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

Answer:

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

Explanation:

The intensity of sound is given by;

$I(dB) = 10Log(\frac{I}{I_o} )$

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

$120 = 10Log(\frac{I}{1*10^{-12}} )\\\\12 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{12}\\\\I = 1*10^{-12} *10^{12}\\\\I = 1*10^0\\\\I =1 \ W/m^2$

The intensity of sound of a whisper

$20 = 10Log(\frac{I}{1*10^{-12}} )\\\\2 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{2}\\\\I = 1*10^{-12} *10^{2}\\\\I = 1*10^{-10}\\\\I =10^{-10} \ W/m^2$

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