery) and I1, I2, and I3
represent the current in the individual branches of the circuit.
Throughout this unit, there has been an extensive reliance upon the analogy between charge flow
and water flow. Once more, we will return to the analogy to illustrate how the sum of the current
values in the branches is equal to the amount outside of the branches. The flow of charge in
wires is analogous to the flow of water in pipes. Consider the diagrams below in which the flow
of water in pipes becomes divided into separate branches. At each node (branching location), the
water takes two or more separate pathways. The rate at which water flows into the node
(measured in gallons per minute) will be equal to the sum of the flow rates in the individual
branches beyond the node. Similarly, when two or more branches feed into a node, the rate at
which water flows out of the node will be equal to the sum of the flow rates in the individual
branches that feed into the node.
The same principle of flow division applies to electric circuits. The rate at which charge flows
into a node is equal to the sum of the flow rates in the individual branches beyond the
node. This is illustrated in the examples shown below. In the examples a new circuit
symbol is introduced - the letter A enclosed within a circle. This is the symbol for an
ammeter - a device used to measure the current at a specific point. An ammeter is capable of
measuring the current while offering negligible resistance to the flow of charge.
Diagram A displays two resistors in parallel with nodes at point A and point B. Charge flows
into point A at a rate of 6 amps and divides into two pathways - one through resistor 1 and the
other through resistor 2. The current in the branch with resistor 1 is 2 amps and the current in the
branch with resistor 2 is 4 amps. After these two branches meet again at point B to form a single
line, the current again becomes 6 amps. Thus we see the principle that the current outside the
branches is equal to the sum of the current in the individual branches holds true.
Itotal = I1 + I2
6 amps = 2 amps + 4 amps
Diagram B above may be slightly more complicated with its three resistors placed in parallel.
Four nodes are identified on the diagram and labeled A, B, C and D. Charge flows into point A at
a rate of 12 amps and divides into two pathways - one passing through resistor 1 and the other
heading towards point B (and resistors 2 and 3). The 12 amps of current is divided into a 2 amp
pathway (through resistor 1) and a 10 amp pathway (heading toward point B). At point B, there
is further division of the flow into two pathways - one through resistor 2 and the other through
resistor 3. The current of 10 amps approaching point B is divided into a 6-amp pathway (through
resistor 2) and a 4-amp pathway (through resistor 3). Thus, it is seen that the current values in the
three branches are 2 amps, 6 amps and 4 amps and that the sum of the current values in the
individual branches is equal to the current outside the branches.
Itotal = I1 + I2 + I3
12 amps = 2 amps + 6 amps + 4 amps
A flow analysis at points C and D can also be conducted and it is observed that the sum of the
flow rates heading into these points is equal to the flow rate that is found immediately beyond
these points.
Equivalent Resistance
The actual amount of current always varies inversely with the amount of overall resistance.
There is a clear relationship between the resistance of the individual resistors and the overall
resistance of the collection of resistors. To explore this relationship, let's begin with the simplest
case of two resistors placed in parallel branches, each having the same resistance value of 4 .
Since the circuit offers two equal pathways for charge flow, only one-half the charge will choose
to pass through a given branch. While each individual branch offers 4 of resistance to any
charge that flows through it, only one-half of all the charge flowing through the circuit will
encounter the 4 resistance of that individual branch. Thus, as far as the battery that is pumping
the charge is concerned, the presence of two 4- resistors in parallel would be equivalent to
having one 2- resistor in the circuit. In the same manner, the presence of two 6- resistors in
parallel would be equivalent to having one 3- resistor in the circuit. And the presence of two
12- resistors in parallel would be equivalent to having one 6- resistor in the circuit.
Now let's consider another simple case of having three resistors in parallel, each having the same
resistance of 6 . With three equal pathways for charge to flow through the external circuit, only
one-third the charge will choose to pass through a given branch. Each individual branch offers 6
of resistance to the charge that passes through it. However, the fact that only one-third of the
charge passes through a particular branch means that the overall resistance of the circuit is 2 .
