Introduction to Waves - Florence High School

advertisement
Introduction to Waves:
Transverse and Longitudinal
Physics
Coach Stephens
Definitions
• ..\Wave Terms and Snakey Spring KEY.pdf
Parts of a Wave
• http://www.youtube.com/watch?feature=play
er_detailpage&v=Z3O2Ju3ULvo#t=0s
Making Waves Using a Slinky
• http://www.youtube.com/watch?v=vJ8pnFhHl8&edufilter=3vKYiNm3XpGng5TQ3-qEBw
• http://www.youtube.com/watch?v=Rbuhdo0A
ZDU&edufilter=3vKYiNm3XpGng5TQ3-qEBw
Calculating Speed, Frequency &
Wavelength
Physics
Coach Stephens
Measuring the Speed of Waves
• ..\Measuring the Speed of Waves.pdf
• ..\Traveling Wave Problems.pdf
• Answer Keys:
– ..\Measuring the Speed of Waves KEY.pdf
– ..\Traveling Wave Problems KEY.pdf
Sound Waves & Sources
Physics
Coach Stephens
Bell Work
• What causes waves?
• What are some properties of waves?
Bell Work Answer
• Oscillations and vibrating objects
• frequency, speed, wavelength, vibrating
source
• transverse and longitudinal (electromagnetic,
sound, and water waves)
Simulations & Videos
• http://phet.colorado.edu/new/simulations/sims.
php?sim=Wave_on_a_String
• http://phet.colorado.edu/new/simulations/sims.
php?sim=Sound
• http://phet.colorado.edu/new/simulations/sims.
php?sim=Fourier_Making_Waves
• http://phet.colorado.edu/new/simulations/sims.
php?sim=Radio_Waves_and_Electromagnetic_Fi
elds.
Vibrations
Vibrational Motion
•
•
•
•
•
•
Wiggling
Back and forth
Vibrating
Shaking
Oscillating
These phrases describe the motion of a variety of
objects. They even describe the motion of matter
at the atomic level. Even atoms wiggle - they do
the back and forth.
Example of Vibrational Motion
• Bobblehead Doll -– A bobblehead doll consists of an oversized
replica of a person's head attached by a spring
to a body and a stand. A light tap causes it to
bobble. The head wiggles; it vibrates; it
oscillates. The back and forth doesn't happen
forever. Over time, the vibrations tend to die off
and the bobblehead stops bobbing and finally
assumes its usual resting position.
– This usual resting position is referred to as the
equilibrium position.
• This means that all forces acting on the object are
balanced.
Forced Vibration
• If a force is applied to the bobblehead, the equilibrium will be disturbed
and it will begin vibrating. We could use the phrase forced vibration to
describe the force which sets the resting bobblehead into motion.
– A short-lived, momentary force begins the motion.
– Each repetition of its motion is a little less vigorous than its previous
repetition.
– The extent of its displacement from the equilibrium position becomes less
and less over time.
– Because the forced vibration that initiated the motion is a single instance
of a short-lived, momentary force, the vibrations ultimately cease.
• The bobblehead is said to experience damping.
– Damping is the tendency of a vibrating object to lose or to dissipate its
energy over time.
– Without a sustained forced vibration, the motion of the bobblehead
eventually ceases as energy is dissipated to other objects.
• A sustained input of energy would be required to keep the back and forth motion
going.
• `After all, if the vibrating object naturally loses energy, then it must continuously be
put back into the system through a forced vibration in order to sustain the vibration.
The Restoring Force
• Why doesn't the bobblehead stop the first time it returns to the
equilibrium position?
– According to Newton's law of inertia, “an object in motion will
remain in motion and an object at rest will remain at rest unless
acted upon by an unbalanced force.”
• An object which is moving will continue its motion if the forces are
balanced.
– So every instant in time that the bobblehead is at the equilibrium
position, the momentary balance of forces will not stop the motion.
• It moves past the equilibrium position towards the opposite side of its
swing.
– As the bobblehead is displaced past its equilibrium position, then a
force capable of slowing it down and stopping it exists.
• This force that slows the bobblehead down as it moves away from its
equilibrium position is known as a restoring force.
• The restoring force acts upon the vibrating object to move it back to its
original equilibrium position.
Vibrational vs. Translational Motion
• Vibrational motion is often contrasted with translational
motion.
– In translational motion, an object is permanently displaced.
• The initial force that is imparted to the object displaces it from its
resting position and sets it into motion.
• Yet because there is no restoring force, the object continues the
motion in its original direction.
• When an object vibrates, it doesn't move permanently out
of position.
