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How to compare decibels?
How many times greater is the intensity of a sound of 110 dB than the intensity of a sound of 90 dB?
Please show me how to do this, I'm not just looking for a free pass on my homework.
1 Answer
- KeplJoeyLv 71 decade agoFavorite Answer
The relationship between the intensity of sound and its dB measurement is given by the equation in
http://hyperphysics.phy-astr.gsu.edu/hbase/sound/i...
dB = 10 log(I/Io) <the log is to base 10>
Dividing by 10 and then taking 10 to the power of both sides,
I/Io = 10^(dB/10), si
I = lo*10^(dB/10).
Therefore the intensity of a 110 dB sound is
Io * 10^(110/10) = 10^11 Io.
And the intensity of a 90 dB sound is
Io * 10^(90/10) = 10^9 Io.
So to find the number of times greater the intensity of a 110 dB is, just divide them. The 110 dB intensity is
(10^11 Io)/(10^9 Io)
= 10^11/10^9 <Io's cancel>
= 10^2 = 100.
