The maximum pressure variation that the human ear can tolerate in loud sound is about $30\text{ Nm}^{-2}$.…
The maximum pressure variation that the human ear can tolerate in loud sound is about $30\text{ Nm}^{-2}$. The corresponding maximum displacement for a sound wave in air having a frequency of $10^3\text{ Hz}$ is (Take, velocity of sound in air is $300\text{ ms}^{-1}$ and density of air is $1.5\text{ kg m}^{-3}$)
$\frac{2\pi}{3} \times 10^{-2}\text{ m}$
$\frac{2 \times 10^{-4}}{\pi}\text{ m}$
$\frac{\pi}{3} \times 10^{-2}\text{ m}$
$\frac{10^{-4}}{3\pi}\text{ m}$
Solution
$(\Delta p)_{\text{max}} = B A k$
$\therefore A = \frac{(\Delta p)_{\text{max}}}{B k}$ ...(i)
Here, $v = \sqrt{\frac{B}{\rho}} = \frac{\omega}{k} \Rightarrow k = \omega \sqrt{\frac{\rho}{B}} = 2\pi f \sqrt{\frac{\rho}{B}}$
Further, $B = \rho v^2$
Substituting in Eq. (i), we get
$A = \frac{(\Delta p)_{\text{max}}}{2\pi f \sqrt{B\rho}} = \frac{(\Delta p)_{\text{max}}}{2\pi f \rho v}$ ($\because B = \rho v^2$)
Substituting all the values, we get
$A = \frac{30}{2\pi \times 10^3 \times 1.5 \times 300} = \frac{10^{-4}}{3\pi}\text{ m}$