Question

A concert loudspeaker suspended high off the ground emits 33.0 W of sound power. A small microphone with a 0.600 cm2 area is 52.0 m from the speaker. What is the sound intensity at the position of the microphone

Answer 1

The sound intensity at the position of the microphone is
`9.71*10^(-4) W/m^(2)`

Step-by-step explanation

Sound intensity is given by the formula

`I=(P)/(A)`

Where
`I` is the sound intensity,
`P` is the power and
`A` is the area.

Since the loudspeaker radiates sound in all directions, we have a spherical sound wave where the radius r is the distance of the microphone from the speaker.

∴
`A` is given by
`4\pi r^(2)` where
`r` is the radius

From the question,
`P` = 33.0W,
`r` = 52.0m

`I=(P)/(A) = (P)/(4\pi r^(2) )`

`I = (33.0)/(4\pi * (52.0)^(2) )`

∴
`I = 9.71*10^(-4) W/m^(2)`

Hence, the sound intensity at the position of the microphone is 9.71 × 10⁻⁴ W/m²