As far as the battery that is pumping the charge is concerned, the presence of three 6- resistors
in parallel would be equivalent to having one 2- resistor in the circuit. In the same manner, the
presence of three 9- resistors in parallel would be equivalent to having one 3- resistor in the
circuit. And the presence of three 12- resistors in parallel would be equivalent to having one 4resistor in the circuit.
This is the concept of equivalent resistance. The equivalent resistance of a circuit is the amount
of resistance that a single resistor would need in order to equal the overall effect of the collection
of resistors that are present in the circuit. For parallel circuits, the mathematical formula for
computing the equivalent resistance (Req) is
1 / Req = 1 / R1 + 1 / R2 + 1 / R3 + ...
where R1, R2, and R3 are the resistance values of the individual resistors that are connected in
parallel. The examples above could be considered simple cases in which all the pathways offer
the same amount of resistance to an individual charge that passes through it. The simple cases
above were done without the use of the equation. Yet the equation fits both the simple cases
where branch resistors have the same resistance values and the more difficult cases where branch
resistors have different resistance values. For instance, consider the application of the equation to
the one simple and one difficult case below.
Case 1: Three 12
1/Req = 1/R1 + 1/R2 + 1/R3
resistors are placed in
1/Req = 1/(12 ) + 1/(12 ) + 1/(12
parallel
)
Using a calculator ...
1/Req = 0.25
Req = 1 / (0.25
-1
-1
)
Req = 4.0
Case 2: A 5.0 , 7.0 ,
and 12 resistor are
placed in parallel
1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/(5.0 ) + 1/(7.0 ) +
1/(12 )
Using a calculator ...
1/Req = 0.42619
Req = 1 / (0.42619
-1
-1
)
Req = 2.3
It's Your Turn to Try It
Voltage Drops for Parallel Branches
It has been emphasized throughout the Circuits unit of The Physics Classroom tutorial that
whatever voltage boost is acquired by a charge in the battery is lost by the charge as it passes
through the resistors of the external circuit. The total voltage drop in the external circuit is equal
to the gain in voltage as a charge passes through the internal circuit. In a parallel circuit, a charge
does not pass through every resistor; rather, it passes through a single resistor. Thus, the entire
voltage drop across that resistor must match the battery voltage. It matters not whether the charge
passes through resistor 1, resistor 2, or resistor 3, the voltage drop across the resistor that it
chooses to pass through must equal the voltage of the battery. Put in equation form, this principle
would be expressed as
Vbattery = V1 = V2 = V3 = ...
If three resistors are placed in parallel branches and powered by a 12-volt battery, then the
voltage drop across each one of the three resistors is 12 volts. A charge flowing through the
circuit would only encounter one of these three resistors and thus encounter a single voltage drop
of 12 volts.
Electric potential diagrams were introduced in Lesson 1 of this unit and subsequently used to
illustrate the consecutive voltage drops occurring in series circuits. An electric potential diagram
is a conceptual tool for representing the electric potential difference between several points on an
electric circuit. Consider the circuit diagram below and its corresponding electric potential
diagram.
As shown in the electric potential diagram, positions A, B, C, E and G are all at a high electric
potential. A single charge chooses only one of the three possible pathways; thus at position B, a
single charge will move towards point C, E or G and then passes through the resistor that is in
that branch. The charge does not lose its high potential until it passes through the resistor, either
from C to D, E to F, or G to H. Once it passes through a resistor, the charge has returned to
nearly 0 Volts and returns to the negative terminal of the battery to obtain its voltage boost.
Unlike in series circuits, a charge in a parallel circuit encounters a single voltage drop during its
path through the external circuit.
