Sound Waves

Sound wave is a spherical wave Example 10.7: 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

Sound Stationary wave Doppler (a) the power of loudspeaker,
intensity effect
(b) the sound intensity at distance 6 m from the

Stretched Air source.
Solution
(a)
strings columns

10.3Sound intensity
(a) Define & use sound intensity = = 4 12

It is a scalar (b) r2 = 6 m
quantity, 0.85 = 4(32)
unit: watt
Equation per squared = = 96.1 = 0.212 −2
metre (W 4 22 4(62)
Definition: the m2)
rate of sound =
energy flow
across unit Sound intensity is defined 10.4Applications of standing waves
area as the sound power per (a) Solve problems related to the fundamental &
perpendicula unit area
r to the overtone frequencies for:
direction of (i) stretched string
the sound = (ii) air columns (open and closed end)
propagation
(b) Use wave speed in stretched string, = √ 
(b) Discuss the dependence of intensity on
amplitude & distance from a point source by Stretched Wave speed on the string:

using graphic illustrations string
Explanation = √
• The power
carried by a Tension, T
sound wave Mass per
unit length, µ
spreads out v depends on
after leaving

a source
• The power

passes

perpendicularly through surface 1 and then

surface 2, which has the larger area To determine the
value of 

The factors influence the
value of sound intensity

amplitude distance
of the from the
sound source of If the length of the If the radius of the
I  A2 sound string is l and its string is r and its
1 density,  thus
 2 mass, m thus



 =  =  2
 =

Page7

Explanation Example 10.8: A stretched wire of length 80.0 cm

 When a string is plucked, a progressive and mass 15.0 g vibrates transversely. Waves travel
transverse wave is produced on the string. along the wire at speed 220 m s1. Two antinodes
This wave is travelling to the both fixed ends
(incident wave) and reflected (reflected wave) can be found in the stationary waves formed in
as shown
between the two fixed ends of the wire.
 The superposition of both waves making
stationary transverse wave and the simplest (a) Sketch and label the waveform of the
pattern of the stationary wave on the string is
shown below stationary wave.

Fundamental Mode/ 1st harmonic (b) Determine
Frequency
(i) the wavelength of the progressive wave
= 
which move along the wire,

 0 =  (ii) the frequency of the vibration of the wire,

(iii) the tension in the wire.
= √ Solution
(a) 1st Overtone/ 2nd
1
0 = 2 √ harmonic
l =  = 0.8 m

Wavelength (b)

(i) 1st Overtone/ 2nd harmonic
= 2 ,  = 2 (ii)

1st Overtone/ 2nd harmonic = 

Frequency 220 = 0.8

= 275
 1 = 
(iii) Mass per unit length

1  = = 0.015 = 0.0188 −1
1 = √ = 2 0 0.8

Wavelength
= √ , = 910
 =
2nd Overtone/ 3rd harmonic

Frequency


 1 = 

3
2 = 2 √ = 3 0
Wavelength

3 2
= 2 ,  = 3
Standing waves occur when both ends of a string
are fixed. In that case, only waves which are
motionless at the ends of the string can persist.
There are nodes, where the amplitude is always
zero, and antinodes, where the amplitude varies
from zero to the maximum value.

Page8

Air column: Close pipe Air column: Open pipe

If the air in a If the air in a
pipe that is open pipe
closed at one
end is (both ends

are open) is
disturbed by
disturbed by a source of sound (e.g. tuning fork), a
progressive longitudinal wave travels along the air a source of sound (e.g. tuning fork), a progressive
column and is reflected at its end to form a longitudinal wave travels along the air column.
stationary longitudinal wave
This wave will superposition with another
Fundamental Mode/ 1st harmonic
Wavelength progressive longitudinal wave produced by the air
 outside the pipe and form a stationary
= 4 ,  = 4
Frequency longitudinal wave.

=  Fundamental Mode/ 1st harmonic

Wavelength
 0 =  = 4

1st Overtone/ 3rd harmonic = 2 ,  = 2
Wavelength
3 4 Frequency

= 4 ,  = 3 1st Overtone/ 2nd harmonic = 
Frequency
3
 0 =  = 2
 3 = 4 , 1 = 3 0
Wavelength

2nd Overtone/ 5th harmonic =    =
Wavelength Frequency
5 4

= 4 ,  = 5  1 =  = 2 (2 )
Frequency = 2 0
5
2nd Overtone/ 3rd harmonic
 5 = 4 , 2 = 5 0
Wavelength
: = 4 = 0
Example 10.9: A tube 80 cm long is closed at one 3 2
= 2   = 3
end. Resonance occurs and the vibrating air column
Frequency
in the tube produces sound of frequency 1134Hz.
3
The fifth overtone mode is found in the air column.  2 =  = 2
= 3 0
(a) Sketch and label the waveform of the air
Example 10.10: A 3.00 m long air column is open at
column.
both ends. The frequency of a certain harmonic is
(b) Calculate
500 Hz and the frequency of the next higher
(i) the speed of sound in air,
harmonic is 557 Hz. Determine the speed of sound
(ii) the fundamental frequency.
Solution in the air column. The end correction may be
(a)
neglected.
Solution

For open pipe: 2 21 = 2 1 − 2

2 = (1) + 1
= 1

2 = ( 1 +) − (2)
2
Rearrange eq (1)

(b) (i) 1 = 2 1 − (3)
11 11
(3) in (2)
= 4   0.8 = 4 
 = 0.291 (2 1 + 1)
2
=  = 0.291(1134) = 330 −1 2 =
(ii) For 5th overtone, n = 11
= 2 ( 2 − 1) = 2(3)(557 − 500) = 342 −1
5 = 0
1134 = 11 0 Page9
0 = 103