UNIT 8: OSCILLATIONS, WAVES AND SOUND
Relationship to Labs: This unit is supported by Lab 7: The Simple Pendulum (kit-based lab). It may also
be supported by Lab 8: Cathode Ray Oscilloscope and the Speed of Sound (RWSL).
HOOKE’S LAW
When you push on a spring, it pushes back. When you stretch a spring, it pulls back towards its unstretched position. When the spring is un-stretched or un-compressed, it is in what is called the
equilibrium position.
F=0
F
F
equilibrium
position
If we will define x = 0 at the equilibrium position, Hooke’s Law states that the force due to the spring is
Fx  kx
where k is the spring constant. This is the restoring force because it is the force that restores the string
to its equilibrium position. The negative sign indicates that the restoring force is always opposite the
displacement from equilibrium. Also note that the restoring force gets larger as the object gets further
from the equilibrium position.
We can use Newton’s 2nd Law to find the acceleration of a mass m attached to the end of the spring
(spring usually has no mass for introductory physics course).
max  kx
ax  
k
x
m
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or
ax  cx
Using calculus, this becomes
d 2x
k
 x
2
dt
m
or
d 2x
 cx
dt 2
where c is a constant. Here the second derivative of the displacement with respect to time (the
acceleration) is proportional to the displacement and the second derivative is in the direction opposite
that of the displacement.
Systems with the relations between the restoring force and the displacement will undergo simple
harmonic motion (SHM) when displaced from equilibrium. A mass attached to a spring that is displaced
and released will oscillate back and forth until friction forces dissipate this motion.
SHM shows up all over the place. This is very convenient because, once we know how to deal with one
system under SHM, we can apply the basic principles to any system undergoing SHM (variable names
may change but the form is the same). This is so useful that, whenever you see an equation of the form
d 2x
 cx ,
dt 2
a  cx
or
Fx  kx ,
you can think “this is SHM”, and apply your SMH relations to the new case.
Consider the function
x  sin

c t


c t

c  t  cx
v
dx
 c cos
dt
a
dv
 c sin
dt

first derivative

second derivative
We can get the same result with x  cos


c t .
It turns out that any object undergoing SHM can have its displacement described by a sine function,
cosine function or combination of the two as a function of the time.
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In general, the displacement is defined using the wave equation
x  A cost   
where ω is the angular speed or angular frequency, A is the amplitude and  is the phase (also phase
angle or phase constant). We will use cosine because mass-spring systems and pendulums usually start
at the maximum displacement from equilibrium and cos t  is at a maximum when t  0 . In reality,
we can start timing (t = 0) whenever we want so we allow for variation in start time with the phase
angle.
The wave equation is a sine function if we let the phase equal –π/2. This is a consequence of the
trigonometric relationship


cos  t    sin t 
2

Because the sine and cosine function have values between –1 and +1, this function will have values
between –A and +A. The amplitude describes the range of x values, the size of the oscillation.
This is the position-time curve for an object oscillating back and forth along the x-axis. Sample wave
equations with different amplitudes and phase angles are shown.
4
x
x  3 cos t
3
x  cost  0.5
2
x  cos t
1
t
0
-1
-1 0
1
2
3
4
5
6
7
8
9
10
-2
-3
-4
The basic curve is shown in blue. If we increase the amplitude, we get the red line. If we change the
phase the curve shifts to the left or right in time.
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The angular frequency is a measure of how rapidly the oscillations occur. The units of angular frequency
are s-1 or radians per second. The plot below shows samples.
x
1.5
x  cost / 2
x  cos2t 
x  cos t
1
0.5
t
0
-1
-0.5
0
1
2
3
4
5
6
7
8
9
10
-1
-1.5
Once again, the basic curve is in blue. The dotted cyan curve is for the oscillation with twice the
frequency as the basic curve. The dashed purple curve is for an oscillation with half the frequency. All
curves have the same magnitude and phase.
The first derivative of x  A cost    is
dx
  A sin t   
dt
The second derivative is
d 2x
  A 2 cost   
2
dt
The substituting the original equation on the left, second derivative is
d 2x
  2 x
dt 2
or
a   2 x
This has the same form as the SHM equation we had for the spring motion
d 2x
k
 x
2
dt
m
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The two equations are identical if we let
2 
k
m

or
k
m
Notice that the angular frequency is not proportional to the amplitude. It doesn’t matter how much
the spring is stretched, the angular frequency of the oscillation is constant.
The period of an oscillation is the time that it takes to complete one full oscillation. Periods for the basic
curve and the one with twice the frequency are shown in the position-time curve below.
x
1.5
T1
1
0.5
t
0
-1
-0.5
0
1
2
3
4
5
6
7
8
9
10
-1
T2
-1.5
The period is related to the angular frequency by
T
2

 2
m
k
Period is measured in seconds.
Continuing with the definitions, we can also define the frequency f is the number of oscillations the
object undergoes per unit time interval. It is also the inverse of the period.
f 
1
1

T 2
k
m
The units of frequency are cycles per second, or Hertz (Hz). Like radians, cycles have are unit-free. In
cases where the motion is due to rotation, we can define frequency in revolutions per second or
revolutions per minute (RPM). The angular frequency is a measure of the angle (in radians) the
oscillating object goes through in 1 s. One cycle for a sine or cosine curve is equivalent to 2 radians.
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The frequency f is then related to the angular frequency by
f 

