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# Sound Intensity?

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Q) Two friends are walking, having a normal conversation; the sound level of the conversation if 60 dB.  When they get near a busy street, they have to double the intensity of the sound to be able to hear each other.  Calculate the new sound level.

I used the formula: L=10log(I/Io)

Then: 120 = 10log(I/10^-12)

Is that correct?

Jan 8, 2018

#1
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120dB is the level you might experience at a rock concert!

You need to double the I value:

L = 10log(2*I/Io)

L = 10log(I/Io) + 10log(2) → 60 + 3.01 ≈ 63dB

Jan 8, 2018
#2
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But I don't really get it. Why did you just add 10log(2)? Is there some sort of formula?

Julius  Jan 8, 2018
#3
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Remember, Julius, that

log ( a  * b)   =  log a + log b

So.....

Let a  =  I/I0       and  Let b   =   2

So

L  =  10 log  [  (2 I / I0 ) ]   =

10  [ log [  I / I0   *  2 ]   ]  =   distribute  the 10  over the above log property

10log (I / I0)  +  10log (2) =

60   +  3.01  ≈

63 dB   Jan 8, 2018
#4
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Okay, I see you're using the product law. But shouldn't division be the quotient law? Where they would be subtracted?

Even though, you said a = I/Io and b = 2, are you allowed to do that since the 2 is like writing 2I?

Julius  Jan 9, 2018
edited by Julius  Jan 9, 2018