Answer
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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.
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.
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