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 a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

Step-by-step explanation:

Given;

rock concert sound intensity level, β₁ = 120 dB

whisper sound intensity level, β₂ = 20 dB

The sound intensity level is given as;

`\beta = 10Log((I)/(I_o) )\\\\`

where;

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

I is the sound intensity

Intensity of sound at rock concert ;

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

The intensity of sound of a whisper;

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

Determine the factor by which the intensity of sound at a rock concert louder than that of a whisper

`(I_(Concert))/(I_(whisper)) = (1)/(10^(-10)) \\\\(I_(Concert))/(I_(whisper)) = 1 * 10^(10)\\\\I_(Concert) = 1 * 10^(10)*I_(whisper)`

Therefore, the intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