The current through a given branch can be predicted using the Ohm's law equation and the
voltage drop across the resistor and the resistance of the resistor. Since the voltage drop is the
same across each resistor, the factor that determines that resistor has the greatest current is the
resistance. The resistor with the greatest resistance experiences the lowest current and the resistor
with the least resistance experiences the greatest current. In this sense, it could be said that
charge (like people) chooses the path of least resistance. In equation form, this could be stated as
I1 = V1 / R1
I2 = V2 / R2
I3 = V3 / R3
This principle is illustrated by the circuit shown below. The product of I•R is the same for each
resistor (and equal to the battery voltage). Yet the current is different in each resistor. The current
is greatest where the resistance is least and the current is least where the resistance is greatest.
Mathematical Analysis of Parallel Circuits
The above principles and formulae can be used to analyze a parallel circuit and determine the
values of the current at and electric potential difference across each of the resistors in a parallel
circuit. Their use will be demonstrated by the mathematical analysis of the circuit shown below.
The goal is to use the formulae to determine the equivalent resistance of the circuit (Req), the
current through the battery (Itot), and the voltage drops and current for each of the three resistors.
The analysis begins by using the resistance values for the individual resistors in order to
determine the equivalent resistance of the circuit.
1 / Req = 1 / R1 + 1 / R2 + 1 / R3 = (1 / 17 ) + (1 / 12 ) + (1 / 11 )
1 / Req = 0.23306
-1
Req = 1 / (0.23306
-1
)
Req = 4.29
(rounded from 4.29063 )
Now that the equivalent resistance is known, the current in the battery can be determined using
the Ohm's law equation. In using the Ohm's law equation ( V = I • R) to determine the current in
the battery, it is important to use the battery voltage for V and the equivalent resistance for R.
The calculation is shown here:
Itot = Vbattery / Req = (60 V) / (4.29063 )
Itot = 14.0 amp
(rounded from 13.98396 amp)
The 60 V battery voltage represents the gain in electric potential by a charge as it passes through
the battery. The charge loses this same amount of electric potential for any given pass through
the external circuit. That is, the voltage drop across each one of the three resistors is the same as
the voltage gained in the battery:
V battery = V1 = V2 = V3 = 60 V
There are three values left to be determined - the current in each of the individual resistors.
Ohm's law is used once more to determine the current values for each resistor - it is simply the
voltage drop across each resistor (60 Volts) divided by the resistance of each resistor (given in
the problem statement). The calculations are shown below.
I1 = V1 / R1
V2 = V 2 / R2
V3 = V 3 / R3
I1 = (60 V) / (17
)
I2 = (60 V) / (12
)
I3 = (60 V) / (11
)
I1 = 3.53 amp
I2 = 5.00 amp
I3 = 5.45 amp
As a check of the accuracy of the mathematics performed, it is wise to see if the calculated
values satisfy the principle that the sum of the current values for each individual resistor is equal
to the total current in the circuit (or in the battery). In other words, is Itot = I1 + I2 + I3 ?
Is Itot = I1 + I2 + I3 ?
Is 14.0 amp = 3.53 amp + 5.00 amp + 5.45 amp ?
Is 14.0 amp = 13.98 amp ?
Yes!!
(The 0.02 amp difference is simply the result of having previously rounded the Itot value from
13.98.)
The mathematical analysis of this parallel circuit involved a blend of concepts and equations. As
is often the case in physics, the divorcing of concepts from equations when embarking on the
solution to a physics problem is a dangerous act. Here, one must consider the concepts that the
voltage drops across each one of the three resistors is equal to the battery voltage and that the
sum of the current in each resistor is equal to the total current. These understandings are essential
in order to complete the mathematical analysis. In the next part of Lesson 4, combination or
compound circuits in which some devices are in parallel and others are in series will be
investigated.
Check Your Understanding
1. As more and more resistors are added in parallel to a circuit, the equivalent resistance of the
circuit ____________ (increases, decreases) and the total current of the circuit ____________
(increases, decreases).
2. Three identical light bulbs are connected to a D-cell as shown below.
P, Q, X, Y and Z represent locations along the circuit. Which one of the
following statements is true?
a. The current at Y is greater than the current at Q.
b. The current at Y is greater than the current at P.
c. The current at Y is greater than the current at Z.
d. The current at P is greater than the current at Q.
e. The current at Q is greater than the current at P.
f. The current is the same at all locations.
