Question

Maximum sound pressure tolerable by the human ear is $30{\text{N/m}}^2$. For a sound of frequency ${10}^3\text{Hz}$ and speed $300\text{m/s}$, the maximum displacement of particles of the medium is
[density of air $=1.5{\text{kg/m}}^3)$]

• A. $\frac{1}{6\pi }\times {10}^{-4}\text{m}$
• B. $\frac{1}{5\pi }\times {10}^{-4}\text{m}$
• C. $\frac{1}{4\pi }\times {10}^{-4}\text{m}$
• D. $\frac{1}{3\pi }\times {10}^{-4}\text{m}$

Answer 1

The correct option is D $\frac{1}{3\pi }\times {10}^{-4}\text{m}$

We know that the relation between maximum pressure in the medium and maximum displacement of the particles in the medium is given by,

$P_o=B{A}_{o}k$

where, $P_o$ is the maximum pressure, $B$ is the bulk modulus, $A_o$ is the maximum displacement and $k$ is the wave number.

Also, speed of sound,

$v=\sqrt{\frac{B}{\rho }}$

$\Rightarrow B=v^2\rho$

Wave number,

$k=\frac{2\pi }{\lambda }=\frac{2\pi f}{v}$

$P_o=v^2\rho \times A_o\times \frac{2\pi }{\lambda }$

$P_o=\frac{2\pi v^2\rho A_{o}f}{v}=2\pi v\rho A_{o}f$

$\Rightarrow A_o=\frac{P_o}{2\pi v\rho f}$

$\Rightarrow A_o=\frac{30}{2\pi \times 300\times 1.5\times {10}^3}=\frac{1}{3\pi }\times {10}^{-4}\text{m}$