Question

Suppose that sound is emitted uniformly in all directions by a public address system. The intensity at a location 26 m away from the sound source is 3.40 10-4 W/m2. What is the intensity at a spot that is 84 m away?

Answer 1

Answer:

Intensity at the 84 m will be $3.257\times 10^{-5}W/m^2$

Explanation:

We have given that intensity at a location of $r_1=$ 26 m away from the sound source is $I_1=3.4\times 10^{-4}W/m^2$

We have to find the intensity at a distance $r_2=84m$

We know that intensity is inversely proportional to the square of distance from the sound source

So $\frac{I_1}{I_2}=\frac{r_2^2}{r_1^2}$

$\frac{3.4\times 10^{-4}}{I_2}=\frac{84^2}{26^2}$

$I_2=0.3257\times 10^{-4}=3.257\times 10^{-5}W/m^2$