Question

A concert loudspeaker suspended high off the ground emits 26.0 W of sound power. A small microphone with a 1.00 cm2 area is 55.0 m from the speaker.

a What is the sound intensity at the position of the microphone?

b How much sound energy impinges on the microphone each second?

Answer 1

Given : P = 26 W , A = 1 cm² , r = 55 m

Solution:

(a) Sound intensity at the position of the microphone

As the microphone is positioned at 55 m from the speaker.

So , r = 55m

Sound intensity is given by: I = P/4r²

I = (26 W)/ (4×3.14×55²) = 6.84 × 10^-4 W/m²

Answer : I = 6.84 ×10^-4 W/m²

(b) Sound energy impinges on the microphone each second

The unit of power (Watt) can be written as Joule per second (J/s)

Now , the area of microphone is given: A = 1 cm² or 1×10^-4m²

The sound energy impinges on microphone per second can be given as:

E = I*A

= (6.84×10^-4)*(1×10^-4) = 6.84 ×10^-8 J/s

Answer: E = 6.84 × 10^-8 J/s