How much more intense is sound at 40dB than sound at 0 dB?

Answer 1

Ok, remember this :

dB = 10dB*log(I / Io)

where : Io = 10^-12 W/m^2

I = intensity of the sound

0 = 10dB*log(I / Io)

I = Io = 10^-12 W /m^2

for 40 dB

40 = 10*log(I / Io)

10^4 = I / Io

I = 10^-8 W / m^2

So the intensity at 40 db, is 10^4 times the intensity at 0 db

Hope that helps

Answer 2

Rule of thumb is each time sound increases by 3dB, the sound energy doubles.

So 40dB/3dB = 13.3
2^13.3 = 10 321 x more = 10 000

So the sound is 10 000 more intense.

Answer 3

Relative intensity in decibels

dB = 10 log (I1/I2)

here dB = 40

40 = 10 log (I1/I2)

(I1/I2)= 4

I(40) = 10^4 I(0)