waves and sound

Question

An audio speaker producing a steady sound at an outdoor concert is 40 ft away from you. If you move to a position where the speaker is 63 ft distant, by what factor will the amplitude of the sound change?

Answer 1

Let's start from the beginning. In physics we quantitatively describe sound by the term intensity. The Intensity of a sound becomes less the further you are away from the source of the sound. This is called the Inverse Square Law; Intensity is inversely proportional to the square of the distance squared.

I ∝ 1/d²

In this problem you are given the distance from the source but are asked to find the relative change in Amplitude. For this we need to use the relationship between Intensity and Amplitude. Intensity is equal to the square of the amplitude.

I = A² so A = √I

We can combine these relationships to solve for Amplitude:

I ∝ 1/d² = A² therefore: A = 1/d

We can set the two distance up as a ratio to determine the change in amplitude:

A_new/A _original = d_original/d_new

where:

d_original = 40 ft → note we don't have to convert to meters because we are simply finding the ratio.

d_new = 63 ft

A_new = is new amplitude compared to

A _original = original amplitude -→because we are finding the relative change in amplitudes, we can set this to one and solve for the factor that the new amplitude will be reduced by (we know it will be reduced because we are moving further from the sound, increasing the distance)

A_new = 40/63 = 0.635 smaller than the original amplitude.

Answer 2

d1=40ft

d2=63ft

=sqrt(40²/63²)

= 40/63 = 0.635

Therefore, the amplitude of the sound will decrease by a factor of 0.635