Question

A point source emits sound waves of amplitude A. The sound intensity at a distance d from the source is I. What is the sound intensity at a distance 0.5D from the source when the source emits waves of amplitude 2A?

Answer 1

Sound intensity (I) is related to the amplitude (A) and the distance (d) from the point source by the inverse square law. The inverse square law states that the intensity of a wave decreases with the square of the distance from the source.

Mathematically, the relationship is expressed as:

$I=\frac{A^2}{d^2}$

Where:

• I is the sound intensity at distance d from the source.
• A is the amplitude of the sound waves emitted by the source.
• d is the distance from the source.

Let's calculate the sound intensity (I') at a distance of 0.5D from the source when the source emits waves of amplitude 2A.

Given: Amplitude of the new waves, A' = 2A (amplitude is doubled) Distance from the source, d' = 0.5D (0.5 times the original distance)

Now, we can use the inverse square law to find the new intensity (I'):

$I^{\prime }=\frac{(A^{\prime }{)}^2}{(d^{\prime }{)}^2}$ $I^{\prime }=\frac{(2A{)}^2}{(0.5D{)}^2}$ I′=4A20.25D2I' = dfrac{4A^2}{0.25D^2}I′=<span clas