– The restoring force acts to slow it down, change its direction
and force it back to its original equilibrium position.
• An object in translational motion is permanently displaced
from its original position.
• But an object in vibrational motion wiggles about a fixed
position - its original equilibrium position.
Other Examples of Vibrational Motion
• Bobblehead dolls are not the only objects that vibrate. It might be safe
to say that all objects in one way or another can be forced to vibrate to
some extent. As long as a force persists to restore the object to its
original position, a displacement from its resting position will result in a
vibration.
• A pendulum is a classic example of an object that is considered to
vibrate. A simple pendulum consists of a relatively massive object hung
by a string from a fixed support. It typically hangs vertically in its
equilibrium position. When the mass is displaced from equilibrium, it
begins its back and forth vibration about its fixed equilibrium position.
• An inverted pendulum (i.e. trees, skyscrapers, tennis ball on a dowel rod
and support, tuning fork) is another example of vibrational motion.
• Another example is a mass on a spring. The mass hangs at a resting
position. If the mass is pulled down, the spring is stretched. Once the
mass is released, it begins to vibrate. It does the back and forth,
vibrating about a fixed position. (i.e. springs inside of a mattress, car
suspension systems, bathroom scales)
Animations and Pictures
Properties of Periodic Motion
• A vibrating object is wiggling about a fixed position.
– Like the mass on a spring, a vibrating object is moving over the same
path over the course of time.
– Its motion repeats itself over and over again.
– If it were not for damping, the vibrations would endure forever (or
at least until someone catches the mass and brings it to rest).
• The mass on the spring not only repeats the same motion, it
does so in a regular fashion.
– The time it takes to complete one back and forth cycle is always the
same amount of time.
– If it takes the mass 3.2 seconds to complete the first back and forth
cycle, then it will take 3.2 seconds to complete the seventh back and
forth cycle.
• In Physics, a motion that is regular and repeating is referred to
as a periodic motion.
– Most objects that vibrate do so in a regular and repeated fashion;
their vibrations are periodic.
The Sinusoidal Nature of a Vibration
• Suppose that a motion detector was placed below a vibrating mass on a
spring in order to detect the changes in the mass's position over the
course of time. And suppose that the data from the motion detector
could represent the motion of the mass by a position vs. time plot. The
graphic below depicts such a graph.
• One obvious characteristic of the graph has to do with its shape. Many
students recognize the shape of this graph from experiences in
Mathematics class. The graph has the shape of a sine wave.
• A second obvious characteristic of the graph may be its periodic nature.
The motion repeats itself in a regular fashion.
• A third obvious characteristic of the graph is that damping occurs with
the mass-spring system. Some energy is being dissipated over the
course of time.
Language Used to Describe the Graph
• By looking at the graph, what does the motion of the wave appear to
be doing?
• If you said “slowing down” you are incorrect. As shown in the chart
below, it took 2.3 seconds to complete the first cycle and 2.3 seconds
to complete the sixth cycle.
Cycle
Letters
Times at Beginning and End of Cycle (s)
Cycle Time (s)
1st
A to E
0.0 sto 2.3 s
2.3
2nd
E tp I
2.3 s to 4.6 s
2.3
3rd
I to M
4.6 s to 7.0 s
2.4
4th
M to Q
7.0 s to 9.3 s
2.3
5th
Q to U
9.3 s to 11.6 s
2.3
6th
U to Y
11.6 s to 13.9 s
2.3
Continued…
•
•
•
•
•
•
The mass will both speed up and slow down over the course of a single cycle. So to say
that the mass is "slowing down" is not entirely accurate since during every cycle there
are two short intervals during which it speeds up.
The time to complete one cycle of vibration is NOT changing.
The extent to which the mass moves above or below the resting position varies over
the course of time.
In the first full cycle of vibration being shown, the mass moves from its resting position
(A) 0.60 m above the motion detector to a high position (B) of 0.99 m cm above the
motion detector. This is a total upward displacement of 0.29 m.
In the sixth full cycle of vibration that is shown, the mass moves from its resting
position (U) 0.60 m above the motion detector to a high position (V) 0.94 m above the
motion detector. This is a total upward displacement of 0.24 m cm.
The table below summarizes displacement measurements for several other cycles
displayed on the graph.
Cycle
Letters Maximum Upward Displacement Maximum Downward Displacement
1st
A to E
0.60 m to 0.99 m
0.60 m to 0.21 m
2nd
E to I
0.60 m to 0.98 m
0.60 m to 0.22 m
3rd
I to M
0.60 m to 0.97 m
0.60 m to 0.23 m
6th
U to Y
0.60 m to 0.94 m
0.60 m to 0.26 m
Continued…
• Over the course of time, the mass continues to vibrate.
