If the intensity of a sound from one source is 100 times that of another sound, how much more is the decibel level of the louder sound than the softer one?

1 Answer
Oct 30, 2017

2 times

Explanation:

Let the intensity of the softer sound be xcolor(white)iW/m^2.
therefore the intensity of the louder sound is 100xcolor(white)iW/m^2.

beta=10color(white)i"dB"*log(I/I_o)

For the softer sound:

beta_1=10color(white)i"dB"*log((xcolor(white)iW/m^2)/(10^-12color(white)iW/m^2))

beta_1=10*log(10^14 x)

For the louder sound:

beta_2=10color(white)i"dB"*log((100xcolor(white)iW/m^2)/(10^-12color(white)iW/m^2))

beta_2=10*log(10^12 x)

How much more ?
=beta_1/beta_2

=(10*log(10^14 x))/(10*log(10^12 x)

=2