PHYSICS
CHAPTER 11
CHAPTER 11:
Sound Wave
(3 Hours)
1
PHYSICS
CHAPTER 11
Learning Outcome:
11.1 Intensity and beats (1 hour)
At the end of this chapter, students should be able to:

Define sound intensity.

Ilustrate and use the dependence of intensity on:
amplitude : I 1 A²
1
I


distance from a point source :
r2

Use the principle of superposition to explain beats.
Use the formulae for beat frequency,


fb  f1  f 2
to solve related problems.
2
PHYSICS
CHAPTER 11
Intensity

is defined as the rate of sound energy flow across unit area,
perpendicular to the direction of the sound propagation.
OR
E
P
OR I 
I
tA
where
A
I : sound intensity
E : sound energy
A : area perpendicu lar to the direction of
sound propagatio n
t : time
P : sound power


It is a scalar quantity.
Its unit is watt per squared metre (W m2).
3
PHYSICS

CHAPTER 11
The factors influence the value of sound intensity are
 the amplitude of the sound i.e.
I  A2

where A : amplitude of the sound
the distance from the source of sound i.e.
1
I 2
r
where
wave
r : distance from the sound source
Therefore
A2
I 2
r

A sound wave flows out from the source in all directions, hence
it is a three-dimensional wave and is said to be a spherical
wave as shown in Figure 11.1.
4
PHYSICS

CHAPTER 11
Figure 11.1
Consider two points at distances r1 and r2 from the source
(power, P is constant), then (refer equation pg 3)
P
I1 
4r12
So
and
P
I2 
4r22
2
1
2
2
I2 r

I1 r
5
PHYSICS
CHAPTER 11
Graphs in Figure 11.1 show the variation of sound intensity.
(a) I
(b) I

0
(c) I
Figure 11.1
0
r2
1
r2
0
(d) I
A
0
6
A2
PHYSICS
CHAPTER 11
Example 11.1 :
A loudspeaker radiates sound waves uniformly in all directions. At
a distance 3 m the intensity of the sound is 0.85 W m-2. Calculate
a. the power of loudspeaker,
b. the sound intensity at distance 6 m from the source.
Solution : r  3 m; I  0.85 W m 2
1
a. From the definition of the sound intensity, thus
P
2
I  and A  4r1
A
P
P
I
0.85 
2
2
4r1
4 3
P  96.1 W
b. Given r2  6 m, thus
96.1
P
I
I
2
2
4 6 
4r2
I  0.212 W m 2
7
PHYSICS
CHAPTER 11
Superposition and beats
Beats




Beats is defined as the periodic variation in amplitude of sound
at a given point due to superposition of two sound waves
having slightly different
frequencies.
This phenomenon occurs because
 the frequencies of both sound waves slightly different.
 the phase difference change over time.
When the beats are occurred, a listener will hear a periodic
rises and falls in loudness (intensity) of the sound wave.
The phenomenon of beats can be used
 to measure the unknown frequency of a note.
 to “tune” an instrument to a given note.
Beat
8
PHYSICS
CHAPTER 11
y
(a)
0
t
y
(b)
0
t
y
A
(c)
0
Beat period (T)
C
B
A,C : Constructive interference
B,D : Destructive interference
t
D
Beat period (T)
Figure 11.2
9
PHYSICS
CHAPTER 11
Explanation of the Figure 11.2
 At point A and C, the two waves are shown to be in phase and
interfere constructively where the amplitude of the resultant
wave is maximum (increases in loudness of the sound).
 At point B and D, the two waves are shown to be anti-phase and
interfere destructively where the amplitude of the resultant wave
is minimum (decreases in loudness of the sound).
 Therefore the phenomenon of the increases and decreases in
loudness are periodic and is called beat.
 The equation of the beat frequency is given by
1 1
OR
fb 

T2 T1
f b : Beat frequency
f1 : frequency of the first source
f 2 : frequency of the second source
f b  f1  f 2
where