– However, the amount of displacement of the mass is decreasing from one
cycle to the next.
– This illustrates that energy is being lost from the mass-spring system.
– If given enough time, the vibration of the mass will eventually cease as its
energy is dissipated.
• In physics (or at least in the English language), "slowing down" means to
"get slower" or to "lose speed".
– Speed, a physics term, refers to how fast or how slow an object is moving.
– To say that the mass on the spring is "slowing down" over time is to say
that its speed is decreasing over time.
– But as mentioned, the mass speeds up during two intervals of every cycle - as the restoring force pulls the mass back towards its resting position the
mass speeds up.
• For this reason, a physicist adopts a different language to communicate
the idea that the vibrations are "dying out".
– We use the phrase "energy is being dissipated or lost" instead of saying the
"mass is slowing down.“
Period & Amplitude
• The key measurements that have been made are
measurements of:
– Period -- the time for the mass to complete a cycle, and
– Amplitude -- the maximum displacement of the mass above
(or below) the resting position.
• An object in periodic motion can have a long period or a
short period.
– The terms fast and slow are not used since physics types
reserve the words fast and slow to refer to an object's speed.
– Instead, we use frequent or infrequent.
– Here in this description we are referring to the frequency, not
the speed.
– An object can be in periodic motion and have a low frequency
and a high speed.
Frequency
• Frequency is another quantity that can be used
to describe the motion of an object in periodic
motion.
– The frequency is defined as the number of complete
cycles occurring per period of time.
– Since the standard metric unit of time is the second,
frequency has units of cycles/second.
– The unit cycles/second is equivalent to the unit Hertz
(abbreviated Hz).
• The unit Hertz is used in honor of Heinrich Rudolf Hertz, a
19th century physicist who expanded our understanding of
the electromagnetic theory of light waves.
How Often Something Occurs
• Frequency is a word we often use to describe how often something occurs.
– You might say that you frequently check your Facebook – you check it often.
– In physics, frequency is used with the same meaning - it indicates how often a
repeated event occurs.
• High frequency events that are periodic occur often, with little time in
between each occurrence - like the back and forth vibrations of the tines of a
tuning fork.
– The vibrations are so frequent that they can't be seen with the naked eye.
– A 256-Hz tuning fork has tines that make 256 complete back and forth vibrations
each second. At this frequency, it only takes the tines about 0.00391 seconds to
complete one cycle.
– A 512-Hz tuning fork has an even higher frequency. Its vibrations occur more
frequently; the time for a full cycle to be completed is 0.00195 seconds.
– In comparing these two tuning forks, it is obvious that the tuning fork with the
highest frequency has the lowest period.
• The two quantities frequency and period are inversely related to each other.
CYU #1
• According to Wikipedia, Tim Ahlstrom of
Wisconsin holds the record for hand clapping.
He is reported to have clapped his hands 793
times in 60.0 seconds.
• What is the frequency and what is the period
of Mr. Ahlstrom's hand clapping during this
60.0-second period?
Answer #1
• In this problem, the event that is repeating itself is the clapping of
hands; one hand clap is equivalent to a cycle.
• The frequency can be thought of as the number of cycles per second.
Calculating frequency involves dividing the stated number of cycles by
the corresponding amount of time required to complete these cycles. In
contrast, the period is the time to complete a cycle. Period is calculated
by dividing the given time by the number of cycles completed in this
amount of time.
• Frequency = cycles per second
= 793 cycles/60.0 seconds
= 13.2 cycles/s
= 13.2 Hz
• Period = seconds per cycle
= 60.0 s/793 cycles
= 0.0757 seconds
CYU #2
• A pendulum is observed to complete 23 full
cycles in 58 seconds. Determine the period
and the frequency of the pendulum.
Answer #2
• The frequency can be thought of as the number of cycles per
second. Calculating frequency involves dividing the stated
number of cycles by the corresponding amount of time
required to complete these cycles. In contrast, the period is the
time to complete a cycle. Period is calculated by dividing the
given time by the number of cycles completed in this amount
of time.
• frequency = 23 cycles/58 seconds
= 0.39655 Hz
= ~0.40 Hz
• period = 58 seconds/23 cycles
= 2.5217 sec
= ~2.5 s
CYU #3
• A mass is tied to a spring and begins vibrating
periodically. The distance between its highest
and its lowest position is 38 cm. What is the
amplitude of the vibrations?
Answer #3
• Answer: 19 cm
• The distance that is described is the distance
from the high position to the low position. The
amplitude is from the middle position to
either the high or the low position.
