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

Intensity at the 84 m will be
`3.257* 10^(-5)W/m^2`

Step-by-step explanation:

We have given that intensity at a location of
`r_1=` 26 m away from the sound source is
`I_1=3.4* 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
`(I_1)/(I_2)=(r_2^2)/(r_1^2)`

`(3.4* 10^(-4))/(I_2)=(84^2)/(26^2)`

`I_2=0.3257* 10^(-4)=3.257* 10^(-5)W/m^2`