Question

A typical loud sound wave of frequency $1{\text{kHz}}$ has a pressure amplitude of about $10{\text{Pa}}$ . Find the maximum value of displacement at any time and place.
Take the density of the air to be $1 \cdot 29{\text{kg}}{{\text{m}}^{ - 3}}$ .
A. $0 \cdot 03\mu {\text{m}}$
B. $3 \cdot 63\mu {\text{m}}$
C. $30\mu {\text{m}}$
D. ${\text{80}}\mu {\text{m}}$

Answer 1

Hint:The maximum displacement of the wave refers to the amplitude of the wave in the y-direction and can also be referred to as the displacement amplitude. Sound is often described as a pressure wave as it consists of compressions and rarefactions i.e., regions of high pressure and low pressure. The change in pressure depends on the bulk modulus and the displacement of the wave and the change in pressure will peak when the displacement is maximum.

Formulae used:
->The wavelength of a wave is given by, $\lambda = \dfrac{v}{f}$ where $f$ is the frequency of the wave and $v$ is the velocity of the wave.
->The pressure amplitude of a sound wave is given by, ${\left( {\Delta P} \right)_{\max }} = BAk$ where $A$ is the amplitude, $B$ is the bulk modulus and $k$ is the wavenumber.
->The bulk modulus of a pressure wave is given by, $B = \rho {v^2}$ where $v$ is the velocity of the wave and $\rho $ is the density of the medium.

Complete step-by-step solution:
Step 1: List the parameters mentioned in the question.
The pressure amplitude of the sound wave is given to be ${\left( {\Delta P} \right)_{\max }} = 10{\text{Pa}}$ .
The frequency of the wave is $f = 1{\text{kHz}}$ and the medium in which it propagates is air.
The density of air is given to be $\rho = 1 \cdot 29{\text{kg}}{{\text{m}}^{ - 3}}$ .
Let $A$ be the maximum displacement amplitude of the sound wave.
We assume the speed of sound in air to be $v = 340{\text{m}}{{\text{s}}^{ - 1}}$ .
Step 2: Express the relation for the pressure amplitude of the sound wave to find the amplitude displacement.
The pressure amplitude of a sound wave is given by, ${\left( {\Delta P} \right)_{\max }} = BAk$ where $B$ is the bulk modulus and $k$ is the wavenumber.
$ \Rightarrow A = \dfrac{{{{\left( {\Delta P} \right)}_{\max }}}}{{Bk}}$ --------- (1)
The bulk modulus is expressed as $B = \rho {v^2}$ and the wavenumber can be expressed as $k = \dfrac{\lambda }{{2\pi }} = \dfrac{v}{{2\pi f}}$
Substituting for $B = \rho {v^2}$ and $k = \dfrac{{2\pi v}}{f}$ in equation (1) we get, $A = \dfrac{{{{\left( {\Delta P} \right)}_{\max }}v}}{{\rho {v^2}2\pi f}}$
$ \Rightarrow A = \dfrac{{{{\left( {\Delta P} \right)}_{\max }}}}{{\rho v2\pi f}}$ ---------- (2)
Substituting for ${\left( {\Delta P} \right)_{\max }} = 10{\text{Pa}}$ , $v = 340{\text{m}}{{\text{s}}^{ - 1}}$ , $\rho = 1 \cdot 29{\text{kg}}{{\text{m}}^{ - 3}}$ and $f = 1{\text{kHz}}$ in equation (2) we get, $A = \dfrac{{10}}{{1 \cdot 29 \times 340 \times 2\pi \times 1000}} = 3 \cdot 63 \times {10^{ - 6}}{\text{m}}$
Thus the displacement amplitude of the sound wave is $A = 3 \cdot 63\mu {\text{m}}$ .
Hence the correct option is B.

Note:- When the values of physical quantities are substituted in any equation we have to make sure that all the quantities were expressed in their respective S.I. units. If this is not so, then necessary conversions must be done. Here the frequency of the wave was expressed in kilohertz. So during its substitution in equation (2), the value of frequency was changed to Hertz as $f = 1000{\text{Hz}}$