3. Three identical light bulbs are connected to a D-cell as shown below.
P, Q, X, Y and Z represent locations along the circuit. At which
location(s), if any, will the current be ...
a. ... the same as at X?
b. ... the same as at Q?
c. ... the same as at Y?
d. ... less than at Q?
e. ... less than at P?
f. ... twice that at Z?
g. ... three times that at Y?
4. Which adjustments could be made to the circuit below that would
decrease the current in the cell? List all that apply.
a. Increase the resistance of bulb X.
b. Decrease the resistance of bulb X.
c. Increase the resistance of bulb Z.
d. Decrease the resistance of bulb Z.
e. Increase the voltage of the cell (somehow).
f. Decrease the voltage of the cell (somehow).
g. Remove bulb Y.
5. A 12-V battery , a 12-ohm resistor and a 4-ohm resistor are
connected as shown. The current in the 12-ohm resistor is ____ that
in the 4-ohm resistor.
a. 1/3 b. 1/2
d. the same as
f. twice
h. four times
c. 2/3
e. 1.5 times
g. three times
6. A 12-V battery , a 12-ohm resistor and a 4-ohm resistor are
connected as shown. The voltage drop across the 12-ohm resistor is
____ that across the 4-ohm resistor.
a. 1/3 b. 1/2
d. the same as
f. twice
h. four times
c. 2/3
e. 1.5 times
g. three times
7. A 12-V battery and a 12-ohm resistor are connected
as shown in circuit. A 6-ohm resistor is added to the
12-ohm resistor to create circuit Y as shown. The
voltage drop across the 6-ohm resistor in circuit Y is
____ that across the resistor in X.
a. larger than
b. smaller than
c. the same as
8. Use your understanding of equivalent resistance to complete the following statements:
a. Two 6- resistors placed in parallel would provide a resistance that is equivalent to one _____resistor.
b. Three 6- resistors placed in parallel would provide a resistance that is equivalent to one
_____- resistor.
c. Three 8- resistors placed in parallel would provide a resistance that is equivalent to one
_____- resistor.
d. Three resistors with resistance values of 2- , 4- , and 6- are placed in parallel. These
would provide a resistance that is equivalent to one _____- resistor.
e. Three resistors with resistance values of 5- , 6- , and 7- are placed in parallel. These
would provide a resistance that is equivalent to one _____- resistor.
f. Three resistors with resistance values of 12- , 6- , and 21- are placed in parallel. These
would provide a resistance that is equivalent to one _____- resistor.
9. Based on your answers to the above question, complete the following statement:
The overall or equivalent resistance of three resistors placed in parallel will be _____.
a. greater than the resistance of the biggest R value of the three.
b. less than the resistance of the smallest R value of the three.
c. somewhere in between the smallest R and the biggest R value of the three.
d. ... nonsense! No such generalization can be made. The results vary.
10. Three resistors are connected in parallel. If placed in a circuit with a 12-volt power supply.
Determine the equivalent resistance, the total circuit current, and the voltage drop across and
current in each resistor.
Combination Circuits
Previously in Lesson 4, it was mentioned that there are two different ways to connect two or
more electrical devices together in a circuit. They can be connected by means of series
connections or by means of parallel connections. When all the devices in a circuit are connected
by series connections, then the circuit is referred to as a series circuit. When all the devices in a
circuit are connected by parallel connections, then the circuit is
referred to as a parallel circuit. A third type of circuit involves the
dual use of series and parallel connections in a circuit; such
circuits are referred to as compound circuits or combination
circuits. The circuit depicted at the right is an example of the use
of both series and parallel connections within the same circuit. In
this case, light bulbs A and B are connected by parallel
connections and light bulbs C and D are connected by series
connections. This is an example of a combination circuit.