2

or
k
m
Problem 1
Find the angular frequency, frequency and period of the simple harmonic motion of a 1.0 kg mass
attached to an ideal spring with spring constant 6.4 N/m if the amplitude of motion is 2.5 cm.
Let’s look at displacement and its derivatives a little closer. We had the displacement as
x  A cost   
The position-time graph shows the simple case x  cos  2t  , A  1 ,   2 and   0 .
-1
5 x
4
3
2
1
0
-1 0
-2
-3
-4
-5
x  cos( 2t )
1
2
3
4
5
6
7
8
9
10
t
As noted earlier, the maximum and minimum values of the displacement are ±A, ±1 in this case.
In the kinematics section, we saw that the instantaneous velocity at any time is the slope of the curve at
that time. We can use calculus to show that
vx  dx
dt
  A sin t   
or
vx   A sin t   
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Notice that the amplitude of the velocity curve is  A . The velocity-time curve for the simple case is
shown below. The maximum and minimum values of the velocity are ±ωA, ±2 in this case. Also note
that the curve is shifted to the left by  / 2   / 4 .
5
4
3
vx  2 sin( 2t )
2
1
0
-1
-1
0
1
2
3
4
5
6
7
8
9
10
-2
-3
-4
-5
Kinematics also showed us that the instantaneous acceleration is the slope of the velocity-time curve.
We can verify this using calculus
2
ax  d x
dt 2
  A 2 sin t   
or
ax   A 2 cos t   
The amplitude of the acceleration-time curve is A 2 . The acceleration-time curve for the simple case
is shown below. The maximum and minimum values of the acceleration are ±ωA2, ±4 in this case. Also
note that the curve is shifted to the left by another  / 2   / 4 .
-1
5
4
3
2
1
0
-1 0
-2
-3
-4
-5
ax  4 cos( 2t )
1
2
3
4
5
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7
8
9
10
7
Plotting displacement, velocity and acceleration together, we see that the maximum or minimum
displacement occurs when the velocity crosses zero. The max/min displacement occurs at the same
time as the min/max acceleration. We can also see that the maximum or minimum velocity occurs
when the acceleration and displacement are zero.
-1
5
4
3
2
1
0
-1 0
-2
-3
-4
-5
1
2
3
4
5
6
7
8
9
max/min velocity occurs when both
acceleration and displacement are
max/min displacement and min/max
acceleration occur when velocity is
zero
zero
10
Problem 2
A 3.0 kg block is connected to a spring with an elastic constant of 12 N/m. If the block is initially
displaced 2.0 m from the equilibrium position and released from rest:
a)
calculate the amplitude, period and angular frequency;
b)
calculate the displacement, velocity and acceleration when t = 0.76 s.
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ENERGY OF A SIMPLE HARMONIC OSCILLATOR
The elastic potential energy of stored in the spring of a system undergoing SHM of the form
x  A cost   
is
U  12 kx 2
 12 kA2 cos2 t   
The kinetic energy of the same system is
K  12 mv 2
 12 m 2 A2 sin 2 t   
We will use the fact that
2 
k
m
m 2  k
or
to combine the equations to find the total energy
E  K U
 12 m 2 A2 sin 2 t     12 kA2 cos 2 t   
 12 kA2 sin 2 t     12 kA2 cos 2 t   

 12 kA2 sin 2 t     cos 2 t   

 12 kA2  constant
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The kinetic, potential and total energies are plotted as functions of time in the figure below.
2.5
E
2
1.5
K
U
1
0.5
0
-1
-0.5
0
1
2
3
4
5
6
7
8
9
10
The mass oscillates back and forth. The energy is converted between elastic potential energy and
kinetic energy (horizontal system). The total energy remains constant.
When we first looked at spring energy in the Energy section, we had
K  U  12 mv 2  12 kx 2  constant
 12 kA2
We can use the energy relationship to find either the position or velocity when we know the total
energy and velocity or position.
Problem 3
A 79 kg student is connected to a spring with a spring constant of 250 N/m. The student is on a
frictionless horizontal surface.
a)
Calculate the total energy of the system if the maximum displacement is 3.0 m.
b)
What is the velocity of the student when the position is 1.0 m.
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SMALL ANGLE APPROXIMATION
The small angle approximation is a useful way to estimate sin  , cos and tan  when then angle 
is less than 0.1 or 0.2 radians. The angle must be in radians, for the approximation.
The approximations are:
sin   ,
cos  1 ,
and
tan  
Let’s try a few values for angles up to 0.25 radians. The percent differences are
%difference 
appoximation  trigonometry function
 100%
trigonometry function
 (radians)
sin 
% diff
cos
% diff
tan
% diff
0.00000
0.00000
0.00
1.00000
0.00
0.00000
0.00
0.05000
0.04998
0.04
0.99875
0.13
0.05004
-0.08
0.10000
0.09983
0.17
0.99500
0.50
0.10033
-0.33
0.15000
1.14494
0.38
0.98877
1.14
0.15114
-0.75
0.20000
0.19867
0.67
0.98007
2.03
0.20271
-1.34
0.25000
0.24740
1.05
0.96891
3.21
0.25534
-2.09
All approximations have small percent differences for these angles, with sin   best.
The approximations can be viewed geometrically.
hypotenuse  r
arc length  r
side opposite  ,
length  r sin 

side adjacent to  , length  r cos 
With small values of  , the right triangle is almost an isosceles triangle with the length side adjacent to
 nearly equal to the length of the hypotenuse.
r cos  r
or
cos  1
We can also see that the length side opposite  nearly equal to the arc length.
r sin  r
or
sin  
The tangent approximation follows from the sine and cosine approximations, tan  
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sin  
 .
cos  1
11
PENDULUM
Consider the case where a ball, mass m, on the end of a string, length L, swinging back and forth.
y
x
T
L
θ
mg
mgcosθ
mgsinθ
s
The object is constrained to move at a constant radius L about the pivot so we can define our
dimensions with the y-direction along the radius. The radius is constant so there is no movement in the
y-direction parallel to the tension. This gives
T  mg cos 
The net force is only that in the x-direction.
Fnet  mg sin  
This causes acceleration along the arc traced by the pendulum
m
d 2s
dt 2
d 2s
dt 2
  mg sin  
or
  g sin  
ma  mg sin  
a   g sin  
We can now use the small angle approximation
sin    
This works when  is in radians and the angle is small. The motion of the pendulum for small angles is
then
d 2s
  g
dt 2
or
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a   g
12
Further, we can use the definition of arc length
s  L
or, with a bit of work
d 2s
dt 2
L
d 2
dt 2
or
a  L
This gives us our final form of motion for the pendulum
L
d 2
dt 2
d 
2
dt 2
  g
L   g
or
g
 
