Question: A sound wave of frequency 300 Hz has an intensity of $\mathbf{1}\mathit{\mu }\mathit{W}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$. What is the amplitude of the air oscillations caused by this wave?

1. Velocity of the sound wave, V = 343 m/s
2. Density of sound wave is, $\rho =1.2{\text{1kg/m}}^{\text{3}}$
3. Frequency of sound, $f=300\text{Hz}$
4. Intensity of sound waves, $l=1\mu W/m^2={10}^{-6}w/m^2$

A sound wave is a sinusoidal wave that consists of intensity, wavelength, and speed. The intensity of a sound wave, at the surface, is the average rate per unit area at which energy is transferred. Also, the intensity of a sound wave depends on displacement amplitude. Using the formula for intensity in terms of density, velocity, angular frequency, and displacement amplitude, find the displacement amplitude.

The intensity of sound wave is given as-

$I=\frac{1}{2}\rho v{\omega }^2{s}_m^2$

Where,I is intensity, $\rho$ is density, $\omega$is velocity and $sm$ is amplitude.

Intensity in terms of displacement amplitude is,

$I=\frac{1}{2}\rho v{\omega }^2{s}_m^2$

Now, find the displacement amplitude for high intensity.

By rearranging the intensity equation,

$s_m^2=\frac{2I}{\rho v{\omega }^2}$

For the given values, the equation becomes-

$s_m^2=\frac{2\times {10}^{-6}W/m^2}{\left(1.2{\text{1kgm}}^{\text{3}}\right)\times 343m/s\times {\left(2\mathrm{\pi }\times 300\mathrm{Hz}\right)}^2}$

$$\begin{gathered} s_m^2=1.36\times {10}^{-15}{\text{m}}^{\text{2}} \\ {s}_m=3.68\times {10}^{\text{- 8}}\text{m} \end{gathered}$$

Hence,the amplitude caused by this wave is $s_m=3.68\times {10}^{\text{- 8}}\text{m}$.