If the beat frequency, fb = 3 Hz means that a listener would hear
3 beats in one second.
10
PHYSICS
CHAPTER 11
Example 11.2 :
Two sound waves with wavelengths 66 cm and 44 cm undergo
superposition. If both of the waves have the same speed of
330 m s1. Calculate a number of beats is produced in one second.
1
Solution : 1  0.66 m; 2  0.44 m; v  330 m s
By applying the equation of wave speed,
v  f
then
v
330
f1  
1 0.66
v
330
f2 

2 0.44
f1  500 Hz
f 2  750 Hz
The beat frequency is given by
f b  f 2  f1  750  500
f b  250 Hz
Therefore the number of beats produce in one second is 250
11
PHYSICS
CHAPTER 11
Example 11.3 :
Two tuning forks A and B make 3 beats per second when sounded
together. If the frequency of tuning fork A is 305 Hz, Determine the
possible values of the frequency for tuning fork B.
Solution : f  3 Hz; f  305 Hz
b
A
By applying the equation of beat frequency,
fb  f A  f B
f B  f A  fb
f B  305  3  308 Hz
OR
f B  305  3  302 Hz
12
PHYSICS
CHAPTER 11
Exercise 11.1 :
1. A tuning fork of unknown frequency makes three beats per
second with a standard fork of frequency 384 Hz. The beat
frequency decreases when a small piece of wax is put on a
prong of the first fork. Calculate the frequency of this fork.
ANS. : 387 Hz
2. y1
Tuning fork A
0
y2
0
0.05 t (s)
Tuning fork B
0.05 t (s)
Figure 11.3
Figure 11.3 shows the displacement-time graph for tuning fork A
and B. Both forks vibrate in phase at t = 0 , and then out of
phase and finally back in phase at time t = 0.05 s later.
Calculate the beat frequency produced when the two tuning
forks are sounded together.
13
ANS. : 20 Hz
PHYSICS
CHAPTER 11
Learning Outcome:
11.2 Application of Stationary waves (1 hour)
At the end of this chapter, students should be able to:

Use diagram to explain the formation of stationary waves
along

stretched string

air columns (open and closed end)
and use the equations to determine the fundamental and
overtone frequencies.

Explain qualitatively the formation of resonance in air
column.
14
PHYSICS
CHAPTER 11
11.2 Stationary waves
11.2.1 Wave on a stretched string
Wave speed on the string
 The equation of the wave speed on the string is given by
where
v

T

v : wave speed on the string
T : tension in the string
 : mass per unit length of the string
Its value depends on

the tension in the string, T
the mass per unit length of the string, 
The S.I. unit for



T : newton (N)

 : kg m1
15
PHYSICS

CHAPTER 11
The value of  is obtained by using two method:

If the length of the string is l and its mass, m thus
m

l

If the radius of the string is r and its density,  thus
where
  A A : cross sectional area
2
 πr
OR
  r
2
16
PHYSICS
CHAPTER 11
Vibrational modes on a string fixed at both ends
Incident wave



Reflected wave
Figure 11.4
When a string is plucked, a progressive transverse wave is
produced on the string.
This wave is travelling to the both fixed ends (incident wave)
and reflected (reflected wave) as shown in Figure 11.4.
The superposition of both waves making stationary
transverse wave and the simplest pattern of the stationary
wave on the string is shown in Figure 11.5.
N
A
Figure 11.5
N
17
PHYSICS





CHAPTER 11
From the Figure 11.5, both ends of the string as a node (N) and
the middle of the string as an antinode (A).
The string forms one segment (loop) and the pattern of this
vibration is called fundamental mode (first harmonic mode).
The frequency of fundamental mode is called fundamental
frequency (f0).
The stationary wave on the string forced the air vibrates and
produces a sound wave in the air.
If the string vibrating in the fundamental mode hence the
sound wave produced in the fundamental tone. Therefore
f 0 (sound wave)  f 0 (string vibration )


This phenomenon is called resonance.
Table 11.1 shows the harmonic series on a string fixed at both
ends.
18
Mode
PHYSICS
Figure
N
N
A
l
Fundamental
(1st harmonic)
N
A
N

