Sound intensity

markosheehan

Hi
I am stuck on this question .
If I knew the sound intensity I could use the formula B=(10dB)log10(I/Io) to determine the dB
But unfortunately I am only given the Watts of the sound from the source so I am unsure what to do.

romsek

Math Team
First we have to get the $200~W$ as $dB$.

I see they refer to the picowatt for this so

$P_0 = 10 \log_{10}\left(\dfrac{200}{10^{-12}}\right) = 143.01 ~dB$

at $100~m$ the loss due to spherical spreading is

$A = 10\log_{10}\left(\dfrac{1}{4\pi 10^2}\right) = -50.99~dB$

Thus at $100~m$ we have

$P =143.01 - 50.99 = 92.02~dB$

So given the choices I'd choose $92~dB$

Last edited:

skeeter

Math Team
$I = \dfrac{Power}{4\pi r^2}$

$I_0 = 10^{-12} \dfrac{w}{m^2}$

markosheehan

First we have to get the $200~W$ as $dB$.

I see they refer to the picowatt for this so

$P_0 = 10 \log_{10}\left(\dfrac{200}{10^{-12}}\right) = 143.01 ~dB$

at $100~m$ the loss due to spherical spreading is

$A = 10\log_{10}\left(\dfrac{1}{4\pi 10^2}\right) = -50.99~dB$

Thus at $100~m$ we have

$P =143.01 - 50.99 = 92.02~dB$

So given the choices I'd choose $92~dB$
I did not realize you could sub the 200 watts in for the intensity in the formula. I thought the units for intensity were always W/M^2

$I = \dfrac{Power}{4\pi r^2}$
$I_0 = 10^{-12} \dfrac{w}{m^2}$