Sound Intensity
Wave Motion of Class 11
Sound Intensity
Consider a harmonic sound wave propagating along a tube of crosssectional area S, as shown in fig(14.21). The quantity p is the excess pressure caused by the wave, and ∂y/∂t is the velocity of an element of the fluid. The instantaneous power supplied by the wave to the element is
P = Fv = pA δy/δ
Using equation (14.36) & (14.37) P = [p_{o }cos(kx  ωt)]S[ωAcos(kx  ωt)] or P = p_{o}AωS cos^{2}(kx − ωt) At any position, say x = 0, the average of cos2ωt over one period is 1/2 ; hence the average power transmitted by the wave is Pav = 1/2 ρS(ωA)2v(14.40) 

Note that this has the same form as equation (14.20) for the power transported by a wave on a string.
The intensity I of a wave is defined as the energy incident per second per unit area normal to the direction of propagation:
I = Power/Area = P/A (14.41)
The SI unit of intensity is W/m^{2.}
From equation (14.37) we know po = Bak
Since k = ω/v and B = ρv2 , therefore po = ρωvA
Thus, I_{av} = p^{2}_{0}/2ρv (14.42)
When expressed in terms of p0, the intensity is proportional to the square of the pressure amplitude and is independent of frequency.
In the particular case of waves emitted by a point source, the radiated energy spreads uniformly over wave fronts that are spherical surfaces. Since the surface area of a sphere of radius r is 4πr2, the intensity at a distance r from a point source radiating with a power P is
I = p / 4πr^{2}(14.43)
That is, I ∝ 1/r2, the intensity decreases as the inverse square of the distance from a point source.
For cylindrical waves I ∝ 1/r
Example 14.16
Measurement of sound waves show that the maximum pressure variations in the loudest sound that the ear can tolerate without pain are of the order of 30 Pa. Find the corresponding maximum displacement, if the frequency is 1000 Hz and v = 350 m/s.
Solution
k = ω/v = 2(3.14)(1000)/350 = 18 m1
B = γp = (1.4)(105)
A = p/Bk = 30/(1.4 x 10^{5})(18) = 1.18 x 10^{5}m = 0.0118 mm
Example 14.17
In the previous example, if the density of air is ρ = 1.22 kg/m3, then find the intensity of a sound wave of the largest amplitude tolerable to the human ear.
Solution
Using,I = p^{2}m/2ρv = (30)^{2} / 2(1.22) (350) W/m2
Intensity Level: The Decibel Scale
The sound intensities that the human ear can hear range from 1012 W/m2 to 1 W/m2. The intensity of a sound is perceived by the ear as the subjective sensation of loudness. However, if the intensity doubles, the loudness does not increase by a factor of 2. Experiments first carried out by A.G. Bell showed that to produce an apparent doubling in loudness, the intensity of sound must be increased by a factor of about 10. Therefore, it is convenient to specify the intensity level β in terms of the decibel (dB), which is defined as
β = 10log 1/I_{o}(14.44)
where I is the measured intensity and I0 is some reference value. If one takes I0 to be 1012W/m2, then the threshold of hearing corresponds to β = 10 log 1 = 0 dB. At the threshold of pain, 1 W/m2, the intensity level is
β = 10 log(^{1/1012 })= 120 dB
A list of the intensity level of various sources is given in Table 17.1.
Intensity Levels (dB)
Threshold of hearing 
0 
Leaves rustling 
10 
Quiet hall 
25 
Office 
60 
Conversation 
60 
Heavy traffic (3 m) 
80 
Loud classical music 
95 
Loud rock music 
120 
Jet engine (20 m) 
130 
Example 14.18
The sound emitted by a source reaches a particular position with an intensity I1. What is the change in intensity level when another identical source is placed next to the first? (There is no fixed phase relation between the sources.)
Solution
If the initial and final intensities are I1 and I2, then the two intensity levels are
β1 = 10 log] I1/I_{o}; β2 = 10 log ^{I2 / I0}
The change in level is
β2 −β1 = 10log^{l1/ l2}
= 10 log 2 = 3 dB
Thus, when the intensity doubles the intensity level changes by 3 dB. The response of the ear roughly corresponds to this logarithmic scale. The smallest change in level that can be detected by the human ear is about 1 dB.
Example 14.19
A speaker emits 0.8 W of acoustic power. Assume that it behaves as a point source which emits uniformly in all directions. At what distance will the intensity level be 85 dB?
Solution
From equation we know that the intensity of waves from a point source decreases as the inverse square of the distance r; that is
I = p / 4πr^{2 }(1)
We must find the intensity corresponding to an 85−dB sound level:
85 = 10 log 1/Io
Thus, log (I/I0) = 8.5
orI = 1012 × 108.5 = 3.16 × 10^{4}W/m^{2}(2)
Using (2) in (1) we find
r_{2} = P/ 4πI
= (0.08W)/4(3.14)(3.16 x 10^{4} W/m^{2}) = 201 m2
Thus, r = 14.1 m.