L
g
L
  
This is simple harmonic motion with the angular frequency and period

g
L
and
T  2
L
g
We can repeat the same set of arguments for a body with mass m, moment of inertia I and distance
from pivot to centre of mass d.
Here
d 2
 mgd 
 

2
dt
 I 
and

mgd
I
Problem 4
Find the angular frequency and period of a swinging log with mass 1200 kg, length 4.2 m (need to divide
the length by 2 to get the distance to the centre of mass) and moment of inertia 7040 kg·m2.
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CIRCULAR MOTION
Consider the uniform circular motion of an object on a merry-go-round with the centre of rotation as
the origin.
The speed of the object is constant v.
The angular speed (angular frequency) of the object is
v

A
y
If the measurement started when the angle between the line
A
θ
x
joining the object to the origin and the x-axis was
 0   ,
the angle at time t is
 t     t
The x-position of the object at time t is
x  A cos  A cost   
This is the same equation we had for linear motion in a simple harmonic oscillator.
The y-position of the object at time t is


y  A sin t     A cos t    
2

This is another equation that works with the linear simple harmonic motion.
Uniform circular motion can be considered a combination of two simple harmonic motions, one each in
the x- and y- directions, with the two differing in phase by  2 or 90°.
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OPTIONAL/REVIEW TOPICS
VIBRATIONS AND WAVES
Vibrations and waves are everywhere. The ground vibrates when a train passes. The crystals inside
electronic watches vibrate at specific frequencies so precise the watches rarely require adjusting.
Waves appear in on the surface of water. Sound waves travel through the air. Light behaves as a wave
under certain conditions. Even matter can be thought of as vibrations or waves in a multidimensional
string.
Some definitions:
oscillation
repetitive motion or variation of some value in time
the value that oscillates could be anything from air pressure to fashion
trends to fish populations
oscillations usually occur about some equilibrium position
vibration
specific form of mechanical oscillation about an equilibrium position
we will look at spring and pendulum vibrations
wave
oscillation that travels in space
oscillation or vibration that moves away from the initial disturbance
period, T
the time for an oscillation to repeat itself
frequency, f
the number of times and oscillation repeats itself in a second
reciprocal of the period, f 
1
T
In this section, we will look at the oscillations in springs, pendulums and objects in circular motion. This
section is a nice recap of earlier dynamics and kinematics relations. We will also look at several
properties of waves, specifically sound waves.
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WAVES
A mechanical wave involves some form of disturbance that propagates though a material. The material
through with the waves propagate is known as a medium. A medium possesses some physical property
that allows adjacent elements within the medium to affect each other. Here are some examples
mechanical wave
medium
physical mechanism
sound
air
pressure
surface waves
water
pressure
pulse in rope
rope
tension
A travelling wave or pulse that causes the elements of the disturbed medium to move perpendicular to
the direction of propagation is called a transverse wave.
A travelling wave or pulse that causes the elements of the disturbed medium to move parallel to the
direction of propagation is called a longitudinal wave.
When a wave travels around an arena, the participants stand and wave as the wave passes but then sit
back down to enjoy the show. You can also line up people shoulder-to-shoulder and lightly push a
person at one end towards the others. The wave will travel through the line (It helps to have a safety
person at the other end to catch the last person in the line.) In both cases, the waves pass through the
group with minimal disturbance to the individuals in the wave.
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Physical waves behave in the same way. Waves propagate through a medium with minimal
disturbances to the medium itself. A wave will pass through you when you are swimming. You will go
up and down as the wave passes but you will not move along with the wave.
The speed of the waves or wave speed is a measure of how quickly the disturbance propagates through
the medium.
Consider the following transverse wave. At time, t = 0, the disturbance is
vt
v
y
y
v
x
t=0
x
t=t
For a disturbance travelling to the right at point x at time t is the same disturbance that was at x – vt at
time t = 0. We can write this mathematically as
yx, t   f x  vt
where the wave function y(x, t) represents the y-coordinate of the element along the wave at position x
at time t.
If the wave is travelling to the left, we can use the same argument to get.
yx, t   f x  vt
Example
A pulse moving to the right along the x axis has the wave function
y  x, t  
10
 x  2t 2  1
where x and y are in meters and t is in seconds.
a)
Find the speed of the wave.
The varying portion of the equation is  x  2t  . Comparing this to the form of the wave
travelling to the right, gives
 x  vt    x  2t 
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This is true for all x and t so we can make the two sides equal by setting
v  2 m/s
The speed is 2 m/s to the right
b)
Find the disturbance of the wave at the following times and locations.
y
x=0m
x=2m
x=4m
x=6m
x=5m
t=0s
10
2
0.6
0.3
0.2
t=1s
t=2s
t=3s
t=4s
You can verify that the wave moves in the positive x-direction. The waves are plotted at the following
link wave propagation spreadsheet.
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SINUSOIDAL WAVES
Sine waves are the simplest example of a periodic continuous wave. In fact there are theories (Fourier),
analyses and functioning devices that exploit the fact that all waves and, to some extent, pulses can be
represented by sums of sine waves.
These waves are very similar to the simple harmonic motion vibrations. Where SHM has displacement
(position) x varying as a function of time t, sine waves vary have the dependent displacement variable y
as a function of both position x and time t. We will also use the frequency f for sine waves instead of the
angular frequency .
Here are graphs of a sine wave versus both x and t. The first is a plot of the displacement in the ydirection as a function of position x at one instant in time.
-1
5
4
3
2
1
0
-1 0
-2
-3
-4
-5
y
λ
A
x
1
2
3
4
5
6
7
8
9
10
The wavelength λ is the distance between any two identical points on adjacent waves. The amplitude A
is the height of the wave.
-1
5
4
3
2
1
0
-1 0
-2
-3
-4
-5
y
T
A
t
1
2
3
4
5
6
7
8
9
10
This is the graph at a single point x in space. The displacement in the y-direction varies with time. The
period T is the time between any two identical points on adjacent waves. The amplitude A is the height
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19
of the wave (it is the same as when y is plotted versus x because it is a measure of the maximum yvalue).
Just like earlier in this section, the frequency f is the inverse of the period.
f 
1
T
Frequency is usually given in hertz (Hz) or cycles per second (s-1).
The equation for the sine wave is
 2
x  vt
y x, t   A sin 
 

