Approximately a third of people with normal hearing have ears that continuously emit a low-intensity sound outward through the ear canal. A person with such spontaneous to acoustic emission is rarely aware of the sound, except perhaps in a noise free environment, but occasionally the emission is loud enough to be heard by someone else nearby. In one observation, the sound wave had a frequency of $\mathbf{\text{1665Hz}}$and a pressure amplitude of$\mathbf{\text{1665Hz}}$. What were (a) the displacement amplitude and (b) the intensity of the wave emitted by the ear?

• Frequency,$\mathrm{f}=1665\text{Hz}$
• Pressure amplitude,$\mathrm{\Delta P}=1.13\times {10}^{-3}\text{Pa}$

Apply the relation of pressure with amplitude, frequency, and speed of sound to find the displacement amplitude. Using the formula for the intensity of sound in terms of density, velocity, frequency, and amplitude, find the intensity.

The expression for the pressure amplitude is given by,

$\mathrm{P}=2\mathrm{\pi f}\mathrm{\rho Av}$

Where,$P$ is pressure,$f$ is frequency, $A$is area,$\rho$ is density and v is velocity.

Pressure amplitude is given as,

$\Delta P=2\pi f\rho Av$

$\therefore \mathrm{A}=\frac{\mathrm{\Delta P}}{2\mathrm{\pi f}\mathrm{\rho v}}$

Substitute $1.13\times {10}^{-3}\text{Pa}$for $\Delta P$, $\text{1665Hz}$for $\mathrm{f}$, $1.225\mathrm{kg}/{\mathrm{m}}^3$for $\mathrm{\rho }$, $340\mathrm{m}/\mathrm{s}$for $v$into the above equation,

$\begin{array}{c}\mathrm{A}=\frac{1.13\times {10}^{-3}\text{pa}}{2\left(3.14\right)\left(\text{1665Hz}\right)\left(\frac{\text{1.225kg}}{{\text{m}}^{\text{3}}}\right)\left(\frac{\text{340m}}{\text{s}}\right)}\\ =0.259\times {10}^{-9}\text{m}\\ \approx 0.26\text{nm}\end{array}$

Hence, the displacement amplitude is $0.26nm$.

The intensity of the sound wave is given as,

$\begin{array}{c}\mathrm{I}=\frac{1}{2}\mathrm{\rho v}{\left(2\mathrm{\pi f}\right)}^2{\mathrm{A}}^2\\ =0.5\left(\frac{\text{1.225kg}}{{\text{m}}^{\text{3}}}\right)\left(\frac{\text{340m}}{\text{s}}\right){(2\times 3.14\times 1665\text{Hz})}^2{(0.26\times {10}^{-9}\text{m})}^2\\ =\frac{\left({\text{1.53×10}}^{\text{-9}}\right)\text{W}}{{\text{m}}^{\text{2}}}\\ \approx 1.5{\text{nW/m}}^{\text{2}}\end{array}$

Hence, the intensity of the wave emitted by the ear is$1.5{\text{nW/m}}^{\text{2}}$ .