What is the sound level for a noise that has an intensity of 9.9 × 10-5 watts/m2?

Question

asked Aug 25, 2019 212k views

2 Answers

Rick Love · Answer 1 · 2019-09-01T02:23:25+0000

Given: Sound Intensity ( I ) =
9.9 * 10^(-5)(watts)/(m^2)

We know, the approximate threshold of human hearing is at 1kHz. In watts/m^2 it's value is
=(9.9 * 10^(-5))/(10^(-12)) ⁻¹² W/m².

So, we can say, Reference sound intensity
(I_0) = 10⁻¹² W/m².

We have formula for "sound intensity level LI in dB" when entering sound intensity.

LI = 10×log (I / Io) in dB

Plugging values of I and Io in formula.

LI = 10 × log (
(9.9 * 10^(-5))/(10^(-12)) )

LI = 10 × log (99000000)

= 10 × 7.99564

LI = 79.96 dB.

Therefore, 79.96 dB is the sound level for a noise that has an intensity of 9.9 × 10-5 watts/m2.