• So just divide the total distance by 2.
CYU #4
• 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.
Answer #4
• Answer: 0.0195 sec
• Here is an example of a problem with a lot of
extraneous information. The period is simply
the reciprocal of the frequency. In this case,
the period is 1/(51.2 Hz) which is 0.0195
seconds.
• Know your physics concepts to weed through
the extra information.
CYU #5
• Frieda the fly flaps its wings back and forth
121 times each second. The period of the
wing flapping is ____ sec.
Answer #5
• Answer: 0.00826 seconds
• The quantity 121 times/second is the
frequency. The period is the reciprocal of the
frequency.
• T=1/(121 Hz) = 0.00826 s
Nature of a Wave
• The nature of a wave was discussed in a previous
lesson 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
introduces a wave into the slinky.
Frequency Review
• 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, in turn would vibrate at a rate of 2 cycles/second.
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.
• If a coil of a 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).
Energy Transport and the
Amplitude of a Wave
Energy Transport
• 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.
Amplitude
•
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.
•
The logic underlying the energy-amplitude relationship is as follows: If you send a
transverse pulse into the first coil of a slinky, that coil is given an initial amount of
displacement.
– The displacement is due to the force applied by the person.
– 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.
Inertial & Elastic Factors
• 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?
– In a situation such as this, the amplitude is dependent upon
two types of factors: an inertial factor and an elastic factor.
– Two different materials have different mass densities.
• 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 and allow a
greater amplitude to travel through it.
Energy-Amplitude Relationship
• 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.
• 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.
CYU #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.
Answer #1
• Amplitude = 16 cm
– (Amplitude is the distance from the rest position to the
crest position which is half the vertical distance from a
trough to a crest.)
• Wavelength = 96 cm
– (Wavelength is the distance from crest to crest, which is
twice the horizontal distance from crest to nearest
trough.)
• Period = 0.42 s
– (The period is the reciprocal of the frequency. T = 1 / f)
CYU #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
Answer #2
• Answer: C (quadrupled)
• The energy transported by a wave is directly
proportional to the square of the amplitude.
So whatever change occurs in the amplitude,
the square of that affect impacts the energy.
This means that a doubling of the amplitude
results in a quadrupling of the energy.
CYU #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
Answer #3
• Answer: D (four times larger)
• The energy transported by a wave is directly
proportional to the square of the amplitude.
So whatever change occurs in the amplitude,
the square of that affect impacts the energy.
This means that a doubling of the amplitude
results in a quadrupling of the energy.
The Speed of a Wave
• A wave is a disturbance that moves along a medium from
one end to the other.
• If you watch an ocean wave moving along the medium
(the ocean water), you 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:
Changing Mediums
• 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.
Echoes
• Reflection phenomena are commonly observed
with sound waves.
• When you yell within a canyon, you often hear the
echo of the yell.
– 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 yell.
• 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?
Answer
• 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?
• 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 in the table on
the next page.
Speed of a Wave Lab – Sample Data
Trial
Tension (N)
Frequency (Hz)
Wavelength (m)
Speed (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
Change in Frequency
• 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.
Change in Rope Tension
• 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
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.
Wave vs. Medium
• 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.
CYU #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?
Answer #1
• They reach the wall at the same time. Don't
be fooled! The amplitude of a wave does not
affect the speed at which the wave travels.
Both Wave A and Wave B travel at the same
speed. The speed of a wave is only altered by
alterations in the properties of the medium
through which it travels.
CYU #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 ?
Answer #2
• They reach the wall at the same time. Don't
be fooled! The wavelength of a wave does not
affect the speed at which the wave travels.
Both Wave C and Wave D travel at the same
speed. The speed of a wave is only altered by
alterations in the properties of the medium
through which it travels.
CYU #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
b. 0.059 second
c. 0.59 second
d. 2.9 second
Answer #3
•
•
•
•
•
Answer: B
GIVEN: v = 340 m/s, d = 20 m and f = 1000 Hz
Find time
Use v = d / t and rearrange to t = d / v
Substitute and solve.
CYU #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
b. one-third
c. the same as
d. three times larger than
Answer #4
• Answer: C
• The medium is the same for both of these
waves ("the same container of nitrogen gas").
Thus, the speed of the wave will be the same.
Alterations in a property of a wave (such as
wavelength) will not affect the speed of the
wave. Two different waves travel with the
same speed when present in the same
medium.
CYU #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?
Answer #5
• GIVEN: v = 340 m/s, t = 0.150 s (down and
back time)
• Find d (1-way)
• If it takes 0.150 s to travel to the object and
back, then it takes 0.075 s to travel the oneway distance to the object. Now solve for time
using the equation d = v • t.
