Question

The intensity of a sound varies inversely with the square of its distance from the source. At a distance of 1 meter, the intensity of a jet engine noise is 10 watts per square meter. An airport cargo worker is 15 meters from the jet engine. What is the sound intensity at this distance?

Answer 1

Answer:

The sound intensity at the distance of 15 meters is 0.044 watts.

Step-by-step explanation:

Given as :

For the distance of 1 meter , the intensity of jet engine noise = 10 watts

∵ The intensity of sound is define as the power carried by sound waves per unit area in the direction perpendicular to that area

So , the intensity of sound is inversely proportional with square of distance

I.e  $I \propto \frac{1}{d^{2}}$   , where d is the distance

or, $I = \frac{k}{d^{2}}$

Now ,

∵ For 1 meter the intensity of sound is 10 watts

∴  Let for 15 meter the intensity of sound is  $I_2$

So , $\frac{I_2}{I_1} =( \frac{d_1}{d_2} )^{2}$

Or, $\frac{I_2}{10} =( \frac{1}{15} )^{2}$

Or,  $I_2$ = $\frac{1}{225}$ × 10  watts

∴    $I_2$ = 0.044  watts

Hence The sound intensity at the distance of 15 meters is 0.044 watts. Answer