When analyzing combination circuits, it is critically important to have a solid understanding of
the concepts that pertain to both series circuits and parallel circuits. Since both types of
connections are used in combination circuits, the concepts associated with both types of circuits
apply to the respective parts of the circuit. The main concepts associated with series and parallel
circuits are organized in the table below.
Series Circuits



The current is the same in every resistor; this
current is equal to that in the battery.
The sum of the voltage drops across the
individual resistors is equal to the voltage
rating of the battery.
The overall resistance of the collection of
resistors is equal to the sum of the individual
Parallel Circuits



The voltage drop is the same across each
parallel branch.
The sum of the current in each individual
branch is equal to the current outside the
branches.
The equivalent or overall resistance of
the collection of resistors is given by the
resistance values,
equation
Rtot = R1 + R2 + R3 + ...
1/Req = 1/R1 + 1/R2 + 1/R3 ...
Each of the above concepts has a mathematical expression. Combining the mathematical
expressions of the above concepts with the Ohm's law equation ( V = I • R) allows one to
conduct a complete analysis of a combination circuit.
Analysis of Combination Circuits
The basic strategy for the analysis of combination circuits involves using the meaning of
equivalent resistance for parallel branches to transform the combination circuit into a series
circuit. Once transformed into a series circuit, the analysis can be conducted in the usual manner.
Previously in Lesson 4, the method for determining the equivalent resistance of parallel are
equal, then the total or equivalent resistance of those branches is equal to the resistance of one
branch divided by the number of branches.
This method is consistent with the formula
1 / Req = 1 / R1 + 1 / R2 + 1 / R3 + ...
where R1, R2, and R3 are the resistance values of the individual resistors that are connected in
parallel. If the two or more resistors found in the parallel branches do not have equal resistance,
then the above formula must be used. An example of this method was presented in a previous
section of Lesson 4.
By applying one's understanding of the equivalent resistance of parallel branches to a
combination circuit, the combination circuit can be transformed into a series circuit. Then an
understanding of the equivalent resistance of a series circuit can be used to determine the total
resistance of the circuit. Consider the following diagrams below. Diagram A represents a
combination circuit with resistors R2 and R3 placed in parallel branches. Two 4- resistors in
parallel is equivalent to a resistance of 2 . Thus, the two branches can be replaced by a single
resistor with a resistance of 2 . This is shown in Diagram B. Now that all resistors are in series,
the formula for the total resistance of series resistors can be used to determine the total resistance
of this circuit: The formula for series resistance is
Rtot = R1 + R2 + R3 + ...
So in Diagram B, the total resistance of the circuit is 10 .
Once the total resistance of the circuit is determined, the analysis continues using Ohm's law and
voltage and resistance values to determine current values at various locations. The entire method
is illustrated below with two examples.
Example 1:
The first example is the easiest case - the resistors placed in parallel have the same resistance.
The goal of the analysis is to determine the current in and the voltage drop across each resistor.
As discussed above, the first step is to simplify the circuit by replacing the two parallel resistors
with a single resistor that has an equivalent resistance. Two 8 resistors in series is equivalent to
a single 4 resistor. Thus, the two branch resistors (R2 and R3) can be replaced by a single
resistor with a resistance of 4 . This 4 resistor is in series with R1 and R4. Thus, the total
resistance is
Rtot = R1 + 4 + R4 = 5 + 4 + 6
Rtot = 15
Now the Ohm's law equation ( V = I • R) can be used to determine the total current in the
circuit. In doing so, the total resistance and the total voltage (or battery voltage) will have to be
used.
Itot = Vtot / Rtot = (60 V) / (15 )
Itot = 4 Amp
The 4 Amp current calculation represents the current at the battery location. Yet, resistors R1 and
R4 are in series and the current in series-connected resistors is everywhere the same. Thus,
Itot = I1 = I4 = 4 Amp
For parallel branches, the sum of the current in each individual branch is equal to the current
outside the branches. Thus, I2 + I3 must equal 4 Amp. There are an infinite number of possible
values of I2 and I3 that satisfy this equation. Since the resistance values are equal, the current
values in these two resistors are also equal. Therefore, the current in resistors 2 and 3 are both
equal to 2 Amp.
