Physics 231
Topic 11: Waves & Sound
Alex Brown
Nov
11-16
2015
MSU Physics
231 Fall
2015
1
Key Concepts: Waves & Sound
Wave Properties
Transverse vs longitudinal waves
Wave periodicity a speed
Interference and Standing Waves
Superposition, constructive & destructive
interference
Sound Waves
Sound Intensity
Musical Instruments & Harmony
The Doppler Effect
Covers chapter 11 in Rex & Wolfson
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Transverse Waves
The wave moves to the right,
but each point makes a
simple harmonic vertical motion
position y
oscillation
position x
Wave motion
Since the oscillation is in the direction perpendicular
(transverse) to the direction of travel, this is called a
transverse wave. Example: water waves
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Describing a Traveling Wave


y
t=0
x
t=T/4
t=2T/4
t=3T/4
t=T
: wavelength = length (m) of one
oscillation.
T: period = time for one oscillation
T=1/f f: frequency (Hz)
While the wave has traveled one
wavelength, each point on the wave
has made one period of oscillation.
v = x/t = /T = f
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y ( x, t )  A sin (kx   t )
v( x, t )   A cos(kx   t ) transverse
2

T
2
k
v


T
(in direction of motion)
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An Example
A traveling transverse wave is seen to have horizontal
distance of 2m between a maximum and the nearest
minimum and a peak maximum to peak minimum height
of 3m. If it moves at 1m/s, what is its:
a) amplitude
b) period
c) frequency
a) amplitude: difference between maximum (or minimum)
and the equilibrium position in the vertical direction
(transversal!) A = 1.5m
b) v = 1m/s, =2*2m = 4m T = /v = 4/1 = 4s
c) f = 1/T = 0.25 Hz
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Quiz
Two speakers sitting next to each other emit sound waves at
two different frequencies. The first emits a sound wave with
a frequency of 1 kHz and a wavelength of 0.3m. The second
sound wave emits a sound wave at 100 Hz with a wavelength of
3m. If started at the same time, which sound wave reaches
your ears first?
A) The first sound wave
B) The second sound wave
C) They arrive at the same time
1 = 0.3m
2 = 3m
f1 = 1000Hz
f2 = 100Hz
v1 = 1 f1 = 0.31000 = 300 m/s
v2 = 2 f2 = 3100
= 300 m/s
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Sea Waves
An anchored fishing boat is going up and down with the
waves. It reaches a maximum height every 5 seconds
and a person on the boat sees that while reaching a
maximum, the previous wave has moved about 40 m away
from the boat. What is the speed of the traveling waves?
Period: 5 seconds (time between
reaching two maxima)
Wavelength: 40 m
v = /T = 40/5 = 8 m/s
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Longitudinal Waves
The wave moves to the
right, but each point
makes a simple harmonic
horizontal motion
wave
oscillation
Longitudinal wave: movement is in the direction of the
wave motion.
Example: sound waves
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Sound: longitudinal waves
A sound wave consist of longitudinal oscillations in the
pressure of the medium that carries the sound wave.
Therefore, in vacuum: there is no sound.
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Relation between amplitude and
intensity
A
x
time (s)
-A
For sound, the intensity I is proportional to the
square of the amplitude of the longitudinal wave
I~A2
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Intensity
Intensity: rate of energy flow through an area
Power (P) J/s
A (m2)
Intensity: I = P/A (J/m2s = W/m2)
Even if you have a powerful sound source (say a speaker),
the intensity will be small when far away.
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Intensity and Distance
Sound from a point source produces a spherical wave.
Why does the sound get fainter further away from the
source?
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Intensity and Distance
The amount of energy passing
through a spherical surface
at distance r from the source
is constant, but the surface
becomes larger.
I = Power/Surface = P/A = P/(4r2)
r=1
r=2
r=3
I = P/(4r2) = P/(4)
I = P/(4r2) = P/(16)
I = P/(4r2) = P/(36)
I r2
1
1/4
1/9
= constant or I1/I2 = r22/r12
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Wave fronts
Sound emitted from a
point source are
‘spherical’. Far away
from that source, the
wave are nearly
‘plane’.
plane waves
spherical waves
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The Speed of Sound
Depends on the how easily the material is
compressed (elastic property) and how much
the material resists acceleration (inertial
property)
v=(elastic property/inertial property)
v=(B/)
B: bulk modulus
: density
The velocity also depends on temperature.
In air: v=331(T/273 K)
so v=343 m/s at room temperature
The speed does not depend on the frequency
- how to we know this?