The wave travels a distance of one wavelength in one period, so we can define the wave speed as
v

T
 f
The sine wave equation is then
 2
y  x, t   A sin 
 
 

 x  t 
T 

 x t
 A sin  2  
  T



We can use our good old angular frequency (or angular speed) ω and introduce the wave number k

2
T
and
k
2

to get
y  x, t   A sin  kx  t 
Finally, we can install a phase constant to get
y  x, t   Asin  kx  t   
The phase constant  can be found using the initial conditions (the description of the wave at t = 0, x =
0). When t = 0 and x = 0, the wave equation simplifies to
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y  0,0   A sin  
 y  0,0  

 A 
  arcsin 
To continue loading on the formulas, we can also express the wave speed using the new terms.
v

k
Problem 5
Estimate the wavelength, amplitude, period, frequency, angular frequency, wave speed, wave number
and phase constant of the following sine wave. Use proper units.
4
y (cm)
3
2
1
t (s)
0
-1
-1 0
1
2
3
4
5
6
7
8
9
10
-2
-3
-4
4
y (cm)
3
2
1
x (m)
0
-2
-1 0
2
4
6
8
10
12
14
16
18
20
-2
-3
-4
A
λ
v
T
f
ω
k

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Problem 6
A sine wave travelling to the right with an amplitude of 8.0 cm, a wavelength of 60.0 cm, and a
frequency of 10.0 Hz.
a)
Find the period, and speed of the wave.
b)
If the initial conditions state that, at x = 0 and t = 0, the vertical position is y = -6.0 cm,
determine the phase constant (phase angle) and write a general expression for the wave
function.
WAVES IN A STRING
The speed of a wave in a string is
v
T

where T is the tension of the string and μ is the mass per unit length of the string.
Problems 7
a)
A uniform cord has a mass of 0.100 kg and a length of 10.0 m. The cord passes over a
pulley and supports a 16.0 N object. Find the speed of a pulse travelling along the cord.
b)
A Bob and Anne are talking on a string phone. The string is 15 m long and has a mass of
0.20 kg. What is the tension on the line if it takes 0.50 s for a message to travel from Anne
to Bob?
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Reflection and Transmission
When a wave encounters a change in medium, in most cases a portion of the wave will be reflected back
in the direction it came and another portion of the wave will be transmitted into the new medium.
When an incident wave is comes to a fixed point, the wave exerts a force on the fixed point and,
according to Newton’s 3rd Law, the fixed point exerts the opposite force on the wave. This creates a
reflection of the original wave that is inverted with respect to the original but otherwise unchanged
(except the direction change).
When an incident wave is comes to a free end, the reflected wave is not inverted with respect to the
original and otherwise unchanged (except the direction change).
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Similar things happen when the medium changes from one type of string to another. In both cases,
there is a transmitted wave that travels into the second string. The transmitted wave always has the
same orientation to the original wave.
If the wave travels from a light string to a heavy string, this is like the case with the fixed end because
the mass of the heavy string will resist moving. The reflected wave is inverted.
light string
heavy string
If the wave travels from a heavy string to a light string, this is like the case with the free end because
the mass of the light string will provide little resistance to the movement of the heavy string. The
reflected wave is not inverted.
heavy string
light string
I like to think of this in terms of elastic collisions. When a light object collides with a stationary heavy
object, the light object will reverse directions (think of the wave movement in the y-direction). When a
heavy object collides with a stationary light object, the heavy object will maintain its original direction
(think of the wave movement in the y-direction). In either case, the originally stationary object will end
up moving in the same direction as the original moving object.
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SOUND
Sound waves are longitudinal waves that travel through a medium. If you hit a tuning fork, the tines will
oscillate back and forth. This movement creates a series of high and low pressure regions in the air
(compression and rarefaction) that travel outwards from the tuning fork (longitudinal waves).
position of high
pressure region
travelling away from
slope is the speed
of sound
tuning fork
position
low pressure
high
pressure
time
All sound waves behave in this manner. Audible sound waves are in the range of frequencies between
20 and 20 000 Hz. Infrasonic waves have frequencies below the audible range (earthquake waves).
Ultrasonic waves have frequencies above the audible range (ultrasound, dog whistles).
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With string we had the wave speed
v
T

This expression has the square root of an elastic term, tension, divided by a density term, mass per unit
length.
The speed of sound in a gas has a similar expression
v
B

This expression has the square root of an elastic term, bulk modulus, divided by a density term, mass per
unit volume.
Both terms in the expression for sound waves depend on temperature. For air, the result is
v   331 m/s  1 
TC
273C
where TC is the temperature in degrees Celsius. The -273°C is absolute zero.
The table below shows the speeds of sound for longitudinal waves in various bulk media.
Medium
v (m/s)
Medium
Gases
v (m/s)
Medium
Liquid
v (m/s)
Solid
helium (0°C)
970
seawater
1 530
steel
6 000
air (0°C)
331
fresh water
1 480
glass
4 000
air (20°C)
343
chloroform
1 000
lead
2 000
Note:
The values shown here are approximate, and are intended for demonstration purposes only.
Check the textbook, on-line, and in reference books to find more accurate values for a wider
array of materials.
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Problems 8
a)
Lightning strikes a tree 3.0 km away. You see the light 10 μs later. If the air temperature is
12°C, how long is it before you hear the sound?
b)
Calculate the bulk modulus of water based on the speed of sound.
As the tuning fork oscillates the pressure waves move outward from it. The pressure varies between the
high and low pressure regions. The air oscillates between compression and rarefaction.
λ
slice view P
of speaker
x
We can write the pressure change from normal air pressure as
P  Pmax sin kx  t 
The pressure wave causes the air particles to move back and forth. The displacement molecules is given
by
s  smax coskx  t 
The pressure is 90° out of phase with the displacement. I recommend you check out the derivations of
these expressions in the text.
ΔP
s
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SOUND INTENSITY
As the wave moves along it carries energy with it. This energy is converted into electrical signals when it
enters a microphone, and into nerve signals when enters our ears. With a bit of work, we can show that
the energy in one wavelength is
E  12  A  smax  
2
where A is the area perpendicular to the direction of wave motion and ρ is the density of the medium.
We can divide this by the period to get the rate of energy transfer (power) in one period of oscillation.
P
1
2
 A  smax  
2
used v 
T
 12  A  smax  v
2

