Question

A small speaker has a capacity of 3 watt. A microphone is placed at a distance $2\,{\text{m}}$ from the speaker. The displacement amplitude of particles of air near the microphone if the frequency of sound emitted by the sound is $1.0\,{\text{kHz}}$, is: (Density of air=$1.2\,{\text{kg/}}{{\text{m}}^3}$ and speed of sound in air=$330\,{\text{m/s}}$)

A. $2.76 \times {10^{ - 4}}\,{\text{cm}}$

B. $4 \times {10^{ - 4}}\,{\text{cm}}$

C. $10 \times {10^{ - 4}}\,{\text{cm}}$

D. $3.8 \times {10^{ - 3}}\,{\text{cm}}$

Answer 1

Use the equation for the intensity of the sound at a distance from the source of the sound to determine the intensity of the sound near the microphone. Also use the other equation for the intensity of the sound. This equation gives the relation between the displacement amplitude, velocity of sound and density of the medium.

Formulae used:

The expression for the intensity $I$ of the sound at a distance from the source is

$I = \dfrac{P}{{4\pi {r^2}}}$ …… (1)

Here, $P$ is the power of the sound source and $r$ is the distance from the source of sound.

The intensity $I$ of the source of sound is given by

$I = 2{\pi ^2}{A^2}{f^2}\rho v$ ……. (2)

Here, $A$ is the displacement amplitude of the particles of the air, $f$ is the frequency of sound from the source, $\rho$ is the density of the medium through which the sound travels and $v$ is the speed of the sound in air.

Complete step by step answer:

We have given that the power of the speaker is $3\,{\text{W}}$ and the distance of the microphone from the speaker is $2\,{\text{m}}$.

$P = 3\,{\text{W}}$

$r = 2\,{\text{m}}$

We can determine the intensity of the sound at the microphone using equation (1).

Substitute $3\,{\text{W}}$ for $P$, $3.14$ for $\pi$ and $2\,{\text{m}}$ for $r$ in equation (1).

$I = \dfrac{{3\,{\text{W}}}}{{4\left( {3.14} \right){{\left( {2\,{\text{m}}} \right)}^2}}}$

$\Rightarrow I = \dfrac{3}{{16\pi }}\,{\text{W/}}{{\text{m}}^2}$

Hence, the intensity of the sound near microphone is $\dfrac{3}{{16\pi }}\,{\text{W/}}{{\text{m}}^2}$.

We can determine the displacement amplitude of the particles of the air near the microphone using equation (2).

Rearrange equation (2) for $A$.

$A = \dfrac{1}{{\pi f}}\sqrt {\dfrac{I}{{2\rho v}}}$

Substitute $\dfrac{3}{{16\pi }}\,{\text{W/}}{{\text{m}}^2}$ for $I$, 3.14 for $\pi$, $1.0\,{\text{kHz}}$ for $f$, for $\rho$ and $330\,{\text{m/s}}$ for $v$ in the above equation.

$A = \dfrac{1}{{\left( {3.14} \right)\left( {1.0\,{\text{kHz}}} \right)}}\sqrt {\dfrac{{\dfrac{3}{{16\left( {3.14} \right)}}\,{\text{W/}}{{\text{m}}^2}}}{{2\left( {1.2\,{\text{kg/}}{{\text{m}}^3}} \right)\left( {330\,{\text{m/s}}} \right)}}}$

$\Rightarrow A = \dfrac{1}{{\left( {3.14} \right)\left( {1.0 \times {{10}^3}\,{\text{Hz}}} \right)}}\sqrt {\dfrac{{\dfrac{3}{{16\left( {3.14} \right)}}\,{\text{W/}}{{\text{m}}^2}}}{{2\left( {1.2\,{\text{kg/}}{{\text{m}}^3}} \right)\left( {330\,{\text{m/s}}} \right)}}}$

$\Rightarrow A = 2.76 \times {10^{ - 6}}\,{\text{m}}$

$\Rightarrow A = 2.76 \times {10^{ - 4}}\,{\text{cm}}$

Therefore, the displacement of amplitude of the air particles near the microphone is $2.76 \times {10^{ - 4}}\,{\text{cm}}$.

Hence, the correct option is (A).

Note: The students may forget to convert the unit of the frequency of sound in the SI system of units. Otherwise, the resultant value of the displacement amplitude will not be the correct. One should use the proper conversion factor to convert the final displacement amplitude into centimeters.