# An electronic point source emits sound isotropically at a frequency of 5000 Hz and a power of 34 watts. A small microphone has an area of 0.79 cm2 and is located 156 meters from the point source. What would be the sound intensity level at the microphone if the point source doubled its power output?

Dec 19, 2014

A point source emitting sound isotropically means the wavefront spreads out spherically and its intensity falls off with distance according to the inverse-square law:

$I \left(r\right) = {P}_{s r c} / {r}^{2}$, where ${P}_{s r c}$ is the power of the source.

The intensity level of sound at a distance $r$ from the source, in deci-bells ($\mathrm{dB}$) is:

(10dB)\ln((I(r))/(I_o)); \qquad \qquad I(r)=P_{src}/r^2
${I}_{o} = 1 \setminus \times {10}^{- 14} W {m}^{- 2}$ is the threshold of hearing intensity.

To find the intensity level of the sound at the microphone all we need is the distance between the source and the microphone. The information about the area of the microphone and the frequency of the sound are just red herrings.

P_{src)=34 W; \qquad r=156 m \qquad => I(r)=1.397\times10^{-3} Wm^{-2}.

Thus the intensity level in deci-bells is :
$\left(10 \mathrm{dB}\right) \setminus {\log}_{10} \left(\frac{1.397 \setminus \times {10}^{- 3}}{1 \setminus \times {10}^{- 14}}\right) = 111 \mathrm{dB}$