# 17.3 Sound intensity  (Page 2/14)

 Page 2 / 14
$\frac{dV}{V}=\underset{\text{Δ}x\to 0}{\text{lim}}\frac{A\left[s\left(x+\text{Δ}x,t\right)-s\left(x,t\right)\right]}{A\text{Δ}x}=\frac{\partial s\left(x,t\right)}{\partial x}.$

The fractional change in volume is related to the pressure fluctuation by the bulk modulus     $\beta =-\frac{\text{Δ}p\left(x,t\right)}{dV\text{/}V}.$ Recall that the minus sign is required because the volume is inversely related to the pressure. (We use lowercase p for pressure to distinguish it from power, denoted by P .) The change in pressure is therefore $\text{Δ}p\left(x,t\right)=\text{−}\beta \frac{dV}{V}=\text{−}\beta \frac{\partial s\left(x,t\right)}{\partial x}.$ If the sound wave is sinusoidal, then the displacement as shown in [link] is $s\left(x,t\right)={s}_{\text{max}}\text{cos}\left(kx\mp \omega t+\varphi \right)$ and the pressure is found to be

$\text{Δ}p\left(x,t\right)=\text{−}\beta \frac{dV}{V}=\text{−}\beta \frac{\partial s\left(x,t\right)}{\partial x}=\beta k{s}_{\text{max}}\text{sin}\left(kx-\omega t+\varphi \right)=\text{Δ}{p}_{\text{max}}\text{sin}\left(kx-\omega t+\varphi \right).$

The intensity of the sound wave is the power per unit area, and the power is the force times the velocity, $I=\frac{P}{A}=\frac{Fv}{A}=pv.$ Here, the velocity is the velocity of the oscillations of the medium, and not the velocity of the sound wave. The velocity of the medium is the time rate of change in the displacement:

$v\left(x,t\right)=\frac{\partial }{\partial y}s\left(x,t\right)=\frac{\partial }{\partial y}\left({s}_{\text{max}}\text{cos}\left(kx-\omega t+\varphi \right)\right)={s}_{\text{max}}\omega \phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right).$

Thus, the intensity becomes

$\begin{array}{cc}\hfill I& =\text{Δ}p\left(x,t\right)v\left(x,t\right)\hfill \\ & =\beta k{s}_{\text{max}}\text{sin}\left(kx-\omega t+\varphi \right)\left[{s}_{\text{max}}\omega \phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\right]\hfill \\ & =\beta k\omega {s}_{\text{max}}^{2}{\text{sin}}^{2}\left(kx-\omega t+\varphi \right).\hfill \end{array}$

To find the time-averaged intensity over one period $T=\frac{2\pi }{\omega }$ for a position x , we integrate over the period, $I=\frac{\beta k\omega {s}_{\text{max}}^{2}}{2}.$ Using $\text{Δ}{p}_{\text{max}}=\beta k{s}_{\text{max}},$ $v=\sqrt{\frac{\beta }{\rho }},$ and $v=\frac{\omega }{k},$ we obtain

$I=\frac{\beta k\omega {s}_{\text{max}}^{2}}{2}=\frac{{\beta }^{2}{k}^{2}\omega {s}_{\text{max}}^{2}}{2\beta k}=\frac{\omega {\left(\text{Δ}{p}_{\text{max}}\right)}^{2}}{2\left(\rho {v}^{2}\right)k}=\frac{v{\left(\text{Δ}{p}_{\text{max}}\right)}^{2}}{2\left(\rho {v}^{2}\right)}=\frac{{\left(\text{Δ}{p}_{\text{max}}\right)}^{2}}{2\rho v}.$

That is, the intensity of a sound wave is related to its amplitude squared by

$I=\frac{{\left(\text{Δ}{p}_{\text{max}}\right)}^{2}}{2\rho v}.$

Here, $\text{Δ}{p}_{\text{max}}$ is the pressure variation or pressure amplitude in units of pascals (Pa) or ${\text{N/m}}^{2}$ . The energy (as kinetic energy $\frac{1}{2}m{v}^{2}$ ) of an oscillating element of air due to a traveling sound wave is proportional to its amplitude squared. In this equation, $\rho$ is the density of the material in which the sound wave travels, in units of ${\text{kg/m}}^{3},$ and v is the speed of sound in the medium, in units of m/s. The pressure variation is proportional to the amplitude of the oscillation, so I varies as ${\left(\text{Δ}p\right)}^{2}.$ This relationship is consistent with the fact that the sound wave is produced by some vibration; the greater its pressure amplitude, the more the air is compressed in the sound it creates.

## Human hearing and sound intensity levels

As stated earlier in this chapter, hearing is the perception of sound. The hearing mechanism involves some interesting physics. The sound wave that impinges upon our ear is a pressure wave. The ear is a transducer    that converts sound waves into electrical nerve impulses in a manner much more sophisticated than, but analogous to, a microphone. [link] shows the anatomy of the ear.

The outer ear, or ear canal, carries sound to the recessed, protected eardrum. The air column in the ear canal resonates and is partially responsible for the sensitivity of the ear to sounds in the 2000–5000-Hz range. The middle ear converts sound into mechanical vibrations and applies these vibrations to the cochlea.

Watch this video for a more detailed discussion of the workings of the human ear.

The range of intensities that the human ear can hear depends on the frequency of the sound, but, in general, the range is quite large. The minimum threshold intensity that can be heard is ${I}_{0}={10}^{-12}\phantom{\rule{0.2em}{0ex}}{\text{W/m}}^{2}.$ Pain is experienced at intensities of ${I}_{\text{pain}}=1\phantom{\rule{0.2em}{0ex}}{\text{W/m}}^{2}.$ Measurements of sound intensity (in units of ${\text{W/m}}^{2}$ ) are very cumbersome due to this large range in values. For this reason, as well as for other reasons, the concept of sound intensity level was proposed.

#### Questions & Answers

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