Question

A point source emits 30.0 W of sound isotropically . A small microphone intercepts the sound in an area of ${\mathbf{\text{0.750cm}}}^{\mathbf{\text{2}}}$, 200 m from the source.

Calculate

(a) the sound intensity there and

(b) the power intercepted by the microphone

Short answer

1. Intensity of sound is $5.97\times {10}^{-5}{\text{\hspace{0.17em}W/m}}^{\text{2}}$.
2. Power intercepted by the microphone is $4.48\times {10}^{-9}\text{\hspace{0.17em}W}$.

Step-by-step solution

Step 1: Given

1. Power,$\mathrm{P}=30\text{W}$
2. Area of microphone,$\mathrm{A}=0.75{\text{0\hspace{0.17em}cm}}^{\text{2}}$
3. Distance from the source,$\mathrm{r}=\text{200\hspace{0.17em}m}$

The concept

Intensity can be found using power and distance from the relation between power, intensity, and area. Using this intensity, find the power intercepted by the microphone at a given distance.

The relation between intensity and power is given as-

$\mathrm{I}=\frac{\mathrm{p}}{4{\mathrm{\pi r}}^2}$

where, I is intensity, p is power and r is radius.

(a) Determining the sound intensity

Now, write intensity in terms of power and area of the sphere as,

$\mathrm{I}=\frac{\mathrm{P}}{4{\mathrm{\pi r}}^2}$

Substituting the values from the given data,

$\begin{array}{c}\mathrm{I}=\frac{\text{30\hspace{0.17em}W}}{\text{4}\left(\text{3.14}\right){\left(\text{200\hspace{0.17em}m}\right)}^{\text{2}}}\\ =5.97\times {\text{10}}^{\text{-5}}{\text{W/m}}^{\text{2}}\end{array}$

Hence, intensity of sound is $5.97\times {10}^{-5}{\text{\hspace{0.17em}W/m}}^{\text{2}}$.

(b) Determining the power intercepted by the microphone

Power can be written in terms of intensity and area of the microphone as,

$\begin{array}{c}\mathrm{P}=\mathrm{IA}\\ =(0.750\times {10}^{-4}{\text{\hspace{0.17em}m}}^2)(5.97\times {\text{10}}^{\text{-5}}\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}})\\ =4.48\times {10}^{-9}\text{\hspace{0.17em}W}\end{array}$

Hence, power intercepted by the microphone is $4.48\times {10}^{-9}\text{\hspace{0.17em}W}$.