Question

A concert loudspeaker suspended high off the ground emits 39 w of sound power. a small microphone with a 1.0 cm2 area is 60 m from the speaker. part a what is the sound intensity at the position of the microphone?

Answer 1

As we know that intensity of sound is defined as power received per unit area

so here the power of source is given as

$P = 39 W$

distance of microphone is given as

$d = 60 m$

now if loudspeaker is considered as spherical source then we will have

$Intensity = \frac{power}{area}$

$Intensity = \frac{39}{4\pi r^2$

here

r = d = 60 m

$Intensity = \frac{39}{4\pi (60^2)}$

$Intensity = 0.86 \times 10^{-3} W/m^2$