PHY 113 C General Physics I
11 AM -12:15 PM TR Olin 101
Plan for Lecture 18:
Chapter 17 – Sound Waves, Doppler Effect
Chapter 18 – Superposition, Standing Waves,
Interference
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PHY 113 C Fall 2013 -- Lecture 18
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PHY 113 C Fall 2013 -- Lecture 18
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 y
2  y

v
2
2
t
x
Reminder!
2
The linear wave equation
Solutions are of the form:
y ( x, t )
position
2
time
Wave variable
Periodic traveling waves are solutions of the Wave Equation
 x t 

y(x, t)  y 0 sin  2π     φ 
 λ T

v=
λ
T
phase factor (radians)
period T = 1/f (s)
wave length (m)
Amplitude
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PHY 113 C Fall 2013 -- Lecture 16
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SOUND WAVES obey the
linear wave equation**
 y
2  y

v
2
2
t
x
2
2
**We’re neglecting attenuation. Attenuation is very important in ultrasonic imaging..
Periodic sound waves, general form
 x


y(x, t)  y 0 sin  2   ft     ; velocity of the wave: v = f

 

phase factor (radians)
frequency f (s-1) determines pitch
wave length (m)
Amplitude (determines
intensity or loudness)
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PHY 113 C Fall 2013 -- Lecture 16
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SOUND: longitudinal mechanical vibrations
transmitted through an elastic solid or a liquid or gas,
with frequencies in the approximate range of 20 to
20,000 hertz. How are sound waves generated? What
is the velocity of sound waves in air?
A simple model with the essential physics (for 113)
STEP ONE
cylinder
piston
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gas
PHY 113 C Fall 2013 -- Lecture 18
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STEP TWO
Questions:
what has moved in the gas?
in what direction have gas molecules moved?
why isn’t the gas uniformly compressed ?
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STEP THREE
Notice that the velocity of the compressed
zone is not the same as the velocity of the
piston.
What happens if the piston oscillates at
1702 piston oscillation
frequency f? UsePHY
animation……….
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What is y(x,t) in the wave
equation of the sound wave?
A sound wave may be
considered as
a displacement wave :
node
antinode
s  s max cos(kx -t)
Or
nodes at  / 2, 3 / 2,...
a pressure wave:
P  Pmax sin(kx  t)
nodes at 0, , 2,...
Since density is proportional
to pressure,
  max sin(kx  t)
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PHY 113 C Fall 2013 -- Lecture 18
Antinode
node
8
The velocity of sound is easy to measure.
v = 343 m/s at 20ºC in air.
How to relate the velocity of sound to the known
properties of gases?
Newton (1643-1727) worked out an analytic
model, assuming isothermal compressions:
v = 298 m/s
Laplace(1749-1827) pointed out that on the
timescale of sound frequencies, compressions
are not isothermal, they’re adiabatic (isentropic)
If we use the linearized fluid equations** in the ideal gas approximation,
we find that the density fluctuations  obey:
 2
2
2  
2 P0
v
0; v 
 343m / s;
2
2

t
x
0
here  is the ratio
cp
cv
.
* * These follow from conservation of mass, momentum, and energy.
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PHY 113 C Fall 2013 -- Lecture 18
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Speed of Sound in a Gas
•Consider an element of the gas
between the piston and the dashed
line.
•Initially, this element is in equilibrium
under the influence of forces of equal
magnitude.
– There is a force from the
piston on left.
– There is another force from
the rest of the gas.
– These forces have equal
magnitudes of PA.
• P is the pressure of the gas.
• A is the cross-sectional area of the
tube.
Section 17.2
Speed of Sound in a Gas, cont.
•After a time period, Δt, the piston has
moved to the right at a constant
speed vx.
•The force has increased from PA to
(P+ΔP)A.
•The gas to the right of the element is
undisturbed since the sound wave
has not reached it yet.
Section 17.2
Impulse and Momentum, cont.
•The change in momentum of the element of
gas of mass m is
p  mv   pvv x At  ˆi
•Setting the impulse side of the equation equal
to the momentum side and simplifying, the
speed of sound in a gas becomes.
v
B

