Question

By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 dB and 20 dB respectively

Answer 1

The intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.

Step-by-step explanation:

The intensity of sound is given by;

`I(dB) = 10Log((I)/(I_o) )`

where;

I is the intensity of the sound

I₀ is the threshold of sound intensity = 1 x 10⁻¹² W/m²

The intensity of sound at a rock concert

`120 = 10Log((I)/(1*10^(-12)) )\\\\12 = Log((I)/(1*10^(-12)) )\\\\(I)/(1*10^(-12)) = 10^(12)\\\\I = 1*10^(-12) *10^(12)\\\\I = 1*10^0\\\\I =1 \ W/m^2`

The intensity of sound of a whisper

`20 = 10Log((I)/(1*10^(-12)) )\\\\2 = Log((I)/(1*10^(-12)) )\\\\(I)/(1*10^(-12)) = 10^(2)\\\\I = 1*10^(-12) *10^(2)\\\\I = 1*10^(-10)\\\\I =10^(-10) \ W/m^2`

Thus, the intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.