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 :

Explanation:

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

Mathematically,    I = [tex]\frac{P}{A}[/tex]

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

Therefore, sound intensity will be calculated as follows.

             I = [tex]\frac{P}{A}[/tex]

               = [tex]\frac{28.0 W}{4 \times 3.14 \times 7 \times 10^{-5} m^{2}}[/tex]

                = [tex]0.318 \times 10^{5} W/m^{2}[/tex]

or,             = [tex]3.18 \times 10^{4} W/m^{2}[/tex]

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