T
The intensity of the wave is the power per unit area. We can find the intensity by dividing by the power
by the area.
I

P
A
1
2
 A  smax  v
2
A
 12  v  smax 
2
We can link in the pressure to get
I
2
Pmax
2 v
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Unless impeded or guided, most sound waves will travel in spherical waves. They travel with equal
intensity in all directions until hitting something. Think of a fire cracker dropped into the air and
exploding. The sound expands from the firecracker in all directions at the same speed so we get a
spherical wave front for the sound.
wave fronts
(crests) moving
outward
λ
For a spherical wave, we can do some averaging to get
Pave
P
 ave2
A
4 r
I
Notice that the intensity at some distance r1 from the source is
I1 
Pave
4 r12
If we look at a different distance r2 from the source, we get
I2 
Pave
4 r22
The ratio of the intensities at the two points is
Pave
I2
4 r12

I1
Pave
4 r12

r12
r22
The ratio of intensities is the inverse of the ratio of the squares of the radii.
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Problem 9
A source emits waves with a power output of 120 W.
a)
Find the intensity 12 m from the source.
b)
Find the distance at which the intensity is 10.0% of that in part a).
Decibels (dB)
The loudest sounds the ear can tolerate have an intensity of 1 W/m2. For 1000 Hz sound, the quietest
sounds the ear can detect have intensity as low as 10-12 W/m2. This is a huge range of intensities from
10-12 W/m2 to 1 W/m2. A much more convenient term is the decibel. The decibel is a comparison to the
minimum level. Intensity described in decibels using
I 

 I0 
  10 log 10 
where I0 = 10-12 W/m2.
Examples
A sound with intensity of 10-12 W/m2 has a decibel level of
 1012 W/m 2 
  10 log 10 1  0 dB
12
2 
 10 W/m 
  10 log 10 
A sound with intensity of 10-11 W/m2 has a decibel level of
 1011 W/m 2 
  10 log 10 10  10 dB
12
2 
 10 W/m 
  10 log 10 
A sound with intensity of 5.0×10-5 W/m2 has a decibel level of
 5.0 105 W/m 2 
  10 log 10 5.0 107  77 dB
12
2
 10 W/m 
  10 log 10 

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
30
When working with decibels we use the properties of the logarithm
log10  AB   log10  A  log10  B 
 A
log10    log10  A  log10  B 
B
Example
Bob’s neighbours were at it again, playing their soft rock too loud. Inside Bob’s house, the sound level
for the neighbours’ music was 72 dB. If the neighbours increased their volume by a factor of 5, what
sound level would Bob hear in his house?
terms:
I 
I1  1  10 log10  1   72 dB, I 2  5I1 ,
 I0 
2  ?, I 0  1012 W/m2
new dB level:
 I2 

 I0 
 2  10 log10 
 5I 
 10 log10  1 
 I0 
I 
 10 log10  5   10 log10  1 
 I0 
 7.0  1
 79 dB
Bob would hear the soft rock at 79 dB.
problem 10
The intensity of sound produced by a mosquito 5.2 cm from your ear is 42 dB.
a)
Find the decibel level, if there are two similar mosquitoes 5.2 cm from your ear.
b)
Find the decibel level after you swat one mosquito and the remaining one moves 52 cm from
your ear.
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DOPPLER EFFECT
If you sit on the side of a road or railroad track and listen to the sound of their horn, you will probably
hear a change in frequency (tone) as the vehicle moves past. This is known as the Doppler effect. To
picture this, think of a moving object that is emitting sound. The high pressure regions (wave crests) are
shown in the figures. A wave front forms a circle centred on the point where the wave was created.
The object moves so the circles for successive circular wave fronts also move forming the patterns
below. The motion is relative, it makes no difference whether you or the source do the actual moving.
wavelength, λ
Stationary Source
observer
source
wave crests
Source Moving Toward Observer
source
wavelength, λ′ < λ
observer
wave crests
Source Moving Away from Observer
wavelength, λ′ > λ
source
observer
wave crests
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As seen in the figures, the wavelength changes (or appears to change in the view of the observer). The
wavelengths shrink/grow by the distance the source moves each period
  vS T
where vS is the speed of the source towards to the stationary observer. The observed wavelength is
     
   vS T

vS
f
The wave is still coming at the observer at the speed of sound v so the observed frequency is
f


v


v
vS
f
v
v
f

vS
f
 v 

f
 v  vS 
moving source
If the source moves towards the observer, the observed frequency increases.
Example
If the speed of sound is 330 m/s, the observed frequency of a 200 Hz signal emitted by a source moving
at 30 m/s towards the observer is
 v 
 330 
f
 f 
  200 Hz   220 Hz
 330  30 
 v  vS 
If the source is moving away from the observer, vS  30 m/s and the observed frequency is


 v 
330
f
  200 Hz   183 Hz
 f  
v

v
330


30


S 



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In the case where the observer moves towards the stationary source, the wavelength is unchanged but
the speed becomes
v  v  vO
where vO is the speed of the observer. The modified frequency is then
f

v

v  vO


v  vO   f 
  v 

v  vO
f
v
moving observer
If the observer moves towards the source, the observed frequency increases. If the observer moves
away from the source, the observe frequency decreases. To someone who is stationary, there is not
change in frequency.
Example
If the speed of sound is 330 m/s, the observed frequency by an observer moving at 30 m/s towards a
stationary source making a 200 Hz signal is
 v  vO
f
 v

