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

Answer:

Sound Intensity at microphone's position is $9.417\times 10^{- 4} W/m^{2}$

The amount of energy impinging on the microphone is $9.417\times 10^{- 8} W/m^{2}$

Solution:

As per the question:

Emitted Sound Power, $P_{E} = 32.0 W$

Area of the microphone, $A_{m} = 1.00 cm^{2} = 1.00\times 10^{- 4} m^{2}$

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:

$I_{s} = \frac{P_{E}}{Area} = \frac{P_{E}}{4\pi d^{2}}$

$I_{s} = \frac{32.0}{4\pi (52.0)^{2}} = 9.417\times 10^{- 4} W/m^{2}$

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 $cm^{2} = 1.00\times 10^{- 4} m^{2}$ is:

$I_{s}(1.00\times 10^{- 4}) = 9.417\times 10^{- 4}\times 1.00\times 10^{- 4} = 9.417\times 10^{- 8} J$