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

Answer:

The sound intensity at the position of the microphone is $9.71\times10^{-4} W/m^{2}$

Explanation

Sound intensity is given by the formula

$I=\frac{P}{A}$

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

$I=\frac{P}{A} = \frac{P}{4\pi r^{2} }$

$I = \frac{33.0}{4\pi \times (52.0)^{2} }$

∴ $I = 9.71\times10^{-4} W/m^{2}$

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