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Quiz
As you move farther from a source of light, the intensity
of the light…
a) remains the same.
b) becomes smaller.
c) becomes larger.
The amount of energy passing
through a spherical surface
at distance r from the source
is constant, but the surface
becomes larger.
I = Power/Surface = P/A= P/(4r2)
Units = Watts/m2
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Intensity
Faintest sound we can hear: I~1x10-12 W/m2 (@ 1000 Hz)
Loudest sound we can stand: I~1 W/m2
(@ 1000 Hz)
sound wave
vibrating
ear drum
Factor of 1012? Loudness works logarithmic…
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Sound - Decibel Level 
=10log (I/I0)
I0=10-12 W/m2
y = log10x = log(x)
inverse of x = 10y
( not this: y=ln(x) x=ey )
log(ab) = log(a) + log(b)
log(a/b) = log(a) - log(b)
log(an)
= n log(a)
PHY 231
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Decibels (units are called dB)
=10 log(I/I0)
I0=10-12 W/m2
For an increase of n dB:
the intensity of the sound is
multiplied by a factor of ?.
n = 2-1= 10 log(I2/I0) – 10 log(I1/I0)
n = 10 log(I2/I1)
(n/10) = log(I2/I1)
10 (n/10) = (I2/I1)
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n (I2/I1)
10 10
20 100
30 1000
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Sound Levels
Sound Sources
Examples with distance
Table of sound levels L and corresponding
sound pressure and sound intensity
Sound Pressure
Level Lp dBSPL
Sound Pressure p
N/m2 = Pa
Sound Intensity I
W/m2
Jet aircraft, 50 m away
140
200
100
Threshold of pain
130
63.2
10
Threshold of discomfort
120
20
1
Chainsaw, 1 m distance
110
6.3
10-1 0.1
Disco, 1 m from speaker
100
2
10-2 0.01
Diesel truck, 10 m away
90
0.63
10-3 0.001
Kerbside of busy road, 5 m
80
0.2
10-4 0.0001
Vacuum cleaner, distance 1 m
70
0.063
10-5 0.00001
Conversational speech, 1 m
60
0.02
10-6 0.000001
Average home
50
0.0063
10-7 0.0000001
Quiet library
40
0.002
10-8 0.00000001
Quiet bedroom at night
30
0.00063
10-9 0.000000001
Background in TV studio
20
0.0002
10-10 0.0000000001
Rustling leaves in the distance
10
0.000063
10-1 1 0.00000000001
Threshold of hearing
0 MSU Physics 0.00002
231 Fall 2015
10-1 2 0.000000000001
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Example
A person living on campus (c) 300 m from the rail track
is tired of the noise of the passing trains. They decide to
move to Abbott (a) (3.5 km from the rail track). If the
Sound level of the trains was originally 70dB
(vacuum cleaner), what is the sound level at Abbott?
Campus (c): c = 70 dB = 10log (Ic/I0)
Ic= 107 I0 = 10-5 W/m2
Ia/Ic = rc2/ra2 = 0.0073
Ia = Ic (rc2/ra2) = 7.3x10-8 W/m2
Sound level: a = 49 dB (normal conversation)
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example
A machine produces sound with a level of 80dB. How
many machines can you add before exceeding 100dB?
1 machine
80 dB=10log(I/I0)
8=log(I/I0)=log(I/1E-12)
108=I/1E-12
I1=10-4 W/m2
N machines
100 dB=10log(I/I0)
10 = log(I/I0)=log(I/1E-12)
1010 = I/1E-12
IN=10-2 W/m2
I1/IN = 10-4/10-2 = 1/100
The intensity must increase by a factor of 100;
one needs to add 99 machines.
Shortcut: x = 20 so the increase in I is 10(x/10) = 100.
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Clicker Quiz
The speed of sound in a material does NOT depend on:
a)
b)
c)
d)
e)
The density of the material
The frequency of the sound
The temperature of the material
The pressure on the material
None of the above
The speed of sound depends on the density of the material:
higher density leads to lower sound speed!
The density and rigidity depend on the temperature of and
the pressure on the material.
Higher frequency means lower wavelength (and vice versa).
These properties are determined by the speed of sound in
the material.
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question
An ambulance is moving towards you with its sirens on. The
pitch of the sound you hear is .......... than the pitch
you would hear if the ambulance were not moving at all.
a) higher
b) the same
c) lower
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Doppler effect: a non-moving
source
vsound
source
f = vsound/
you

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doppler effect: a source moving towards you
the distance between
the wave front is
shortened
v  v sound
vsource
source
v s  v source
vs
v vs
     
f
f
f
observer
 v
v
f
 f 

 v  vs



Prime (’): heard observable
The frequency becomes larger: higher tone
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Doppler Effect: a source moving away from you
the distance between
the wave front becomes longer
vsource
observer
source
vs
v vs
     
f
f
f
 v 
v

f
 f 

 v  vs 
v s is negative in this case
The frequency becomes lower: lower tone
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Doppler Effect: you moving towards the source
Equivalent to increasing the velocity
vsound
f 
f '
source
observer
v