2
  2l
l
A
overtone
(2nd harmonic)
 l
l
N A N A
N A
N
and
v
3
l
2
2l

3
l
Table 11.1

T

nv
f0 
2l
f1 
N
1st
2nd overtone
(3rd
harmonic)
Frequency
CHAPTER 11
v
v  f f 0 
Wavelength
v

nv
f1 
l
f1  2 f 0
f2 
v
 3nv
f2 
f2  3 f0
2l
19
PHYSICS

CHAPTER 11
In General,
nv
f 
2l
where
n T
f 
2l 
OR
f  nf 0
n  1,2,3,... (refer to a number of segment
or harmonic)
l = length of the string
T = tension of the string
µ = mass per unit length


All harmonics are allowed in vibrational mode of a string
fixed at both ends.
The expression above valid if the length of the string is
constant.
20
PHYSICS
CHAPTER 11
Example 11.4 :
A stretched wire of length 80.0 cm and mass 15.0 g vibrates
transversely. Waves travel along the wire at speed 220 m s1. Two
antinodes can be found in the stationary waves formed in between
the two fixed ends of the wire.
a. Sketch and label the waveform of the stationary wave.
b. Determine
i. the wavelength of the progressive wave which move along the
wire,
ii. the frequency of the vibration of the wire,
iii. the tension in the wire.
Solution : v  220 m s 1 ; l  0.800 m; m  15.0 103 kg
a.
N A N A N
0.800 m
21
PHYSICS
CHAPTER 11
Solution : v  220 m s 1 ; l  0.800 m;
b. i. From the figure in (a), thus
l 
m  15.0 103 kg
  0.800 m
ii. By applying the formula of wave speed, thus
v  f
220  0.800 f
f  275 Hz
iii. The mass per unit length of the wire is
m 15.0 103
 
l
0.800
  1.88 10 kg m
2
Therefore the tension in the wire is
v
T

T
220 
1.88 102
T  910 N
22
1
PHYSICS
CHAPTER 11
11.2.2 Wave in an air column
Closed Pipe (air column with one end closed)
Reflected wave
Incident wave
Figure 11.7


If the air in a pipe that is closed at one end is disturbed by a
source of sound (e.g. tuning fork), a progressive longitudinal
wave travels along the air column and is reflected at its end to
form a stationary longitudinal wave is shown in Figure 11.7.
The simplest pattern of the stationary wave was produced have
the node at the closed end while the antinode is at the open end
as shown in Figure 11.8.
23
PHYSICS
CHAPTER 11
N
A
l
Figure 11.8
where



l : length of the air column
When the frequency of the tuning fork coincides with the
fundamental frequency, f0 of the air column, resonance takes
place.
A sound of high intensity is produced at this frequency.
Table 11.9 shows the harmonic series in an air column with one
end closed (closed pipe).
24
PHYSICS
CHAPTER 11
Mode
Figure
Wavelength
N
A
l
Fundamental
(1st harmonic)
4
  4l
l
A
N
A
N
1st
overtone
(3rd harmonic)
l
N
A
A
N
2nd overtone
(5th harmonic)
l

N
A
3
l
4
4l

3
5
l
4
4l

5
Table 11.9
Frequency
v  f
f0 
where
v

v : speed of sound
v
f0 
4l
f1 
v

3v
f1 
4l
f1  3 f 0
f2 
v

5v
f2 
4l
f2  5 f0
25
PHYSICS

CHAPTER 11
In General,
nv
f 
4l
OR
f  nf 0
where n  1,3,5,... (odd numbers)
 Only odd harmonics are allowed in vibrational modes of an air
column in closed pipe.
 Examples of vibrational modes an air column in closed pipe for
musical instruments are flute and recorder.
Open Pipe (air column with both ends open)
Figure 11.10

If the air in a open pipe (both ends are open) is disturbed by a
source of sound (e.g. tuning fork) as shown in Figure 11.10, a
progressive longitudinal wave travels along the air column. 26
PHYSICS


CHAPTER 11
This wave will superposition with another progressive
longitudinal wave produced by the air outside the pipe and form
a stationary longitudinal wave.
The simplest pattern of the stationary wave was produced have
the antinode at the both open ends while the node is at the
middle of the pipe as shown in Figure 11.11.
N
A
A
l

