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Sound Intensity and sound proofing

  • Thread starter shaka23h
  • Start date
  • #1
38
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A recording engineer works in a soundproofed room that is 36.2 dB quieter than the outside. If the sound intensity in the room is 2.45 x 10-10 W/m2, what is the intensity outside?

It seems so ez yet I'm getting the wrong answer

ok so here is what I am able to do

the difference of 36.2dB is

I set 36.2dB to 10 log I/I0

and divide through by 10 on both sides so I get

log 10 ^ 3.52 = log I/I0

I get that I0 = 1x 10^-12 so

10^-12 x 10 ^3.62 would give me a I = 10 x -8.38 Is this the intensity difference? so when I add this value to my given value of inside the room I should get the value outside the room? But when I did this I got eh wrong answer.

Thanks for all ur help

Jason
 

Answers and Replies

  • #2
38
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anyone have any suggestions?
 
  • #3
OlderDan
Science Advisor
Homework Helper
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You have a typo in one place where you have 3.52 instead of 3.62, but in your calculation you have 3.62 so that is not your problem. Your problem is that the reference intensity in this problem is not the threshold of human hearing. The reference in this problem is the known intensity at one place and you are trying to find the intensity at another place where the dB difference between the two places is known.
 

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