• d = v • t = (340 m/s) • (0.075 s) = 25.5 m
CYU #6
• TRUE or FALSE:
• Doubling the frequency of a wave source
doubles the speed of the waves.
Answer #6
• FALSE!
• The speed of a wave is unaffected by changes
in the frequency.
CYU #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.
Answer #7
• GIVEN: v = 342 m/s, t = 0.82 s (2-way)
• Find d (1-way)
• If it takes 0.82 s to travel to the canyon wall
and back (a down-and-back time), then it
takes 0.41 s to travel the one-way distance to
the wall. Now use d = v • t
• d = v • t = (342 m/s) • (0.41 s) = 140 m
CYU #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.
Answer #8
• GIVEN: d (1-way) =25 m, t (2-way)=25 s
• Find v.
• If the pool is 25 meters long, then the backand-forth distance is 50 meters. The wave
covers this distance in 25 seconds. Now use v
= d / t.
• v = d / t = (50 m) / (25 s) = 2 m/s
CYU #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.
Answer #9
• If the wave travels 2.5 meters in one second
then it will travel 5.0 meters in 2.0 seconds. If
Wilbert gets drenched every time the wave
has traveled 5.0 meters, then he will get
drenched every 2.0 seconds.
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.
Production of a Wave
• 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 one-fourth 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 backand-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.
Speed Equation
• 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•
Check Your Understanding
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
Wavelength
Frequency
Zinc, 1-in. dia. coils
1.75 m
2.0 Hz
Zinc, 1-in. dia. coils
0.90 m
3.9 Hz
Copper, 1-in. dia.
coils
1.19 m
2.1 Hz
Copper, 1-in. dia.
coils
0.60 m
4.2 Hz
Zinc, 3-in. dia. coils
0.95 m
2.2 Hz
Zinc, 3-in. dia. coils
1.82 m
1.2 Hz
Speed
Answers
• Multiply the frequency by the wavelength to
determine the speed.
•
•
•
•
•
Row 1: speed = 3.5 m/s Row 2: speed = 3.5 m/s
Row 3: speed = 2.5 m/s
Row 4: speed = 2.5 m/s
Row 5: speed = 2.1 m/s
Row 6: speed = 2.2 m/s
CYU #2
• As the wavelength of a wave in a uniform
medium increases, its speed will _____.
• a. decrease
• b. increase
• c. remain the same
Answer #2
• Answer: C
• In rows 1 and 2, the wavelength was altered
but the speed remained the same. The same
can be said about rows 3 and 4 and rows 5
and 6. The speed of a wave is not affected by
the wavelength of the wave.
CYU #3
• As the wavelength of a wave in a uniform
medium increases, its frequency will _____.
• a. decrease
• b. increase
• c. remain the same
Answer #3
• Answer: A
• In rows 1 and 2, the wavelength was increased
and the frequency was decreased. Wavelength
and frequency are inversely proportional to
each other.
CYU #4
• 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.
Answer #4
• Answer: A
• Whenever the medium is the same, the speed
of the wave is the same. However, when the
medium changes, the speed changes. The
speed of these waves were dependent upon
the properties of the medium.
CYU #5
• 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
Answer #5
• Answer: B
• Frequency and wavelength are inversely
proportional to each other. The wave with the
greatest frequency has the shortest
wavelength. Twice the frequency means onehalf the wavelength. For this reason, the
wavelength ratio is the inverse of the
frequency ratio.
CYU #6
• 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.
Answer #6
• Answer: A
• Doubling the frequency will not alter the wave
speed. Rather, it will halve the wavelength.
Wavelength and frequency are inversely
related.
CYU #7
• 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?
Answer #7
• Answer: f = 70 Hz and wavelength = 5.0 m
• The frequency is given and the wavelength is
the v/f ratio.
CYU #8
• 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.
Answer #8
• The wavelength is 8.6 meters and the period is 6.2
seconds.
• The frequency can be determined from the period. If T =
6.2 s, then
• f =1 /T = 1 / (6.2 s) f = 0.161 Hz
• Now find speed using the v = f • wavelength equation.
• v = f • wavelength = (0.161 Hz) • (8.6 m) v = 1.4 m/s
CYU #9
• 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.
Answer #9
• The diagram is helpful. The wavelength must
be 8 meters (see diagram).
• The period is 3 seconds so the frequency is 1 /
T or 0.333 Hz.
• Now use speed = f • wavelength Substituting
and solving for v, you will get 2.67 m/s.
Download