 330  30 
 f   330   200 Hz   218 Hz



If the observer is moving away from a stationary source at vO  30 m/s and the observed frequency is
 330   30  
 v  vO 
f
f 
  200 Hz   182 Hz

330
 v 


The general expression for the Doppler frequency shift is
 v  vO 
f
f
 v  vS 
If the observer and source are moving towards each other, the observed frequency increases. If the
observer and source are moving apart, the observed frequency decreases.
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Shock Waves
If the source is travelling faster than the speed of sound, the spherical waves still start at the source, but
no waves travel in front of the source because it outruns the sound. The wave crests of sound emitted
at different times build up to form a shock front. We can see shock waves that build up on the surface
of water behind a boat that travels faster than the surface wave speed.
shock front
source
Reflections
One of the first things that people do when they enter a very large, empty, enclosed or partially encloses
space is yell “Hello” (or a similar term in other languages), and then listen for the echo. When a sound
wave strikes a rigid structure like a large wall or a cliff, it changes direction and continues to on in the
opposite direction (assuming the sound was initially travelling perpendicular to the wall).
Problem 11
As Bob drove towards a large cliff, he blasted his horn and listened for the echo. If the horn emits an
annoying 620 Hz tone, what is the frequency of the echo? Assume that the speed of sound in air is 330
m/s.
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INTERFERENCE OF WAVES
The superposition principle states:
If two or more travelling waves are moving through a medium, the resultant value of the wave function
at any point is the algebraic sum of the values in the wave functions of the individual waves.
In this figure we have two waves, one moving to the left and one moving to the right. When the two
waves meet, there displacements add due to superposition.
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In the following figure, we have two transverse waves with displacements in the opposite direction
travelling towards each other. As per the superposition principle, we still add the displacements, but
due to the opposite displacements, the waves tend to cancel each other out when they meet.
Both cases show the consequence of superposition principal and that two travelling waves can pass
through each other without being destroyed or even altered.
Check out http://phet.colorado.edu/new/simulations/sims.php?sim=Wave_on_a_String. Set the
tension (mid-way or less) to slow the pulse. Try sending pulse in the same direction and in opposite
directions. More on this applet later…
The combination of separate waves in the same region is called interference. When two waves with
displacements in the same direction interfere, the superposition is larger than either single wave. This is
constructive interference. When two waves with displacements in opposite direction interfere, the
superposition is reduces the displacement. This is destructive interference.
Consider the interference of the two waves
y1  A sin kx  t 
and
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y2  A sin kx  t   
37
Both waves are travelling in the same direction with speed
v

k
The superposition of the two waves results in the wave function
y  y1  y2
 A sin  kx  t   A sin  kx  t   
Here we can use the relation
 a b   a b 
sin a  sin b  2 cos
 sin 

 2   2 
with a  kx  t   and b  kx  t to get
a b  a b

 ,
 kx  t 
2
2
2
2
and

  
y  2 A cos   sin  kx  t  
2
2 
This figure below shows the typical case where the phase difference is no special value. There is a mix
of constructive interference when both waves have the same sign and destructive interference when
they have opposite sign. The phase is a constant so the amplitude of the combined signal is
 
Ac  2 A cos 
2
with the combined wave


y  Ac sin  kx  t   .
2

y1 + y2
y1
x
y2
The phase is  is zero in the following figure. y1 and y2 are identical so y1 + y2 is twice y1 or y2.
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y1 + y2
y1 and y2
x
We get complete constructive interference whenever the phase term is 0, 2π, 4π … any even multiple of
π. If we look at the interference equation, constructive interference occurs when
 
cos    1
2
and
Ac  2 A .
In this case below,    so y1 and y2 are exact opposites. We get complete destructive interference so
y1 + y2 is zero.
y1
y1 + y2
y2
x
Complete destructive interference occurs whenever the phase term is π, 3π, 5π … any odd multiple of π.
Looking at the interference equation, we get destructive interference when
 
cos    0
2
and
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Ac  0 .
39
Problem 12
Find the amplitude due to the superposition of two waves
y1  4.0 sin 2.0x  4.0t 
and
y2  4.0 sin 2.0x  4.0t   
when
a)
the phase difference is   0.1 ,
b)
the phase difference is   3.0 .
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40
INTERFERENCE OF SOUND WAVES
Imagine two speakers, each playing sound of the same frequency of sound with the same phase.
speaker 1
1
speaker 2
2
An observer is a distance r1 from speaker 1 and r2 from speaker 2. r1 and r2 are called the path lengths
for the two sources. If the two path lengths are different, the sound will take different amounts of time
to reach the observer.
sound from
speaker 1
sound from
speaker 2
observer
We can use the wavelengths as the units for the measure of the distance between sources and the
r
r
observer. Speaker 1 is 1 from the observer while speaker 2 is 2 from the observer. The difference in


path lengths is
r2


r1


r

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The term
r

is a fraction of a wavelength. Let’s say this term is 0.50, one speaker is half of a
wavelength closer to the observer than the other. We can convert this term to the phase difference by
multiplying by 2.

A value of
r

2r

 2  0.5   
 0.50 corresponds to a phase difference of . The value Δr is known as the path length
difference.
Examples
In the diagram below, the phase difference is


2r

2  4.5  4.0 

.

