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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? 

Julius  Jan 8, 2018
 #1
avatar+26745 
+3

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

Alan  Jan 8, 2018
 #2
avatar+946 
+1

Thanks for the reply! 

 

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
avatar+87294 
+1

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

 

 

cool cool cool

CPhill  Jan 8, 2018
 #4
avatar+946 
0

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

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