Question

A concert loudspeaker suspended high off the ground emits 32.0 W of sound power. A small microphone with a 1.00 cm^2 area is 52.0 m from the speaker. Part complete What is the sound intensity at the position of the microphone?How much sound energy impinges on the microphone each second?

Answer 1

Sound Intensity at microphone's position is
`9.417* 10^(- 4) W/m^(2)`

The amount of energy impinging on the microphone is
`9.417* 10^(- 8) W/m^(2)`

Solution:

As per the question:

Emitted Sound Power,
`P_(E) = 32.0 W`

Area of the microphone,
`A_(m) = 1.00 cm^(2) = 1.00* 10^(- 4) m^(2)`

Distance of microphone from the speaker, d = 52.0 m

Now, the intensity of sound,
`I_(s)` at a distance away from the souce of sound follows law of inverse square and is given as:

`I_(s) = (P_(E))/(Area) = (P_(E))/(4\pi d^(2))`

`I_(s) = (32.0)/(4\pi (52.0)^(2)) = 9.417* 10^(- 4) W/m^(2)`

Now, the amount of sound energy impinging on the microphone is calculated as:

If
`I_(s)` be the Incident Energy/
`m^(2)/s`

Then

The amount of energy incident per 1.00
`cm^(2) = 1.00* 10^(- 4) m^(2)` is:

`I_(s)(1.00* 10^(- 4)) = 9.417* 10^(- 4)* 1.00* 10^(- 4) = 9.417* 10^(- 8) J`