Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Chapter 12: Sound
•Sound Waves
•The Speed of Sound
•Amplitude & Intensity of Sound Waves
•Standing Sound Waves
•Beats
•The Doppler Effect
•Shock Waves
•Echolocation
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
1
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§12.1 Sound Waves
Sound waves are longitudinal. They can be represented
by either variations in pressure (gauge pressure) or by
displacements of an air element.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
2
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The middle of a
compression (rarefaction)
corresponds to a pressure
maximum (minimum).
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
3
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§12.2 The Speed of Sound Waves
The speed of sound in different materials can be determined
as follows:
v
B
In thin solid rods v 
Y
In fluids


B is the bulk modulus of the
fluid and  its density.
Y is the Young’s modulus of
the solid and  its density.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
4
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
In ideal gases
v  v0
T
T0
Here v0 is the speed at a temperature T0 (in kelvin) and v
is the speed at some other temperature T (also in kelvin).
For air, a useful approximation to the above expression is
v  331  0.606TC  m/s
where Tc is the air temperature in C.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
5
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Materials that have a high restoring force (stiffer) will have a
higher sound speed.
Materials that are denser (more inertia) will have a lower
sound speed.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
6
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 12.8): A copper alloy has a Young’s
Modulus of 1.11011 Pa and a density of 8.92 103 kg/m3.
What is the speed of sound in a thin rod made of this alloy?
1.11011 Pa
v

 3500 m/s
3
3

8.9 10 kg/m
Y
The speed of sound in this alloy is slightly less than the
value quoted for copper (3560 m/s) in table 12.1.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
7
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 12.1): Bats emit ultrasonic sound
waves with a frequency as high as 1.0105 Hz. What is the
wavelength of such a wave in air of temperature 15.0 C?
v  331  0.606TC  m/s
The speed of sound in
air of this temperature
is 340 m/s.
v
340 m/s
3
 

3
.
4

10
m
5
f 1.0 10 Hz
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
8
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 12.10): A lightning flash is seen in the
sky and 8.2 seconds later the boom of thunder is heard.
The temperature of the air is 12.0 C.
(a) What is the speed of sound in air at that temperature?
v  331  0.606TC  m/s
The speed of sound in
air of this temperature
is 338 m/s.
(b) How far away is the lightning strike?
d  vt  338 m/s 8.2 s  2800 m  2.8 km
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
9
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
The speed of light is 3.00105 km/s. How long does it
take the light signal to reach the observer?
d
2.8 km
-6
t 

9
.
3

10
sec
5
v 3.0 10 km/s
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
10
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§12.3 Amplitude & Intensity of
Sound Waves
For sound waves:
Ip
Is
2
0
2
0
p0 is the pressure amplitude and
s0 is the displacement amplitude.
The intensity of sound waves also follow an inverse
square law.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
11
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Loudness of a sound is measured by the logarithm of the
intensity.
The threshold of hearing is at an intensity of 10-12 W/m2.
Sound intensity level is defined by
I
  10dB log
I0
dB are decibels
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
12
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 12.12): The sound level 25 m from a
loudspeaker is 71 dB. What is the rate at which sound
energy is being produced by the loudspeaker, assuming it to
be an isotropic source?
I
 71 dB
Given:   10dB log
I0
Solve for I, the intensity of a sound wave:
I
log  7.1
I0
I
 107.1
I0



I  I 0107.1  10 12 W/m 2 107.1  1.3 10 5 W/m 2
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
13
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
The intensity of an isotropic source is defined by:
P
I
4r 2
P  I 4r 2
 (1.3 10 5 W/m 2 )4 25 m 
2
 0.10 Watts
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
14
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: Two sounds have levels of 80 dB and 90 dB.
What is the difference in the sound intensities?
I
1  10dB log  80 dB
I0
Subtracting:
I
 2  10dB log  90 dB
I0

I2
I1 
 2  1  10 dB  10 dB log  log 
I0
I0 


I2 
10 dB  10 dB log 
I1 

I2
 101
I1
I 2  10 I1
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
15
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§12.4 Standing Sound Waves
Consider a pipe open at both ends:
The ends of the pipe are open to the atmosphere. The
open ends must be pressure nodes (and displacement
antinodes).
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
16
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The distance between two adjacent antinodes is ½. Each
pair of antinodes must have a node in between.
The fundamental mode (it has the fewest number of
antinodes) will have a wavelength of 2L.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
17
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The next standing wave pattern to satisfy the conditions at
the ends of the pipe will have one more node and one more
antinode than the previous standing wave. Its wavelength
will be L.
The general result for standing waves in a tube open at
both ends is
2L
n 
n
v
where n=1, 2, 3,…
nv
fn 

 nf1
n 2 L
f1 is the fundamental frequency.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
18
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
19
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Now consider a pipe open at one end and closed at the
other.
As before, the end of the pipe open to the atmosphere
must be a pressure node (and a displacement
antinode).
The closed end of the pipe must be a displacement
node (and a pressure antinode).
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
20
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
One end of the pipe is a pressure node, the other a
pressure antinode. The distance between a consecutive
node and antinode is one-quarter of a wavelength.
Here, the fundamental mode will have a wavelength of 4L.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
21
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The next standing wave to satisfy the conditions at the ends
of the pipe will have one more node and one more antinode
than the previous standing wave. Its wavelength will be
(4/3)L.
The general result for standing waves in a tube open at
one end and closed at the other is
4L
n 
n
v
where n=1, 3, 5,…. n (odd values only!!)
nv
fn 

 nf1
n 4 L
f1 is the fundamental frequency.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
22
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
23
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 12.22): An organ pipe that is open at
both ends has a fundamental frequency of 382 Hz at
0.0 °C. What is the fundamental frequency for this pipe at
20.0 °C?
At Tc = 0.0 °C, the speed of sound is 331 m/s.
At Tc = 20.0 °C, the speed of sound is 343 m/s.
The fundamental frequency is
v
v
f1  
1 2 L
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
24
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
The ratio of the fundamental frequencies at the two
temperatures is:
f1, 20
f1, 0
f1, 20
v20
v20
2
L


