Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at a location 27 m away from the sound source is 3.0 × 10-4 $\frac{W}{m^{2}}$ $W/m^2$ . What is the intensity at a spot that is 87 m away?

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Answer 1 · 2015-11-03T19:14:06

$2.89 \times 10^{- 5} {W/m}^{2}$ $2.89xx10^(-5)"W/m"^2$

I have assumed that the inverse square law is followed here:

$I \propto \frac{1}{d^{2}}$ $Iprop(1)/(d^2$

$:.I=(k)/(d^2)$

For the 1st case we can write:

$I_1=(k)/(27^2)$

$:.k=3xx10^(-4)xx27^2=0.2187"W"$

For the 2nd case:

$I_2=(k)/(87^2)=(0.2187)/(7569)$

$I_2=2.89xx10^(-5)"W/m"^2$

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at a location 27 m away from the sound source is 3.0 × 10-4 W/m^2Wm2W/m^2. What is the intensity at a spot that is 87 m away?