(v  vo )

vo  vobserver
 v  vo 
f f 

 v 

If you move away from source use vo < 0
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Doppler Effect: In General
source
you
 v  vo
f   f 
 v v s



vo =vobserver: positive if moving towards to source
vs = vsource: positive if moving towards the observer
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question
An ambulance is moving towards you with its sirens on. The
frequency of the sound you hear is …… than the frequency
you would hear if the ambulance were not moving at all.
a) higher
b) the same
c) lower
 v  vobserver 

f   f 
 v  vsource 
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applications of doppler effect: weather
radar (radio waves – electromagnetic)
Both humidity (reflected intensity) and speed of clouds
(doppler effect) are measured.
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example
A police car using its siren (frequency 1200Hz) is driving
west towards you over Grand River with a velocity of 25m/s.
You are driving east over grand river, also with 25m/s.
a) What is the frequency of the sound from the siren that
you hear? b) What would happen if you were also driving west
(behind the ambulance)? v = vsound = 343 m/s
a)
b)
 v  vo 

f   f 
 v  vs 
 343  25 
f   1200

 343  25 
(1200)(1.16)  1389 Hz
 v  vo
f   f 
 v  vs



 343  25 
 
f   1200
 343  (25) 
(1200)(1)  1200 Hz
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
 1 
  sin   where M is the Mach number and  is the mach angle
M 
speed of the plane
M 
speed of sound
1
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Interference: two wave same frequency
Constructive interference:
maxima line up. Waves are “in
phase”
Time (t)
Destructive interference:
maxima lines up with minimum.
Waves are “out of phase” by ½
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Interference
Two traveling pulse waves pass through each other
without affecting each other. The resulting displacement
is the superposition of the two individual waves.
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Interference: two different frequencies (beats)
Amplitude of the “beat” changes with time, so the intensity
of the sound changes as a function of time. fbeat = |fA-fB|
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An Example: two speakers
Two speakers are placed 10m apart, facing each other. Each
speaker is playing a pure tone (ie, 1 frequency) with the same
amplitude. A student notices that the first speaker is making a
tone of 340 Hz and that at 6m from this speaker, there is a
minimum in sound intensity. What are the possible frequencies for
the second speaker? (vsound = 340 m/s)
1
d 2  ( N  ) 2  4
2
N  0,1,2,3...
2  8, 8 / 3, 8 / 5...
f2 
v
2
 42.5, 127.5, 212.5....
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Interference: Standing Wave
If two waves travel in opposite directions
and v1=v2, the superposition of the two
waves produces a standing wave:
maxima and minima always appear at the
same location
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String: standing Wave
A string fixed at two ends can
support different constructive
resonances.
Requires that there is
constructive interference: path
length difference between
NODES must be ½.
Node = point in the resonance
with zero amplitude.
 n  1
L

 2 
 2 
 
L
 n 1
n  0,1,2,3...
n  0,1,2,3....
(n  1) is the harmonic number
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Tube: standing Wave
 2 
 L
 n 1
 
n  0,1,2,3....
(n  1) is the harmonic number
 4 
 L
 2n  1 
 
n  0,1,2,3....
(2n  1) is the harmonic number
 2 
 L
 n 1
 
n  0,1,2,3....
(n  1) is the harmonic number
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Standing Wave
Just like with sound, the velocity of the
standing wave depends on the density of
the material.
Linear mass density of a string:
μ = mass/length
Also depends on the string’s tension: T
𝑣=
𝑇
𝜇
Power transmitted
by a wave on a string
1
2 2
P   v A
2
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An Example
A 1-m-long piano wire has a mass of 1 gram and is under a tension of
160 N.
(a) Find the wave speed for this string.
(b) If you want to tune this wire to make middle C (f = 256 Hz) the
fundamental frequency, what should the wire tension be?
v
T

  2L
 400
v  Lf 
T

T  (2 Lf ) 2   262
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y ( x, t )  A sin (kx   t )
v( x, t )   A cos(kx   t )
2

T
2
k
v


T
(in direction of motion)
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Ruben’s tube with propane gas in a tube with a length of L=2.1 m
Resonance frequencies are observed at 742 and 680 Hz
Find the speed of sound in propane and the first harmonic
frequency.
From the equation below v = 2L*62 = 260 m/s and f1 = 62 Hz
( n 1)
f ( n 1)
 2 

n  0,1,2,3....
L
 n 1
v
v


(n  1)
n  0,1,2,3...
n 2 L
f ( n 1)  f ( n 1)
v

 742  680  62
2L
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