Question

A sound wave of wavelength $40\text{cm}$ travels in air. If the difference between the maximum and minimum pressure at a given point is $1\times {10}^{-3}{\text{Nm}}^{-2}$, find the amplitude of vibration of the particles of the medium. The bulk modulus of air is $1.4\times {10}^5{\text{Nm}}^{-2}$.

• A. $2.2\times {10}^{-8}\text{m}$
• B. $2.2\times {10}^{-9}\text{m}$
• C. $2.2\times {10}^{-10}\text{m}$
• D. $2.2\times {10}^{-11}\text{m}$

Answer 1

The correct option is C $2.2\times {10}^{-10}\text{m}$

Given,
Wavelength, $\lambda =40\text{cm}=0.4\text{m}$

Pressure difference, $\Delta P=1\times {10}^{-3}{\text{Nm}}^{-2}$

Bulk modulus, $B=1.4\times {10}^5{\text{Nm}}^{-2}$

Pressure amplitude, $P_0=\frac{\Delta P}{2}=\frac{1\times {10}^{-3}}{2}$

$P_0=0.5\times {10}^{-3}{\text{Nm}}^{-2}$

Displace amplitude $\left(S_0\right)$ is given as :

$S_0=\frac{P_0}{Bk}=\frac{P_0\times \lambda }{B\times 2\pi }[\because k=\frac{2\pi }{\lambda }]$

$S_0=\frac{0.5\times {10}^{-3}\times 0.4}{1.4\times {10}^5\times 2\times 3.14}$

$\therefore S_0=2.2\times {10}^{-10}\text{m}$

Hence, option $\left(\text{C}\right)$ is correct.

Why this question ? Concept : Displacement amplitude is proportional to the pressure amplitude of the sound wave.