I2 = I3 = 2 Amp
Now that the current at each individual resistor location is known, the Ohm's law equation ( V =
I • R) can be used to determine the voltage drop across each resistor. These calculations are
shown below.
V1 = I1 • R1 =
A p •
)
V1 = 20 V
V2 = I2 • R2 = (2 Amp) • (8 )
V2 = 16 V
V3 = I3 • R3 = (2 Amp) • (8 )
V3 = 16 V
V4 = I4 • R4 = (4 Amp) • (6 )
V4 = 24 V
The analysis is now complete and the results are summarized in the diagram below.
Example 2:
The second example is the more difficult case - the resistors placed in parallel have a different
resistance value. The goal of the analysis is the same - to determine the current in and the voltage
drop across each resistor.
As discussed above, the first step is to simplify the circuit by replacing the two parallel resistors
with a single resistor with an equivalent resistance. The equivalent resistance of a 4- and 12resistor placed in parallel can be determined using the usual formula for equivalent resistance of
parallel branches:
1 / Req = 1 / R1 + 1 / R2 + 1 / R3 ...
1 / Req = 1 / (4 ) + 1 / (12 )
1 / Req = 0.333
-1
Req = 1 / (0.333
-1
Req = 3.00
)
Based on this calculation, it can be said that the two branch resistors (R2 and R3) can be replaced
by a single resistor with a resistance of 3 . This 3 resistor is in series with R1 and R4. Thus,
the total resistance is
Rtot = R1 + 3 + R4 = 5 + 3 + 8
Rtot = 16
Now the Ohm's law equation ( V = I • R) can be used to determine the total current in the
circuit. In doing so, the total resistance and the total voltage (or battery voltage) will have to be
used.
Itot = Vtot / Rtot = (24 V) / (16 )
Itot = 1.5 Amp
The 1.5 Amp current calculation represents the current at the battery location. Yet, resistors R1
and R4 are in series and the current in series-connected resistors is everywhere the same. Thus,
Itot = I1 = I4 = 1.5 Amp
For parallel branches, the sum of the current in each individual branch is equal to the current
outside the branches. Thus, I2 + I3 must equal 1.5 Amp. There are an infinite possibilities of I2
and I3 values that satisfy this equation. In the previous example, the two resistors in parallel had
the identical resistance; thus the current was distributed equally among the two branches. In this
example, the unequal current in the two resistors complicates the analysis. The branch with the
least resistance will have the greatest current. Determining the amount of current will demand
that we use the Ohm's law equation. But to use it, the voltage drop across the branches must first
be known. So the direction that the solution takes in this example will be slightly different than
that of the simpler case illustrated in the previous example.
To determine the voltage drop across the parallel branches, the voltage drop across the two
series-connected resistors (R1 and R4) must first be determined. The Ohm's law equation ( V = I
• R) can be used to determine the voltage drop across each resistor. These calculations are shown
below.
V1 = I1 • R1 = (1.5 Amp) •
)
V1 = 7.5 V
V4 = I4 • R4 = (1.5 Amp) • (8 )
V4 = 12 V
This circuit is powered by a 24-volt source. Thus, the cumulative voltage drop of a charge
traversing a loop about the circuit is 24 volts. There will be a 19.5 V drop (7.5 V + 12 V)
resulting from passage through the two series-connected resistors (R1 and R4). The voltage drop
across the branches must be 4.5 volts to make up the difference between the 24 volt total and the
19.5-volt drop across R1 and R4. Thus,
V2 = V3 = 4.5 V
Knowing the voltage drop across the parallel-connected resistors (R1 and R4) allows one to use
the Ohm's law equation ( V = I • R) to determine the current in the two branches.
