Acoustic Sound Intensity

A room that has walls that absorb sound has an open window of dimensions 2 x 1 m. There is a loudspeaker outside this room situated at a distance of 84 m. The sound that enters the window of the room has an intensity of 56 dB. What is the acoustic power output of the loudspeaker if the acoustic output is uniform in all directions?

Acoustic Intensity

The decibel rating of a wave intensity is by DEFINITION

$dB =10*log (I/I0)$

where the log is in the base 10 and $I0 =10^-12W/m^2$ is the hearing threshold. With the given data we have:

$56 =10*log(I/I0)$

$I = I0*10^5.6 =10^{-12}*10^{5.6} =10^{-6.4} =3.981*10^{-7} W/m^2$

The intensity is by definition given by the equation Intensity = Power/Area

Since the loudspeaker is at a distance of 84 m away from the window and is radiating uniformly, the wave front of the sound will be a sphere having a radius of 84 m. The surface area of this sphere is

$S =4*\pi*R^2 =88668.3 m^2$

We find the acoustic power by multiplying the intensity with the area of the wave. Hence the acoustic power of the speaker is

$P = I*S =3.981*10^{-7}*88668.3 =0.0353 W$

(check: The acoustic power of a music concert is about 0.1 W, from wiki). To repeat this is acoustic power and has nothing to do with the electric power output of the loudspeaker. While the acoustic power is 0.1 W the electric power could be some tens of Watts.

A 56 dB sound intensity is only a bit lower than a normal conversation (which has a “standard” sound intensity of 60 dB). Just imagine that you have a music concert which is at a distance of 100 meters away from you. You will hear it as a normal conversation.


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