Question

A point isotropic sonic source of sound power 1 milliwatts emits sound of frequency $170Hz$ in all directions.
A. Find the distance of a point from the source where the loudness level is$60dB$. [Take ${v_S} = 340m{s^{ - 1}}$v]
B. It is known that at some moment $t = 0$, the displacement of air particles at a certain point $4m$ away from the source is $A$. The amplitude of oscillation at this point is known to be $2A$. Find the displacement of air particles in terms of $A$, at a point $55m$ away from the source at the moment $t = 0$. Consider displacement of particles to be positive if it is away from the source.

Answer 1

Hint: Power $\left( P \right)$ and intensity $\left( I \right)$ of a sound wave are related as $I = \dfrac{P}{{4\pi {r^2}}}$. The loudness of sound is $\beta = 10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)$.
To calculate the displacement, use the general displacement formula $S = {S_{\max }}\cos (kx - \omega t + \phi )$.
Where, ${S_{\max }}$ is the maximum displacement of air particles or amplitude.
$k = \dfrac{{2\pi }}{\lambda }$
$\omega = 2\pi f$
$\phi $ is the initial phase
$\lambda $ is the wavelength of the sound wave and it is defined as $\lambda = \dfrac{v}{f}$

Complete step by step solution:
(A). It is given that,
Power of the sound source is $P = 1milliwatts = {10^{ - 3}}W$.
Frequency of the sound, $f = 170 Hz = 170{\sec ^{ - 1}}$.
Velocity of the sound, $v = 340m{s^{ - 1}}$
Loudness level $\beta = 60dB$
We know that the threshold Intensity of sound for human hearing is ${I_0} = {10^{ - 12}}W/{m^2}$.
Also, $\beta = 10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)$.
Substitute the values of $\beta $ and ${I_0}$ in the above formula. We got
$ \Rightarrow 60 = 10{\log _{10}}\left( {\dfrac{I}{{{{10}^{ - 12}}}}} \right)$
Further simplifying,
$ \Rightarrow 6 = {\log _{10}}\left( I \right) - {\log _{10}}\left( {{{10}^{ - 12}}} \right)$
Further calculating the logarithmic values, we got
$ \Rightarrow {\log _{10}}\left( I \right) = 6 - 12$
$ \Rightarrow {\log _{10}}\left( I \right) = -6$
$ \Rightarrow I = {10^{ - 6}}$
We know that the intensity of sound is related to its power and it is given as,
$I = \dfrac{P}{{4\pi {r^2}}}$.
Where, $r$ is the distance of a point from the source.
Substitute all the values of $I$ and $P$ in the above formula of intensity.
$ \Rightarrow {10^{ - 6}} = \dfrac{{{{10}^{ - 3}}}}{{4\pi {r^2}}}$
Further calculating, we got
$ \Rightarrow {r^2} = \dfrac{{{{10}^3}}}{{4\pi }}$
$r = 8.92m$
Hence, the distance of the point from the source is $8.92m$.
(B) Let’s consider the displacement of air particles at a distance $x$ from the source at any time $t$ is given as,
 $S = {S_{\max }}\cos \left( {kx - \omega t + \phi } \right)$ …… (1)
Where,
${S_{\max }}$ is the maximum displacement of air particles or amplitude.
$k = \dfrac{{2\pi }}{\lambda }$
$\omega = 2\pi f$
$\phi $ is the initial phase
$\lambda $ is the wavelength of the sound wave and it is given by
$\lambda = \dfrac{v}{f}$
Substitute the required values in the above formula
$ \Rightarrow \lambda = \dfrac{{340}}{{170}}m$
 $\lambda = 2m$
So, $k = \dfrac{{2\pi }}{2} = \pi $
It is given that for $x = 4m$ and $t = 0$, the displacement is $S = A$ and amplitude ${S_{\max }} = 2A$
From equation (1) we got,
$A = 2A\cos (4\pi + \phi )$
Further calculating for $\phi $,
$ \Rightarrow \phi = \dfrac{\pi }{3}$
For a distance at $x = 55m$ and time at $t = 0$, the displacement of the particle is given as,
$S = 2A\cos \left( {55\pi + \dfrac{\pi }{3}} \right)$
Further calculating
$ \Rightarrow S = 2A\left( { - \dfrac{1}{2}} \right)$
Or $S = - A$
Since it is asked to consider the displacement as positive.
$S = A$
Hence, the displacement of air particles is $A$ at a distance $55m$ from the source.

Note:
The lowest possible sound that the typical human ear can detect has an intensity of ${10^{ - 12}}W/{m^2}$. Hence, the loudness at this intensity is $0dB$. The particles in air vibrate parallel to the direction of propagation of sound waves. So, the sound wave is known as longitudinal wave.