please see the attached question 11. An enclosed chamber with sound a walls has
ID: 1304018 • Letter: P
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please see the attached question
11. An enclosed chamber with sound a walls has a 2.0 x 1.0 opening for an outside window. A loudspeaker is located outdoors, 84 m away and facing the window. The intensity level of the sound entering the window space from the loudspeaker is Assume the acoustic output of the loudspeaker is uniform in all that acoustic energy directions and upon the ground is completely absorbed and therefore is not reflected into the window. The threshold of hearing is 1.0 x 10-12 W/m2. What is the acoustic powe output of the loudspeaker?3Explanation / Answer
The decibel rating of a sound intensity is by DEFINITION
dB =10*log (I/I0)
where the log is base 10 and I0 =10^-12W/m^2 is the hearing threashhold.
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 definiton
Intensity = Power/Area
Since the loudspeaker is 84 m away from the window and is radiating uniformly, the wave front of the sound will be a sphere having radius 84m.
S =4*pi*R^2 =88668.3 m^2
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 rock concert is about 0.1 W.
http://en.wikipedia.org/wiki/Sound_power#Table_of_selected_sound_sources)
A 56 dB sound intensity is only a bit higher than a normal conversation. Imagine you have a rock concert 100 meters away. You will hear it as a normal conversation.
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