– The bulk modulus of the material is B.
– The density of the material is 
•
Section 17.2
Velocity of sound:
Recall:
For a string,
So, for sound,
v
v



tension

mass per unit length
B


Bulk modulus

density
elastic property
inertial property
elastic property
inertial property
Definition of B: stress = modulus x strain
stress = p ; strain = V / V
P
B
, definition of bulk modulus
V V
Intensity of sound waves;
Power
decibels (dB)
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PHY 113 C Fall 2013 -- Lecture 18
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SUPERPOSITION of WAVES
( Chap. 18)
So far, we’ve considered only one travelling wave. However,
a sum of two or more travelling waves will also satisfy the
wave equation because the wave equation is linear
The Superposition Principle:
If waves y1(x,t) and y2(x,t) are traveling through the
same medium at the same time, the resultant wave
function y3(x,t) = y1(x,t) + y2(x,t).
(This is not always true! If y1 or y2 become large enough,
response of medium becomes nonlinear and the superposition
principle is violated). Examples: music amplifiers; lasers.)
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PHY 113 C Fall 2013 -- Lecture 18
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CASE 1. Superposition of 2 sinusoids, f1 = f2, A1=A2,
traveling in the same direction, but different phase
y1  y0 sin  kx  t 
; y 2  y0 sin(kx  t  )
  
   
use trig identity: sin() + sin() = 2sin 
 cos 

2
2






y1  y 2  2sin(kx  t  ) cos  
2
2
Constructive interference:  = 0, ,…
Destructive interference:  = /2,3/2…
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PHY 113 C Fall 2013 -- Lecture 18
1803 superposition of two
sinusoids
15
Case 2: 2 pulses traveling in opposite directions:
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PHY 113 C Fall 2013 -- Lecture 18
1801
16
Case 3. Standing waves. Two sinusoidal waves, same
amplitude, same f, but opposite directions
 x

yright ( x, t )  y0 sin  2π  ft  

 λ
 x

yleft ( x, t )  y0 sin  2π  ft  

 λ
Use trig identity again:
sin A  sin B  2sin  1 2  A  B   cos  1 2  A  B  
get
 2πx 
y right (x, t)  yleft (x, t)  2y 0 sin 
 cos  2πft 
 λ 
Standing wave:
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Check that the standing wave satisfies the Wave
Equation.
wave equation:
2y
x
 2πx 
ystanding (x, t)  A sin 
 cos  2πft 
λ


Check:
Check:
 2 ystanding
t 2
 2 ystanding
x 2
2y
t
t
2
2 y
v
0
2
2
 2πx 
2
   2πf  A sin 
 cos  2πft 
λ


2
 2πx
2 y
v
  A sin 
2
2
 λ
x
2
 2π 
 2πx 
 
A
sin


 cos  2πft 
 λ 
 λ 
2

2π



2
2
 cos  2πft    2πf   v 
   0

 λ  

Equality satisfied iff fλ  v. Okay? Yes.
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PHY 113 C Fall 2013 -- Lecture 18
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How to generate standing waves
1. Aim two loud speakers at one another.
2. Have two boundaries causing one wave to be
reflected back on itself. Each boundary is a node.
The reflected wave travels in the opposite direction.
1808 Standing waves
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PHY 113 C Fall 2013 -- Lecture 18
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Standing waves between reflecting walls
 2πnx 
Possible spatial shapes : A sin 