 1.04
v0
v0
2L
 1.04 f1, 0  396 Hz
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
25
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
How long is this organ pipe?
Using either
set of v and f1.
v
f1 
2L
v
L
 0.43 m
2 f1
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
26
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§12.5 Beats
When two waves with nearly the same frequency are
superimposed, the result is a pulsation called beats.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
27
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Two waves of
different
frequency
Superposition
of the above
waves
The beat frequency is f  f1  f 2
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
28
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
If the beat frequency exceeds about 15 Hz, the ear will
perceive two different tones instead of beats.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
29
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§12.6 The Doppler Effect
When a moving object emits a sound, the wave crests
appear bunched up in front of the object and appear to be
more spread out behind the object. This change in wave
crest spacing is heard as a change in frequency.
The results will be similar when the observer is in motion
and the sound source is stationary and also when both the
sound source and observer are in motion.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
30
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
31
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The Doppler Effect formula
 vo
1
v
fo  
 1  vs

v



 fs



fo is the observed frequency.
fs is the frequency emitted by the source.
vo is the observer’s velocity.
vs is the source’s velocity.
v is the speed of sound.
Note: take vs and vo to be positive when they move in the
direction of wave propagation and negative when they are
opposite to the direction of wave propagation.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
32
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 12.39): A source of sound waves of
frequency 1.0 kHz is stationary. An observer is traveling at
0.5 times the speed of sound.
(a) What is the observed frequency if the observer
moves toward the source?
fo is unknown; fs= 1.0 kHz; vo = -0.5v; vs = 0; and v is
the speed of sound.
 vo
1
v
fo  
 1  vs

v


  0.5v 

1

v  f  1.5 f  1.5 kHz
 fs  
s
0

 1




v 


Copyright © 2008 – The McGraw-Hill Companies s.r.l.
33
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(b) Repeat, but with the observer moving in the other
direction.
fo is unknown; fs= 1.0 kHz; vo = +0.5v; vs =0; and v is
the speed of sound.
 vo
1
v
fo  
 1  vs

v


  0.5v 

1

v  f  0.5 f  0.5 kHz
 fs  
s
0

 1




v 


Copyright © 2008 – The McGraw-Hill Companies s.r.l.
34
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§12.7 Shock Waves
If a plane were traveling at the
speed of sound , what would
the wave crests looks like?
They would be bunched up in
front of the aircraft and an
observer (to the right) would
measure =0.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
35
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
If the source moves with a speed greater than that
of sound, then the wave crests pile up on top of
each other forming a cone-shaped shock wave.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
36
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§12.8 Echolocation
Sound waves can be sent out from a transmitter of some
sort; they will reflect off any objects they encounter and can
be received back at their source. The time interval between
emission and reception can be used to build up a picture of
the scene.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
37
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 12.47): A boat is using sonar to
detect the bottom of a freshwater lake. If the echo from a
sonar signal is heard 0.540 s after it is emitted, how deep is
the lake? Assume the lake’s temperature is uniform and at
25 C.
The signal travels two times the depth of the lake so
the one-way travel time is 0.270 s. From table 12.1,
the speed of sound in freshwater is 1493 m/s.
depth  vt
 1493 m/s 0.270 s   403 m
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
38
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 12.49): A bat emits chirping sounds of
frequency 82.0 kHz while hunting for moths to eat. If the bat
is flying toward a moth at a speed of 4.40 m/s and the moth
is flying away from the bat at 1.20 m/s, what is the frequency
of the wave reflected from the moth as observed by the bat?
Assume T = 10.0 C.
v  331  0.606TC  m/s
The speed of sound in
air of this temperature
is 337 m/s.
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
39
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
The flying bat emits sound of f =82.0 kHz that is
received by a moving moth. The frequency observed
by the moth is:
 vo
1
v
fo  
 1  vs

v

 1.2 m/s



 1
337 m/s
 fs  

 1   4.4 m/s


337 m/s




82.0 kHz   82.8 kHz



Copyright © 2008 – The McGraw-Hill Companies s.r.l.
40
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
Some of the sound received by the moth will be reflected
back toward the bat. The moth becomes the sound
source (f = 82.8 kHz) and the bat is now the observer.
 vo
1
v
fo  
 1  vs

v


  4.4 m/s

1
337 m/s
 fs  

 1   1.2 m/s


337 m/s




82.8 kHz   83.6 kHz



Copyright © 2008 – The McGraw-Hill Companies s.r.l.
41
Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Summary
•Sound is a longitudinal wave.
•The speed of sound depends on material properties
such as “stiffness”, density, and temperature.
•Sound Intensity Level
•Standing Waves in Pipes (both ends open & one end
open/one end closed)
•The Doppler Effect
•Shock Waves
•Echolocation
Copyright © 2008 – The McGraw-Hill Companies s.r.l.
42