Calculating Power at source from Intensity at distance R

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The discussion focuses on calculating the acoustic power output of a loudspeaker based on the intensity level of sound entering an enclosed chamber through a window. Given that the intensity level is 42 dB and the threshold of hearing is 1.0 × 10^-12 W/m², the intensity at the window is calculated to be approximately 1.5849 × 10^-8 W/m². The area of the window is 2 m², and the distance from the loudspeaker to the window is 46 m. Using the formula for a point source, the acoustic power output of the loudspeaker is determined to be around 4.2 × 10^-4 W. This calculation demonstrates the relationship between sound intensity, area, and distance in acoustic power assessments.
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Homework Statement


An enclosed chamber with sound absorbing walls has a 2.0 m × 1.0 m opening for an outside window. A loudspeaker is located outdoors, 46 m away and facing the window. The intensity level of the sound entering the window space from the loudspeaker is 42 dB. Assume the acoustic output of the loudspeaker is uniform in all directions and that acoustic energy incident upon the ground is completely absorbed and therefore is not reflected into the window. The threshold of hearing is 1.0 × 10-12 W/m2. The acoustic power output of the loudspeaker is closest to

Homework Equations


A=10log(I/I_0)

I_0=1*10^-12 W/m^2

The Attempt at a Solution



We know

42=10log(I/1*10^-12)
10^4.2 (1*10^-12)=I
I=1.5849*10^-8 W/m^2

we also know that it is received in an area of 2m^2

I don't know how to find the ratio for the source. ie distance=0

If you treat it like a point source then... i don't know.
 
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Figured it out. (4 pi r^2)=P/I where r=46.

P=4.2*10^-4
 
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