I2 = V2 / R2 = (4.5 V) / (4 )
I2 = 1.125 A
I3 = V3 / R3 = (4.5 V) / (12 )
I3 = 0.375 A
The analysis is now complete and the results are summarized in the diagram below.
Developing a Strategy
The two examples above illustrate an effective concept-centered strategy for analyzing
combination circuits. The approach demanded a firm grasp of the series and parallel concepts
discussed earlier. Such analyses are often conducted in order to solve a physics problem for a
specified unknown. In such situations, the unknown typically varies from problem to problem. In
one problem, the resistor values may be given and the current in all the branches are the
unknown. In another problem, the current in the battery and a few resistor values may be stated
and the unknown quantity becomes the resistance of one of the resistors. Different problem
situations will obviously require slight alterations in the approaches. Nonetheless, every
problem-solving approach will utilize the same principles utilized in approaching the two
example problems above.
The following suggestions for approaching combination circuit problems are offered to the
beginning student:





If a schematic diagram is not provided, take the time to construct one. Use schematic symbols
such as those shown in the example above.
When approaching a problem involving a combination circuit, take the time to organize yourself,
writing down known values and equating them with a symbol such as Itot, I1, R3, V2, etc. The
organization scheme used in the two examples above is an effective starting point.
Know and use the appropriate formulae for the equivalent resistance of series-connected and
parallel-connected resistors. Use of the wrong formulae will guarantee failure.
Transform a combination circuit into a strictly series circuit by replacing (in your mind) the
parallel section with a single resistor having a resistance value equal to the equivalent resistance
of the parallel section.
Use the Ohm's law equation ( V = I • R ofte a d appropriately. Most a swers will be
determined using this equation. When using it, it is important to substitute the appropriate
values into the equation. For instance, if calculating I2, it is important to substitute the V2 and
the R2 values into the equation.
For further practice analyzing combination circuits, consider analyzing the problems in the
Check Your Understanding section below.
Check Your Understanding
1. A combination circuit is shown in the diagram
at the right. Use the diagram to answer the
following questions.
a. The current at location A is _____ (greater than,
equal to, less than) the current at location B.
b. The current at location B is _____ (greater than,
equal to, less than) the current at location E.
c. The current at location G is _____ (greater than,
equal to, less than) the current at location F.
d. The current at location E is _____ (greater than, equal to, less than) the current at location G.
e. The current at location B is _____ (greater than, equal to, less than) the current at location F.
f. The current at location A is _____ (greater than, equal to, less than) the current at location L.
f. The current at location H is _____ (greater than, equal to, less than) the current at location I.
2. Consider the combination circuit in the diagram
at the right. Use the diagram to answer the
following questions. (Assume that the voltage
drops in the wires themselves in negligibly small.)
a. The electric potential difference (voltage drop)
between points B and C is _____ (greater than,
equal to, less than) the electric potential difference
(voltage drop) between points J and K.
b. The electric potential difference (voltage drop)
between points B and K is _____ (greater than,
equal to, less than) the electric potential difference
(voltage drop) between points D and I.
c. The electric potential difference (voltage drop) between points E and F is _____ (greater than,
equal to, less than) the electric potential difference (voltage drop) between points G and H.
d. The electric potential difference (voltage drop) between points E and F is _____ (greater than,
equal to, less than) the electric potential difference (voltage drop) between points D and I.
e. The electric potential difference (voltage drop) between points J and K is _____ (greater than,
equal to, less than) the electric potential difference (voltage drop) between points D and I.
f. The electric potential difference between points L and A is _____ (greater than, equal to, less
than) the electric potential difference (voltage drop) between points B and K.
3. Use the concept of equivalent resistance to determine the unknown resistance of the identified
resistor that would make the circuits equivalent.
4. Analyze the following circuit and determine the values of the total resistance, total current,
and the current at and voltage drops across each individual resistor.
5. Referring to the diagram in question #4, determine the ...
a. ... power rating of resistor 4.
b. ... rate at which energy is consumed by resistor 3.
Physb.pdf