$\displaystyle B=log(\frac{I}{I_{0}})$

If B=58.4 dB and $\displaystyle I_{0}=10^{-12} \frac{W}{m^2}$

then:

$\displaystyle 58.4=log(\frac{I}{10^{-12}})$

log(I/

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- May 16th 2012, 06:16 PMlamp23Sound intensity
$\displaystyle B=log(\frac{I}{I_{0}})$

If B=58.4 dB and $\displaystyle I_{0}=10^{-12} \frac{W}{m^2}$

then:

$\displaystyle 58.4=log(\frac{I}{10^{-12}})$

log(I/ - May 16th 2012, 06:57 PMMoebiusRe: Sound intensity
If you are wanting to know $\displaystyle I $ it is as follows:

$\displaystyle 58.5=log(\frac{I}{10^{-12}}) $

$\displaystyle 10^{58.5}=\frac{I}{10^{-12}} $

$\displaystyle (10^{58.5})(10^{-12})=I $

$\displaystyle I=3.16 \times 10^{46} $ - May 20th 2012, 05:24 PMlamp23Re: Sound intensity
I meant to delete this. Can I not delete it after a set amount of time?