Figure 11.11
Table 11.3 shows the harmonic series in an air column with
both ends open (open pipe).
27
PHYSICS
CHAPTER 11
Mode
Figure
Wavelength
N
A
A
Fundamental
(1st harmonic)
2
  2l
l
A
N
l

A
N
 l
overtone
(2nd harmonic)
l
A
N
A
N
2nd
overtone
(3rd harmonic)
l
A
N
A
v  f
3
l
2
2l

3
Table 11.12
f0 
where
v

v : speed of sound
v
f0 
2l
f1 
A
1st
Frequency
v

v
v
f1   2 
l
 2l 
f1  2 f 0
f2 
v

3v
f2 
2l
f2  3 f0
28
PHYSICS

CHAPTER 11
In General,
nv
f 
2l


OR
f  nf 0
where n  1,2,3,... (real numbers)
All harmonics are allowed in vibrational modes of an air
column in open pipe.
Example of vibrational modes an air column in open pipe for
musical instruments is clarinet.
End correction, c
 is defined as the distance from the antinode in stationary
wave to the open end of a pipe.
 It occurs because the antinode does not form exactly at the
open end of the pipe.
29
PHYSICS

CHAPTER 11
Closed pipe
N
A
l
l c 

c
4
  4l  c

v
f0 
4l  c 
After considering the end correction, the general equation of
the frequencies for closed pipe is given by
nv
f 
4l  c 
where
n  1,3,5,...
30
PHYSICS

CHAPTER 11
Open pipe
N
A
c
A
l

l  2c 
2
c
  2l  2c

v
f0 
2l  2c 
After considering the end correction, the general equation of
the frequencies for open pipe is given by
nv
f 
2l  2c 
where
n  1,2,3,...
31
PHYSICS
CHAPTER 11
Example 11.5 :
A tube 80 cm long is closed at one end. Resonance occurs and the
vibrating air column in the tube produces sound of frequency 1134
Hz. The fifth overtone mode is found in the air column.
(ignore the end correction)
a. Sketch and label the waveform of the air column.
b. Calculate
i. the speed of sound in air,
ii. the fundamental frequency.
Solution : l  0.80 m; f  1134 Hz
5
a. N
A
N
A
A N
A
N
N
A
A
N
0.80 m
32
PHYSICS
CHAPTER 11
Solution :
b. i. From the figure in (a), thus
11
l 
4
11
0.80  
4
  0.291 m
The speed of sound is
v  f 5
v  0.2911134
v  330 m s 1
ii. For 5th overtone of closed pipe, thus n  11
The fundamental frequency is given by
f  nf 0
f 5  11 f 0
1134  11 f 0
f 0  103 Hz
33
PHYSICS
CHAPTER 11
Example 11.6 :
A 3.00 m long air column is open at both ends. The frequency of a
certain harmonic is 500 Hz and the frequency of the next higher
harmonic is 557 Hz. Determine the speed of sound in the air
column. The end correction may be neglected.
Solution : l  3.00 m; f  500 Hz; f  557 Hz
1
2
By applying the formula for open pipe, thus
n1v
(1)
f1 
2l
n 2v
and n 2  n1  1
f2 
2l

n1  1v
f2 
2l
2lf1
Rearrange the eq. (1) : n 1 
v
(2)
(3)
34
PHYSICS
CHAPTER 11
Solution : l  3.00 m; f1  500 Hz; f 2
By substituting the eq. (3) into eq. (2), thus
 557 Hz
 2lf1 
 1v

v


f2 
2l
v  2l  f 2  f1 
v  23.00557  500
1
v  342 m s
35
PHYSICS
CHAPTER 11
Example 11.7 :
A uniform tube of length 60.0 cm stands vertically with its lower end
dipping into water. When the length above water is 14.8 cm and
again when it is 48.0 cm, the tube resounds to a vibrating tuning
fork of frequency 512 Hz. Calculate
a. the speed of sound,
b. the lowest frequency to which the tube will resound when it is
open at both end.
Solution : l  0.600 m; l  0.148 m; l  0.480 m;
t
f  512 Hz
1
2
a. The distance, l1 is the first distance
of the air in the tube resonate with
the tuning fork at fundamental mode.
Hence the next distance, l2 is refer
to the first overtone mode.
l1
Fundamental
mode
l2
1st overtone
36
PHYSICS
CHAPTER 11
Solution :
a.
lt  0.600 m; l1  0.148 m; l2  0.480 m;
f  512 Hz
l1  c 

