The decibel level of a sound

Question

The decibel level of a sound is given by dB=10log(E/E0), where E0=10^−12watts/m^2 and E is the intensity of the sound.

(a) Find the intensity of a 120-decibel sound.

E=  watts per square meter.

(b) How many times more intense is the sound from (a) than a 70-decibel sound?

A sound at 120 decibels is  times more intense than a sound at 70 decibels.

(c) The loudness (perceived volume by the human ear) of a sound of d decibels doubles each time the intensity of a sound increases by a factor of ten. To the human ear, how many times louder does a 120-decibel sound seem than a 70-decibel sound?

A sound at 120 decibels sounds  times louder than a sound at 70 decibels.

Answer 1

Given:

E₀ = 10^-12 W/m²

a.) dB = 120 dB

dB = 10*log(E/E₀)

dB/10 = log(E/E₀)

10^(dB/10) = E/E₀

E = E₀*10^(dB/10)

E = (10^-12 W/m²) * 10^(120/10) = 1 W/m²

b.) First, calculate the new intensity level.

dB = 70 dB

E = E₀*10^(dB/10)

E = (10^-12 W/m²) * 10^(70/10) = 10^-5 W/m²

Next, to compare them, divide the old intensity level by the new intensity level.

1/10^-5 = 10⁵ times more intense.

c.) For every 10 times increase in intensity the loudness of a sound increases by a factor of two.

In this problem, our intensity increases by 10⁵. This means that the loudness increases by 2⁵, which is 32 times louder.