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Chapter 19 Sound waves
19-1 Properties of Sound waves
19-2 Traveling sound waves
19-3* The speed of sound
19-4 Power and intensity of sound waves
19-5 Interference of sound waves
19-6* Standing longitudinal waves
19-7* Vibrating system and sources of sound
19-8 Beats
19-9 The Doppler effect
19-1 Properites of sound waves
When we discuss sound waves, we normally
mean longitudinal wave in the frequency range 20
Hz to 20,000 Hz, the normal range of human hearing.
For simplification, we will consider the sound
wave in 1D case.
In Fig19-2, as the piston
(活塞) moves back and
forth, it alternately
compresses and expands
(使稀薄)) the air next to it.
This disturbance travels
down the tube as a sound
wave.
( x, t )  0  ( x, t )
P( x, t )  P0  P( x, t )
0
Under
certain
conditions
 m
x
p m
P0
x
Sm
x
It is convenient to use density
and pressure to describe the v
properties of fluids.
x
um
x
Fig19-2
19-2 Traveling sound waves
1) Let us assume that the pistol is driven so that the
density and pressure of air in the tube will vary as a
sine function.
   m sin( kx  t )
(19-1)
P  Pm sin( kx  t )
(19-2)
2) What’s the relationship between  and P ?
From the definitions of bulk modulus(体模量/膨胀系数)
(Eq(15-5)) B   P and density   m , when m is fixed, we
v
(v )
have
v
Δρ  
ρ0
m
m Δv
Δv
ΔP
Δ
v




ρ
(
)

ρ

Δ
P
v2
v v
v
B
B
or  m  Pm
0
(19-3)
B
3) How to find the displacement of an element of gas
inside the tube?
The undisturbed density of x is
x
x=0
x
x'
m
A is the corss0 
s(x+x,t)
sectional area.
Ax
s(x,t)
x'  x''-x'  [ x  x  s(x  x,t)]  [ x  x  s(x,t)]
s( x  x, t )  s ( x, t )
 δx[1 
]
x’ x’’ x
x
s
 δx[1 
ρ
x
]
ρ0
δm

 ρ0 (1-s/x), if s/x  1.
Ax' 1  s/x
    0  0 s / x
(19-6)
Combine Eqs. (19-1) and (19-6), we have:
 m
s
( x, t )


sin( kx  t )
x
0
0
s( x, t )  s m cos( kx  t )
(19-8)
 m Pm
sm 

k 0
kB
s
u x ( x,t) 
 um sin (kx  ωt ) (19-9)
t
vPm
um 
B
μ x (x,t) is the velocity of oscillation of an element in fluids.
v=  /k is the velocity of sound wave.
19-3* The speed of sound
As in the case of the transverse mechanical wave,
the speed of a sound wave depends on the ratio of
an elastic property of the medium and an inertial
property. For a 3D fluid,
B
(19-4)
v
0
vair  343m / s, 20 C
Note:1) B is the bulk modulus,  0 is the mass density.
2) Use Newton’s law for a system of particles.
(  Fext , x  Macm, x)
19-4 Power and intensity of sound waves
u x ( x,t)  um sin (kx  ωt )
As the wave travels, each fluid element exerts a
force on the fluid element ahead of it. If the
pressure increase in the fluid element is P ,
Fx  A  P  A  Pm sin( kx  t )
The power delivered by the element is:
P  u x Fx  A  Pmum sin 2 (kx  t )
A(Pm ) 2
(19-18)
Pav 
2 v
Average over any number of full cycles.
Intensity I:
Pav (Pm ) 2
I

A
2 v
(19-19)
The response of the ear to sound of increasing
intensity is approximately logarithmic.
One can define a logarithmic scale of intensity
called the “sound level SL”
I
SL  10 log
I0
(19-20)
Where I 0 is a reference intensity, which is
chosen to be 10 12 w / m 2 (a typical value for the
threshold of human hearing(听觉阈)).
The unit of the sound level is “decibels” (dB).
A sound of intensity I 0 (听觉阈)has a sound level
of 0 dB.
The sound at the upper range of human hearing,
called the threshold of pain (痛觉阈) has an
intensity of 1w / m 2 and a SL of 120 dB.
几种声音近似的声强、声强级和响度
声源
声强W/m2
声强级dB
响度
引起痛觉的声音
1
120
钻岩机或铆钉机
10-2
100
震耳
交通繁忙的街道
10-5
70
响
通常的谈话
10-6
60
正常
耳语
10-10
20
轻
树叶的沙沙声
10-11
10
极轻
引起听觉的最弱声音
10-12
0
Sample problem 19-2
Spherical sound waves are emitted uniformly in all
directions from a point source, the radiated power
P being 25 w. What are the intensity and the sound
level of the sound wave at a distance r=2.5m from
the source?
Solution:
P
25w
2
I