4
3
l2  c 
4

(2) – (1): l2  l1 
2
Thus
(1)
(2)
0.480  0.148 
v  f
v  5120.664
v  340 m s 1

2
  0.664 m
37
PHYSICS
Solution :
CHAPTER 11
lt  0.600 m; l1  0.148 m; l2  0.480 m;
f  512 Hz
b. From the eq. (1), thus
0.664
0.148  c 
4
c  0.018 m
Therefore the lowest frequency of the open pipe is
v
f0 
2lt  2c 
340
f0 
20.600  20.018
f 0  267 Hz
38
PHYSICS
CHAPTER 11
Exercise 11.2 :
1. a. What is the frequency of the sound emitted by an openended organ pipe 1.7 m long when sounding its fundamental
frequency? (speed of sound in air = 340 m s1)
b. Two open-ended organ pipes are sounded together and 8
beats per second are heard. If the shorter pipe is of length
0.80 m, what is the length of the other pipe? You may ignore
any end corrections. (speed of sound in air = 320 m s1)
ANS. : 100 Hz; 0.83 m
2. When a tuning fork of frequency 256 Hz is sounded together
with a wire fixed at both ends emitting its fundamental
frequency, 6 beats are heard every second. When the prongs of
the tuning fork are lightly loaded, 4 beats are heard every
second.
a. Calculate the fundamental frequency of the wire.
b. If the length of the wire is 25 cm and the mass per unit length
of the wire is 9.0  103 kg m1, determine the tension in the
wire.
39
ANS. : 250 Hz; 140 N
PHYSICS
CHAPTER 11
Learning Outcome:
11.3 Doppler effect (1 hour)
At the end of this chapter, students should be able to:

Explain Doppler Effect for sound waves.

Write and use the Doppler Effect equation for relative
motion between source and observer.

Sketch and explain graph of apparent frequency against
distance travelled.
40
PHYSICS
CHAPTER 11
11.3 Doppler effect

is defined as the change in the apparent (observed )
frequency of a wave as a result of relative motion between
the source and the observer.
41
PHYSICS
CHAPTER 11
11.3.2 Equation of Doppler effect

In general,
where
 v  vO 
v : speed of sound
 fS v : speed of source
f O  
 v  vS  vS : speed of observer
O
f O : apparent (observed) frequency
fS : source (true) frequency
Moving source and stationary observer (vO= 0)
a. A source moves toward the stationary observer.
v
O
vO  0
vS
S
 fS 
vS in thesame directionwith v
thus use minus sign (-)
 v 
 f S
f O  
 v  vS 
 f O  fS
42
PHYSICS
CHAPTER 11
b. A source moves away from the stationary observer.
vS
v
S
 fS 
O
vO  0
vS in oppositedirectionwith v
thus use plus sign ()
 v 
 f S
f O  
 v  vS 
 f O  fS
Moving observer and stationary source (vS = 0)
a. An observer moves toward the stationary source.
vO
O
v
 fS 
S
vS  0
vO in oppositedirectionwith v
thus use plus sign ()
 v  vO 
fO  
 fS
 v 
 f O  fS
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PHYSICS
CHAPTER 11
b. An observer moves away from the stationary source
vO
v
O
 fS 
S
vS  0
vO in thesame directionwith v
thus use minussign (-)
 v  vO 
fO  
 fS
 v 
 f O  fS
The rules of using the general equation for Doppler effect.
RULES
vS OR vO in the same direction
with v (speed of sound)
vS OR vO opposite direction
with v (speed of sound)


