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)((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)(66* 10^(-9))/(P^(o))`

`P_(o)` =
`(66* 10^(-9))/(10^(1.5))`

`P_(o)` = 2.08 ×
`10^(-9)` Pa