0
.
32
w
/
m
4r 2 4 (2.5m) 2
I
0.32w / m 2
SL  10 log
 10 log 12
 115dB
2
I0
10 w / m
19-5 Interference of sound waves
Fig19-6 shows two loudspeakers
driven from a common source.
At point P the pressure variation
due only to speaker s1 is P1 and
that due to s 2 alone is P2 . The
total pressure disturbance at
point P is P  P1  P2
.
Fig 19-6
s2
r1
source
s1
The type of interference that occurs at point P
depends on the phase difference between the
waves.
r2
P
y ( x, t )  y1 ( x, t )  y2 ( x, t )
P  sin( kr1  t )  [2 ym cos( / 2)] sin( kx  t   ' )
The phase difference:
 | kr1  kr2 | k | r1  r2 |
 L

2

2

L, L  r1  r2
(19-22)
When L  m ( m=0,1,2,…...) (19-23),
The intensity reaches a maximum value, forming
constructive interference.
1
When L  (m  ) destructive interference occurs.
2
The intensity has a minimum value.
19-6* Standing longitudinal waves
Fig 19-7
We assume a train of
sine waves travels down
a tube( Fig19-7).
1) If the end is open,
the wave at the end will
behave as a pressure
node (波节);
2) If the end is closed, a
pressure antinode (波腹)
will form at the end.
2L
n 
n
, n=1,2,3…
4L
n 
n
See动画库\波动与光学夹\2-16纵驻波
(a)
open
end
(b)
(c)
close
end
(d)
, n=1,3,5...
Notes:
a).For open end, the longitudinal pressure wave is reflected

with a phase change of180 , because the pressure at the
open end must at the value P0 , same as the environment’s.
In this case, it likes the string fixed at both ends.
b).For the closed end, the pressure can vary freely.
c).The superposition of the original and reflected
waves gives a pattern of standing waves.
d). Resonance can happen, when the driving frequency
matches one of the natural frequency of the system, which
are determined by the length of the tube (L).
19-7* Vibrating system and sources
of sound
We have already studied the propagation of the
sound wave, and now to understand the nature of
the sound we must study the vibration system that
produces it.
We can classify musical instruments into three
categories: those based on vibration string; those
based on vibration column of air, and more
complex system including plates, rods, and
membranes.
(a)The vibrating system has a large number of
natural vibrational frequencies. We write these in
ascending order, so that f1  f 2  f 3       .
The lowest frequency, f1 is called the “fundamental
frequency(基频)”, and the corresponding mode of
oscillation is called the “fundamental mode”. The
higher frequencies are called “overtones(泛音)”, f 2
with being the first overtone, f 3 the second
overtone, and so on. In some systems:
f n  nf1
(b) Why do some vibrating systems produce
pleasant sounds while others produce harsh (刺耳的)
or discordant (不和谐) sounds?
When several frequencies are heard simultaneously,
a pleasant sensation results if the frequencies are in
the ratio of small whole numbers(整数), such as 3:2
or 5:4.
19-8 Beats (节拍)
• Previously, we have considered the “interference in space”.
• Now we shall discuss “interference in time”.
• We consider two waves which have nearly the same
frequency.
P1 (t )  Pm sin 1t
(1 ~ 2 )
P2 (t )  Pm sin  2 t
We have chosen the phase constants to be zero, and
same amplitudes.
The resultant pressure is
(1   2 )
(1   2 )
P(t )  P1 (t )  P2 (t )  [2Pm cos
t ] sin
t
2
2
Set  av  1  2
 amp
2
1   2

