Question

If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, what must have been the reference pressure?

Answer 1

Given:

I = 30dB

P = 66 × $10^{-9}$ Pa

Solution:

Formula used:

I = $20\log_{10}(\frac{P}{P_{o}})$           (1)

where,

I = intensity of sound

P = absolute pressure

$P_{o}$ = reference pressure

Using Eqn (1), we get:

$30 = 20\log _{10}\frac{66\times 10^{-9}}{P^{o}}$

$P_{o}$ = $\frac{66\times 10^{-9}}{10^{1.5}}$

$P_{o}$ = 2.08 × $10^{-9}$ Pa