 2L 
n  1,2,3,4.....
 2πx 
Standing wave form: A sin 
 cos  2πft 



2L
nv
 n 
fn 
n  1, 2,3, 4...........
n
2L
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PHY 113 C Fall 2013 -- Lecture 18
1810 Nodes, in string
Demo:
standing transverse waves in a string
iclicker question after string demo:
Which of the following statements are true
about the string motions described above?
A. The human ear can directly hear the
string vibrations.
B. The human ear could only hear the
string vibration if it occurs in vacuum.
C. The human ear can only hear the string
vibration if it produces a sound wave in
air.
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PHY 113 C Fall 2013 -- Lecture 18
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Standing waves in air:
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Standing longitudinal waves in a gas-filled pipe:
Do Eric’s supersized fireplace log demo.
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6 Strings (primary elastic elements)
frets
Sound board
Sound hole
Bridge(couples string
vibration to sound
board)
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Coupling standing wave resonances in materials
with sound:
iclicker question:
Suppose an “A” is played on a guitar string.
The standing wave on the string has a frequency
(fg), wavelength (g) and speed (cg) . Which
properties of the resultant sound wave are the
same as wave on the guitar string?
A. Frequency fs
B. Wavelength s
C. Speed cs
D. A-C
E. None of these
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PHY 113 C Fall 2013 -- Lecture 18
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Comments about waves in solids and fluids:
 Solids can support both transverse and
longitudinal wave motion. Earthquake
signals
 Fluids, especially gases, can support only
longitudinal wave motion
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PHY 113 C Fall 2013 -- Lecture 18
1820 harmonics
27
The Doppler effect . Funny things happen if wave source and
wave detector are moving with respect to one another.
Case 1. Source stationary, observer moving toward source.
Let v' = speed of wave relative to observer
Observer, vO , f’ = ?
Source.vs=0, freq f
v '  v  vO but wavelength  is unchanged
so for source, v=f , but for observer, v' = f '
f '  v '/ 
f '
v  vO

 v  vO 
f f

v


'
Case 2. It’s easy to see that if observer moves away from source
 v  vO 
f'  f 

v


What happens as observer goes past the source?
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PHY 113 C Fall 2013 -- Lecture 18
1708 Doppler, motorcycle & car 28
Case 3. The Doppler effect for stationary observer, source moving
toward observer A. Let v = velocity of sound, f = frequency of source
During each vibration of source,
source moves distance vST  vS / f.
vS
Therefore observed wavelength    
f
'
But f '  v /  ' for Observer A
v
v
so f 

vS v vS


f
f
f
'
 v 
f  
 f
 v  vS 
'
Case 4. Souce moving away from observer B
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PHY 113 C Fall 2013 -- Lecture 18
 v 
f '  
 f
 v  vS 
1709 Doppler, 2 observers, one source
29
Summary of sound Doppler effect :
v  vO
fO  f S
v  vS
toward
away
Doppler effect for electromagnetic waves :
fO  f S
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v  vR
v  vR
Relative velocity of source
toward observer
PHY 113 C Fall 2013 -- Lecture 18
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Case 4. Source and observer both moving.
Example:
v  vO
fO  f S
v  vS
o
s
s
o
Velocity of sound:
v = 343 m/s
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toward
away
In this case :
v  vO
fO  f S
v  vS
f O  f S  vO  vS
fO
vO  v  (v  vS )  40m/s
fS
PHY 113 C Fall 2013 -- Lecture 18
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Doppler effect for electromagnetic waves
For sound waves, model and equations used
velocity of source or observer with respect to the medium
For electromagnetic (EM) waves (light, radio waves),
there is no medium and the velocity of light is c.

The model and equations have to be different (Einstein).
c  v relative
f f
c  v relative
'
Applications: wavelengths of spectral lines from
stars, e.g. H atom, are slightly shifted.
Doppler radar: make sketch on board
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PHY 113 C Fall 2013 -- Lecture 18
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iclicker question
Is Doppler radar described by the equations given above for
sound Doppler?
(A) yes
(B) no
Is “ultra sound” subject to the sound form of the Doppler
effect?
(A) yes
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(B) no
PHY 113 C Fall 2013 -- Lecture 18
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sin A  sin B  2 sin1 2  A  B cos1 2  A  B 
“Standing” wave:
 2πx 
y right (x, t)  y left (x, t)  2y 0 sin 
 cos  2πft 
 λ 
1810 Nodes & antinodes in string
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