This is a case where we have complete destructive interference. The sound from one speaker cancels
out the sound from the other speaker. The observer hears nothing.
r 

2
r1 = 4.0
1
2
r2 = 4.5
An application of this is noise-cancelling headphones which determine the frequency and phase of the
incoming noise, and then create sound with the appropriate frequency and phase to cancel out the
noise from the surroundings.
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In the next diagram, the phase difference is


2r

2  4.5  3.5 

.
 2
This is a case where we have complete constructive interference. The sound is louder than the sound
for a single e speaker.
r = λ
r1 = 3.5
1
2
r2 = 4.5
Below is a more general case. If r1 is 3.7λ away from the observer and r2 is 4.0λ away the observer, the
phase angle will be



2r

2  r2  r1 

2  4.0  3.7 

 0.6
 1.9 radians
1
2
The interference at the observer is the same situation as earlier. We get constructive interference when
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43
 
cos   1
2
  0, 2 , 4 …
or
This is satisfied when
2r

 0, 2 , 4 ...
or
r  n
where n = 0, 1, 2, …
If speakers are playing one frequency, the sound level is higher when the path length difference
between you and the speakers is an integer multiple of the wavelengths.
We can use similar arguments to show that we get destructive interference when
r  2n  1

2
where n = 0, 1, 2, …
If speakers are playing one frequency, the sound level is lower or zero when the path length difference
between you and the speakers is an integer plus
1
2
multiple of the wavelengths.
These properties are difficult to notice in everyday life because we rarely hear a single frequency from a
pair of speakers for any significant length of time.
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Problem 13
Bob stands in front of two speakers, 12.00 m from each speaker. The speakers produce the same
frequency sound with the same phase, so Bob is at a point with constructive interference. Staying 12.00
m from speaker 1, Bob finds the next point of constructive interference when he moves to a point 12.70
m from speaker 2.
1
12.00 m
12.00 m
12.70 m
12.00 m
2
a) What is the wavelength of the sound?
b) What is the frequency if the speed of sound is 345 m/s?
The applet http://phet.colorado.edu/new/simulations/sims.php?sim=Wave_Interference shows the
sound coming from one speaker or from two speakers with interference. With two speakers, set the
frequency to something medium low. Notice that there are regions where the waves are grey all of the
time and other regions where the waves alternate between black and white. You can place detectors in
these regions to show you the pressure oscillation in these areas. The grey regions are the nodal regions
where there is destructive interference and the sounds cancel out. The black and white regions are
where there is constructive interference so the sound is loud. You can also move the detectors near and
far from the speakers and note the change in intensity (pressure). Try varying some of the other
parameters or check out the other sound applet.
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45
STANDING WAVES
If the two speakers in the previous derivations are facing each other, their waves will be travelling in
opposite directions in the space between the speakers.
1
2
Ignoring phase, we can write the wave functions as
y1  A sin  kx  t 
y2  A sin  kx  t 
and
Note that the speaker 1 wave function has a speed to the right and the speaker 2 wave function has a
speed to the left. This is reflected in the signs in front of the angular frequencies.
The combined wave function is
y  y1  y2  Asin kx  t   Asin kx  t 
We can use the trig rule
 a b   a b 
sin a  sin b  2 cos
 sin 

 2   2 
with a  kx  t and b  kx  t to get
a b
ab
 kc,
 t
2
2
and
y  2 A sin kxcost 
This is the wave function of a standing wave. It is “standing” because there are points that have zero
amplitude; they don’t move in either of the x or y directions. The wave appears to be fixed at these
nodes. The amplitudes at the nodes satisfy
2 A sin  kx   0
The positions of the nodes occur when
kx  0,  , 2 , .... n
where n  0, 1, 2, ...
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46
or, using k 
2

x  0,  ,  , 3 , ... n
2
2
2
where n = 0, 1, 2 …
The cos t  term varies between -1 and 1 for all points. The maximum displacement (amplitude) at
any position x is
ymax  x   2 A sin  kx 
The positions that have the largest maximum displacement are those where the sine term is ±1.
sin  kx   1
The maximum sine values occur when kx  12  , 1 12  , 2 12  , ...
Using
k
2

,
we get
x   , 3 , 5 , ...  n ,
4
4
4
4
n = 1, 3, 5 …
These points are called antinodes.
The distance between adjacent antinodes is  .
2
The distance between adjacent nodes is  .
2
The distance between a node and an adjacent antinode is  .
4
Looking at the wave function again, the whole function is zero if cost   0 .
By definition
  2 T
cost   0
when
t  nT ,
4
n = 1, 3, 5 …
t   2, 3 2, 5 2, ...
or
Similarly, the function has maximum amplitude when
cost   1 when
t  nT ,
2
n = 0, 1, 2 …
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This figure shows the positions of
nodes
a standing wave at different times
t=0
t
T
4
t
T
2
multiple times
as the time blurs the
image it the wave
appears to be
stationary
antinodes
Problem 14
Two waves travel in opposite directions with wave functions
y1  0.10 sin 4.0x  3.0t  and
y2  0.10 sin 4.0x  3.0t 
where x and y are in meters.
a)
Find the amplitude of the simple harmonic motion of an element at x = 1.5 m.
b)
If x = 0 m is a node, find position of the next node and antinode.
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48
Many systems have normal modes which are patterns of oscillation favoured by the natural system.
Some are shown below.
String with Fixed Ends – Nodes at Ends
L
L
n = 1, λ1 = 2L, f1 = v
n = 2, λ2 = L, f2 = v
2L
L
L
fixed ends
n = 3, λ3  2 3 L , f13 = 3v
fn 
nv
2 L 1
 nf1 , n 

2L
n
n
2L
The frequency for n = 1 is the fundamental frequency. Other natural frequencies are integer multiples
of the fundamental frequency. The frequencies form a harmonic series with higher n frequencies
known as harmonics. Positions with no motion are nodes. Positions with lots of motion are loops.
Air Column, Both Ends Open – Antinodes at Ends
L
n = 1, λ1 = 2L, f1 = v
open ends
L
n = 2, λ2 = L, f1 = v
2L
fn 
nv
 nf1 ,
2L
n 
2 L 1

n
n
L
Air Column, One End Open
L
n = 1, λ1 = 4L, f1 = v
L
n = 2, λ2 = 4 3 L , f1 = 3v
4L
4L
one open, one closed end
fn 
 2n  1 v 
4L
 2n  1 f1,
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n 
1
4L

