# Thread: Sound Level Intensity (Logarithm)

1. ## Sound Level Intensity (Logarithm)

Hey Guys, I have a simple question about sound level intensity (Log.)... so here it goes.

2 friends are walking and having normal conversation; the sound level of the conversation is 60 dB . When they get near a busy street, they double their intensity of the sound to be able to hear each other. What is the new sound level ?

Thanks

PS: You cannot just double the sound level intensity..

2. Originally Posted by MordernWar2
Hey Guys, I have a simple question about sound level intensity (Log.)... so here it goes.

2 friends are walking and having normal conversation; the sound level of the conversation is 60 dB . When they get near a busy street, they double their intensity of the sound to be able to hear each other. What is the new sound level ?

Thanks

PS: You cannot just double the sound level intensity..

Sound Level (dB): $\displaystyle L = 10 \log (\frac{I}{10^{-12}})$

Are you not able to use the formula to see what happens when you double intensity

3. Since a normal conversation is $\displaystyle 60dB$:

$\displaystyle 60 = 10 \log (\frac{I}{I{o}})$
$\displaystyle 6 = log(\frac{I}{I{o}})$
$\displaystyle 10^6 = (\frac{I}{I{o}})$
$\displaystyle 10^6{I{o}} = I$

Now that questions states that the intensity doubled near a busy street:

$\displaystyle dB = 10 \log (\frac{2I}{I{o}})$

Since $\displaystyle I = 10^6{I{o}}$:

$\displaystyle dB = 10 \log (\frac{2{10^6{I{o}}}}{I{o}})$
$\displaystyle dB = 10 \log (2*10^6)$

***I'm not sure if the next step is correct, though essentially you would cancel out log and base 10***

$\displaystyle dB = 10 (2*6)$
$\displaystyle dB = 10(12)$
$\displaystyle dB = 120$

Hope that helps!

4. Originally Posted by Sigra
Since a normal conversation is $\displaystyle 60dB$:

$\displaystyle 60 = 10 \log (\frac{I}{I{o}})$
$\displaystyle 6 = log(\frac{I}{I{o}})$
$\displaystyle 10^6 = (\frac{I}{I{o}})$
$\displaystyle 10^6{I{o}} = I$

Now that questions states that the intensity doubled near a busy street:

$\displaystyle dB = 10 \log (\frac{2I}{I{o}})$

Since $\displaystyle I = 10^6{I{o}}$:

$\displaystyle dB = 10 \log (\frac{2{10^6{I{o}}}}{I{o}})$
$\displaystyle dB = 10 \log (2*10^6)$

***I'm not sure if the next step is correct, though essentially you would cancel out log and base 10***

$\displaystyle dB = 10 (2*6)$
$\displaystyle dB = 10(12)$
$\displaystyle dB = 120$

Hope that helps!
No, that last bit isn't correct. Instead evaluate $\displaystyle dB = 10 \log (2*10^6)$ directly