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PHYSICS
CHAPTER 11
11.3.3 The variation of apparent frequency and
sound intensity when a source moves
towards, passes through and moves away
from the stationary observer.
Apparent frequency
fO
Moves away
Moves toward
ft
fS
fa
r
r
Stationary Observer
Figure 11.13
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PHYSICS
CHAPTER 11
where

f t : frequencywhen movestoward
f a : frequencywhen movesaway
fS : source (true)frequency
From the Figure 11.13:
 When the source moves towards the observer, the apparent
frequency heard by the observer is greater than the
source (true) frequency and its value is constant.
 This value is change when the source passes through the
observer where the apparent frequency, fO is equal to the
source frequency, fS.
 When the source moves away from the observer, the
apparent frequency is less than the source frequency
and its value is constant.
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PHYSICS
CHAPTER 11
Example 11.9 :
Wall T
vS
S
O(stationary)
Figure 11.14
Figure 11.17 shows O is a stationary observer. Source S moves
toward the observer O at a speed of 20 m s1 and away from the
wall T. The frequency of the source is 1000 Hz. Determine
a. the wavelength in front and behind the source if there is no wall T,
b. the apparent frequency heard by the observer O directly from the
source,
c. the apparent frequency heard by the observer O caused by the
reflection on the wall T,
d. the beat frequency detected by the observer O due to the sound
heard directly from the source S and the sound reflected on the
wall T,
(Given the speed of sound is 330 m s-1)
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PHYSICS
CHAPTER 11
 20 m s 1; vO  0; fS  1000 Hz
vS
v
v
Solution : vS
a.
B
S
F
O(stationary )
The wavelength in front the moving source, F is
v  vS
F 
fS
330  20
F 
1000
F  0.31 m
and the wavelength behind the moving source, B is
v  vS
B 
fS
330  20
B 
1000
B  0.35 m
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PHYSICS
CHAPTER 11
Solution : vS  20 m s 1 ; vO  0; f S  1000 Hz
b. By applying the general equation of Doppler effect,
 v  vO 
 fS
f O  
 v  vS 
thus
 v 
 fS
f O  
 v  vS 
 330 
fO  
1000
 330  20 
f O  1065 Hz
v 330
OR
fO 

F 0.31
f O  1065 Hz
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PHYSICS
CHAPTER 11
Solution : vS
c.
Wall T
 40 m s 1; vO  0; fS  1000 Hz
vS
v
S
O(stationary )
When the sound wave hits the wall, the apparent frequency
received by the wall is
 v 
v
330
 fS
f O  
OR
fO 

B 0.35
 v  vS 
 330 
fO  
1000
 330  20 
f O  943 Hz
Thus the frequency of the reflected sound is 943 Hz.
50
PHYSICS
Solution : vS
c.
Wall T
(source)
CHAPTER 11
 20 m s 1; vO  0; fS  1000 Hz
v
O(stationary )
When the reflected wave reaches the stationary observer, the
wall becomes the stationary source. Thus
f  f (wall as a stationary source)
O
T
f O  943 Hz
d. By applying the equation of beat frequency,
f b  f1  f 2
fb  f ( b)  f (c)
 1065  943
f b  122 Hz
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PHYSICS
CHAPTER 11
Exercise 11.3 :
1. Two observers X and Y are provided with sources of sound of
frequency 500 Hz. X remains stationary and Y moves away
from X at a velocity of 1.8 m s1. Determine the beats per
second are observed by X and by Y if the velocity of sound
being 330 m s1.
ANS. : 2.71 Hz; 2.73 Hz
2. An observer is moving between two similar sources of sound
along the line joining the two sources. The observer hears
beats of frequency 4.0 Hz. If the frequency of each source is
500 Hz and the speed of sound is 340 m s1. Calculate the
speed of the observer.
ANS. : 1.36 m s1
3. A whistle giving out 500 Hz moves away from a stationary
observer in a direction towards and perpendicular to a flat wall
with a velocity of 1.5 m s1. Determine the beats frequency will
be heard by the observer. (The speed of sound is 336 m s1)
ANS. : 4.4 Hz
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PHYSICS
CHAPTER 11
THE END…
Next Chapter…
CHAPTER 12 :
Elastic properties of matter
53