# Calculating Power at source from Intensity at distance R

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1. Oct 20, 2015

### dmoney123

1. The problem statement, all variables and given/known data
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

2. Relevant equations
A=10log(I/I_0)

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

3. 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 dont know how to find the ratio for the source. ie distance=0

If you treat it like a point source then.... i dont know.

2. Oct 20, 2015

### dmoney123

Figured it out. (4 pi r^2)=P/I where r=46.

P=4.2*10^-4