2
(19-32)
(19-33)
(19-34)
P(t )  2Pm cos(ampt ) sin( avt )
(19-35)
P(t )  2Pm cos(ampt ) sin( avt )
In Fig19-13, the ear
would perceive a tone
at a frequency av .
 av  1   2
1 ~ 2
 av ~ 1 ~ 2
P1 (t )
t
P2 (t )
(a)
Since 1 ~ 2 , the
amplitude frequency
 amp is small. The
amplitude | 2Pm cos(ampt ) |
(b)
fluctuates slowly.
t
t
Fig 19-13
P(t )  2Pm cos(ampt ) sin( avt )
A beat--- that is, a maximum intensity—occurs,
whenever cos( amp t ) equals +1 or -1 ,since the
intensity depends on the square of the amplitude.
Each of these values occurs once in each cycle of
the envelope, thus
beat  2amp  1  2 (19-36)
Sample Problem 19-5
A violin string that should be tuned to concert A (440Hz) is
slightly mistuned. When the violin string is played in its
fundamental mode along with a concert A tuning fork, 3
beats per second are heard. (a) What are the possible
values of the fundamental frequency of the string? (b)
Suppose the string were played in its first overtone
simultaneously with a tuning fork with 880Hz. How many
beats per second would be heard? (c) When the tension of
the string is increased slightly, the number of beats per
second in the fundamental mode increases. What was the
original frequency of the fundamental?
19-9 The Doppler effect
In a paper written in 1842, Doppler (1803~1853)
called attention to the fact that the color of a
luminous body must be changed by relative motion
of the body and the observer. This “Doppler effect”
as it is called, applies to waves in general.
See动画库\波动与光学夹\2-21Doppler Effect A.exe
1.Moving observer, source at rest
Suppose the source and observer move along the
line joining them.
Let us adopt a reference frame at rest in the medium
through which the sound travels.
Fig19-14 shows a source of sound S at rest and an
observer O moving toward the source at a speed v0 .


v0
vs  0
*
S

0
Fig 19-14
If the frequency of wave is f, what is the
actually one f ’ heard by the ear?
vt
An observer at rest in the medium would receive  waves
in time t, where v is the speed of sound in the medium and 
is the wavelength.
Because of the motion toward the source, the observer
receives v0 t additional waves in the same time t.

'
The frequency f that is actually heard is
f 
'
vt  v0 t

t
  v  v0  v  v0 f

v
(19-38)
* When the observer is in motion away from
v  v0
the source,
'
f 
f
(19-39)
v
2. Moving source, observer at rest
In this case, the wavelength is shortened from 
to  '    vsT .
See动画库\波动与光学夹\2-20多普勒效应 3

v0  0

vs
1 2
3
4
5
……
6 S1
S7
'
Fig 19-15
 '    vsT
The frequency of the sound heard by the observer
Is given by vt / '
v
'
f 
(
)f
(19-40)
t
v  vs
* If the source moves away form the observer, the
frequency heard is
v
'
f (
)f
(19-41)
v  vs
3. If both source and observer move through the
transmitting medium,
v  v0
f (
)f
v  vs
'
(19-44)
Where the upper signs (+ numerator, -denominator)
correspond to the source and observer moving toward the
other and the lower signs in the direction away from the
other.
4. If a source of sound is moved away from an
observer and toward a wall, the observer hears
two notes (音符) of different frequency. The note
heard directly from the receding source is lowered
in pitch by the motion. The other note is due to the
waves reflected from the wall, and this is raised in
pitch. The superposition of these two wave trains
produces beats.
A similar effect occurs if a wave from a stationary
source is reflected from a moving object. The beat
frequency can be used to deduce the speed of the
object. This is the basic principle of radar monitors,
and it is also used to track satellites.

How about the wavefront if v is larger
s
than v, sound speed?

vt
vs
4 56
3
12
.
P2
P1
vs t

Wavefront when vs  v .

Wavefront when vs  v.
See动画库\波动与光学夹\2-30冲击波
Doppler cooling in Bose-Einstein Condensate (BEC)
Predicted in 1924...
A. Einstein
S. N. Bose
...Created in 1995，Won Nobel Prize in 2001
E. A. Cornell
•BEC: A given number of
particles approach each other
sufficiently closely and move
sufficiently slowly they will
together convert to the lowest
energy quantum state.
W. Ketterle
C. E. Wieman
What’s the conditions to observe BEC?
Extremely low temperature;
The atoms are still in gas state
•Doppler cooling (Laser cooling)
10-6 K obtained
•Evaporative cooling
10-9 K obtained
Sample problem 19-6
The siren （警报器）of a police car emits a pure tone at a
frequency of 1125 Hz. Find the frequency that you
would perceive in your car.
(a) your car at rest, police car moving toward you at 29 m/s;
(b) police car at rest your moving toward it at 29 m/s
(c) you and police car moving toward one another at 14.5 m/s
(d) you moving at 9 m/s, police car chasing behind
you at 38 m/s
Solution: Using Eq(9-44)
(a) Here v0  0 v s  29m / s
v
343
f 
f (
) f  1229 Hz
v  vs
343  29
'
v0  29m / s
(b) v s  0
v  v0
343  29
'
f 
f 
(1125 Hz )  1220 Hz
v
343
(c) vs  v0  14.5m / s
v  v0
343  14.5
f 
f 
1125  1224 Hz
v  vs
343  14.5
'
(d) v0  9m / s
vs  38m / s
343  9
f 
f  1232 Hz
343  38
'