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?
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
Jennifer's answer is correct, but you have to use -8
To determine the factor by which the pressure has changed when the sound level has fallen by 8.00 decibels, we need to understand the relationship between decibels and pressure.
Decibels are a logarithmic unit used to measure the intensity or relative level of sound. When we say that the sound level has fallen by 8.00 decibels, it means that the sound intensity has decreased by 8.00 decibels compared to the reference level.
Now, the question states that a decibel represents an increase in pressure of 12.2%. This means that for every increase of 1 decibel, the pressure increases by 12.2%.
To find the factor by which the pressure has changed, we'll use the formula:
Factor = 10^(dB/10)
In this case, we want to find the factor when the sound level has fallen by 8.00 decibels. Substituting the given value:
Factor = 10^(-8.00/10)
Calculating this expression:
Factor = 10^(-0.80)
Using logarithmic properties:
Factor = 0.1585
Therefore, the pressure has decreased by a factor of 0.1585 when the sound level has fallen by 8.00 decibels.