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

Answer:

#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#