Sound Intensity level problem

  • Thread starter LeakyFrog
  • Start date
  • #1
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Homework Statement


What fraction of the acoustic power of a noise would have to be eliminated to lower its sound intensity level from 90 to 70dB?


Homework Equations


I = Power/Area
B = 10 * log *(I/I0)

I0 = 10-12W/m2

The Attempt at a Solution


The answer I got for this is that the new power is 1% of the original. Which I don't think is true. A nudge in the right direction would be appreciated
 

Answers and Replies

  • #2
2,049
316
You can just solve [itex] B = 10 * log (I/I_0) [/itex] for I with B=90 and B is 70

remember to use the base 10 logarithm and not the natural logarithm
 
  • #3
22
0
You can just solve [itex] B = 10 * log (I/I_0) [/itex] for I with B=90 and B is 70

remember to use the base 10 logarithm and not the natural logarithm

Thanks. I guess it is 1% of the original power after all. Seems kind of strange that it would have to drop so dramatically for only 20dB.
 

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