Sound Wave Problem?

Question

A person wears a hearing aid, which makes a sounds louder by 10 dB.

This means the sound intensity is increased by how many times?

This means the amplitude of the soundwave is increased by how many times?

Answer 1

The sound intensity, I, (acoustic intensity) is defined as the sound power P per unit area A.

The db measure of sound POWER is 10*log(P1/P2); if sound power goes up by 10db, then10* log(P1/P2) = 10, log(P1/P2) = 1, and P1/P2 = 10, and the sound intensity is 10x higher.

The amplitude of the wave increases as the square root of the power, so the amplitude is increased by √[10] times.

Also, in terms of amplitude, db = 20*log(A1/A2), so for A1/A2 = √[10], db = 20*log√[10] = 20*0.5*log(10) = 10db. Note that the db as defined for power, 10*log(P1/P2) comes out the same as the db defined for amplitude, 20*log(A1/A2)