Question:

There is a speaker emitting sound with a uniform power of 250 W. At what distance will the intensity be just below the threshold of pain, which is 1.00 W/m\(^2\)?

Answer:

The sound intensity is defined by

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

Namely, intensity is power going through certain area.

The sound emitted uniformly in space, so it is spherically propagated. The area of sphere is \(4\pi r^2\). Thus, the intensity becomes

\[ I=\frac{P}{4\pi r^2} \]

Solve for the distance.

\[ r^2 = \frac{P}{4\pi I} \]

\[   r = \sqrt{\frac{P}{4\pi I}}\]

Plug in the numbers.

\[ r = \sqrt{\frac{250 (\mathrm{W})}{4\pi \times 1.00 (\mathrm{W/m}^2)}} \]

Therefore, the distance is

\[ r = 4.46 \mathrm{m}\]

This is the safety distance to protect your ears.

As we discussed above, intensity, power and energy are different although they are almost the same in terms of our daily life conversation.