Question

A surface receiving sound is moved from its original position to a position three times farther away from the source of the sound. The intensity of the received sound thus becomes A. Nine times higher. B. Nine times lower. C. Three times higher. D. Three times lower.

Answer 1

The intensity of the sound will be B. Nine times lower

Explanation:

The intensity of a sound follows an inverse square law, which means that it is inversely proportional to the square of the distance from the source:

$I\propto \frac{1}{r^2}$

where

I is the intensity

r is the distance from the source

In this problem, the siund has an intensity of I when the receiver is placed at a distance r from the source.

Later, the receiver is placed three times farther away, so the new distance is

r' = 3r

Therefore, the new intensity of the sound will be:

$I'\propto \frac{1}{r'^2}=\frac{1}{(3r)^2}= \frac{1}{9} (\frac{1}{r^2})= \frac{1}{9}I$

Therefore, the intensity of the sound received will be nine times lower.

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