Question

A concert loudspeaker suspended high off the ground emits 28.0 W of sound power. A small microphone with a 0.700 cm2 area is 55.0 m from the speaker. What is the sound intensity at the position of the microphone?

Answer 1

Explanation:

It is known that wave intensity is the power to area ratio.

Mathematically,    I = $\frac{P}{A}$

As it is given that power is 28.0 W and area is $7 \times 10^{-5} m^{2}$.

Therefore, sound intensity will be calculated as follows.

             I = $\frac{P}{A}$

               = $\frac{28.0 W}{4 \times 3.14 \times 7 \times 10^{-5} m^{2}}$

                = $0.318 \times 10^{5} W/m^{2}$

or,             = $3.18 \times 10^{4} W/m^{2}$

Thus, we can conclude that sound intensity at the position of the microphone is $3.18 \times 10^{4} W/m^{2}$.