Decibels Rating

**krzyrice** · March 3rd 2010, 04:11 PM

Sound A measures 30 decibels and sound B is 5 times as loud as sound A. What is the decibel rating of sound B to the nearest integer?

Alright so I know the formula for decibels is 10 log (I/I0). But how would I solve this problem? So would sound be be 150 decibels then since it is 5 times as loud as sound A? And what would I do with the decibels afterward to fit it in the equation. The equation seems to only allow inputs such as 10^-16 and such.

Please help me with this. Thank you very much.

**skeeter** · March 3rd 2010, 04:39 PM

Originally Posted by krzyrice

Sound A measures 30 decibels and sound B is 5 times as loud as sound A. What is the decibel rating of sound B to the nearest integer?

Alright so I know the formula for decibels is 10 log (I/I0). But how would I solve this problem? So would sound be be 150 decibels then since it is 5 times as loud as sound A? And what would I do with the decibels afterward to fit it in the equation. The equation seems to only allow inputs such as 10^-16 and such.

Please help me with this. Thank you very much.

decibels is a logarithm. the relationship is not linear.

sound level in dB, $L = 10\log\left(\frac{I}{I_0}\right)$ where $I_0 = 10^{-12}$ watts per square meter ... the sound intensity at the threshold of human hearing.

for a 30 dB sound level ...

$30 = 10\log\left(\frac{I}{I_0}\right)$

$3 = \log\left(\frac{I}{I_0}\right)$

change to an exponential equation ...

$\frac{I}{I_0} = 10^3$

$\frac{I}{10^{-12}} = 10^3$

$I = 10^{-9}$ ... the intensity of a 30 dB sound source.

now increase this intensity by a factor of 5 ...

$I = 5 \cdot 10^{-9}$

find the new sound level in dB ...

$L = 10\log\left(\frac{5 \cdot 10^{-9}}{10^{-12}}\right) \approx 37$ dB

ok ... that's the long way to do it. here is a shortcut.

first, note that $30 = 10\log\left(\frac{I}{I_0}\right)$ .

increase the intensity by a factor of 5, and find the new sound level , $L$ ...

$L = 10\log\left(\frac{5I}{I_0}\right)$

using the laws of logs ...

$L = 10\log(5) + 10\log\left(\frac{I}{I_0}\right)$

$L = 10\log(5) + 30 \approx 37$ db

**krzyrice** · March 3rd 2010, 04:49 PM

Thank you so much skeeter for helping me on this!

Oh by the way is the sound intensity at the threshold of human hearing I0 = 10^-16? Well that is what my book says.

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