 2n  1  2n  1
49
Let’s look at string oscillations between two fixed points (a classic demonstration is to have the string
between a pair of small electric motors that are spinning). You can use your finger to suppress certain
modes of vibration. With no finger placement, the string will usually vibrate in the first harmonic (the
natural frequency).
f1 = v
2L
Lightly placing your finger at the middle node for the second harmonic suppresses the first harmonic,
but allows the string to still oscillate in the second harmonic (the double-frequency overtone).
f2 = v
L = 2f1
Placing your finger at one of the nodes for the third harmonic suppresses both the first and second
harmonics but allows the string to still oscillate in the third harmonic (the triple-frequency overtone).
f3 = 3v
2 L = 3f1
Problem 15
A 0.72 m long string vibrates with a frequency of 780 Hz. The string has two nodes between the ends.
a)
Find the fundamental frequency and the other harmonics with frequencies lower than 780
Hz.
b)
Find the wave speed in the string.
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Resonance
If you wet your finger and move it around the rim of a wine glass at the right speed, the wine glass will
oscillate, producing an audible sound. When you introduce a wave into a string, you have to apply a
force to move one end of the string back and forth. If a periodic force is applied to such a system, the
amplitude of the resulting motion is greatest when the frequency of the applied force is equal to one
of the natural frequencies of the system. The frequencies of the applied force that match the natural
frequency of the system are known as resonance frequencies.
The frequency of your voice is centered on the resonant frequency for your vocal tract.
Some items vibrate with a natural frequency when struck. In a tuning fork, the natural frequency is
enhanced and purified (stripped of side harmonics). A bell is another device designed for a specific
natural frequency. When you drop coins on a hard floor, you can often tell if it is a penny or a loony by
the sound of its natural frequency.
Resonance and natural frequencies are generally avoided in most engineering projects that are not
designed to vibrate. Check the internet for the amazing videos of the Tacoma Narrows Bridge, which
collapsed rather violently during strong winds in 1940.
You can also revisit http://phet.colorado.edu/new/simulations/sims.php?sim=Wave_on_a_String. With
the tension set on high, frequency at 50 and damping off, you can set up a standing wave by applying
the oscillation driver for a few seconds (if the amplitude gets too big, you can use the damping to reduce
it). The green beads (?) mark the nodes of the standing wave. Try setting up other standing waves with
and without a fixed end.
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51
BEATS
Let’s look at the interference of waves of different frequencies. With different frequencies, there are no
stationary spatial properties. Ignoring the spatial part, the wave functions are
y1  A cos2f1t 
y2  A cos2f 2t 
and
Under superposition, the combined wave function is
y  y1  y2  A cos2f1t   A cos2f 2t 
Here, we use the trig rule
 a b  a b 
cos a  cos b  2 cos
 cos

 2   2 
to get

f  f 
f f 


y   2 A cos  2 t 1 2   cos  2 t 1 2 
2 
2 



The term in square brackets is regarded as the resultant amplitude. This amplitude varies with the beat
frequency
f beat 
f1  f 2
2
The figure below shows two waves with slightly different frequencies and the superposition of the two
waves. Notice how the amplitude varies as the two waves go into and out of phase. If the frequencies
of the two waves are very close you can actually hear the amplitude going loud and soft with the beat
frequency.
y1, and y2
y
amplitude
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Problem 16
When Bob howls at a frequency near 230 Hz, his dog Fido will howl along with him. Calculate the beat
frequency if Bob is howling at 233 Hz and Fido is howling at 228 Hz.
As always, check out problems in the text as well as the problems in the assignments. By practicing
problems related to oscillations and waves, you will develop an understanding of the types of questions,
important factors and general method of solving. Also look for examples in real life, listen for the
change in frequency of a train whistle as the train passes you. Recognise natural frequencies and simple
harmonic motion around you.
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53
SUMMARY
Simple Harmonic Motion
d 2x
 cx ,
dt 2
x  sin

a  cx

c t ,
Fx  kx ,
or
v  c cos


c t ,
a  c sin


c  t  cx
Wave Equation
x  A cost   
ω is the angular speed or angular frequency, A is the
amplitude and  is the phase
v   A sin t   
a   2 A cos t      2 x
2 
T
f 
k
,
m
2

 2
1
1

T 2

k
m
for a mass m on a spring with spring constant k
m
k
period is measured in seconds.
k


m 2
frequency in Hz
Small Angle Approximation
cos  1, sin   , tan   
for θ < 0.1 radians
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Pendulum (small angles)
L
g
, T  2
g
L

for pendulum length L, acceleration due to gravity g
Circular Motion



x  A cos t    , y  A sin t     A cos t     … note the  90 phase difference
2
2

Travelling Wave
v

 f
T
wave speed

x t 
 2
y  x, t   A sin 
 x  vt    A sin  2      A sin  kx  t 
 

   T 
displacement as a function
of position and time
k
2

, v

k
wave number k
Sound
v
T
speed of sound in a string, T is tensions, μ is linear density

v   331 m/s  1 
I 2 r12

I1 r22
TC
273C
speed of sound in air at temperature TC
ratio of intensities at different distances from sources
I 

I
 0
  10 log 10 
decibel sound intensity, where I0 = 10-12 W/m2.
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 v  vO 
f
f
 v  vS 
Doppler frequency, vS speed of source towards observer, vO
speed of the observer towards the source
Interference
Depending on the path length difference (pd) between the two sources and the observer, you get

constructive interference  pd  2n  or destructive interference pd   2n  1 
f beat 
f1  f 2
2
 n = 0, 1, 2…
beat frequency
Standing Wave
fn 
nv
2 L 1
 nf1 , n 

2L
n
n
fixed ends, cavity/string length L, speed v, n = 0, 1, 2…
fn 
nv
2 L 1
 nf1 , n 

2L
n
n
open ends (free ends), n = 0, 1, 2…
fn 
 2n  1 v 
4L
 2n  1 f1,
n 
1
4L

one end fixed/one end free, n = 0, 1, 2…
 2n  1  2n  1
NANSLO Physics Core Units and Laboratory Experiments
by the North American Network of Science Labs Online,
a collaboration between WICHE, CCCS, and BCcampus
is licensed under a Creative Commons Attribution 3.0 Unported License;
based on a work at rwsl.nic.bc.ca.
Funded by a grant from EDUCAUSE through the Next Generation Learning Challenges.
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56