Asked by Steeve

Taking a decibel to be an increase in pressure of 12.2%, by what factor has the pressure changed when the sound level has fallen by 8.00 decibels?

Answers

Answered by Jennifer
1 decibel = A*log(P/P0) = A*log(1.122)

Where P0 is the initial pressure, P is the pressure

1 decibel = 0.05*A
A = 20

L = 20*log(P/P0)

8 = 20*log(P/P0)

Solve for P/P0
Answered by Ada
Jennifer's answer is correct, but you have to use -8
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