What is the relationship between sound intensity level and power?

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SUMMARY

The discussion centers on calculating the fraction of acoustic power that must be eliminated to reduce sound intensity from 90 dB to 70 dB. The relevant equations include I = Power/Area and B = 10 * log10(I/I0), with I0 defined as 10^-12 W/m². The conclusion reached is that the new power level is indeed 1% of the original power, highlighting the significant decrease in power required for a 20 dB reduction in sound intensity level.

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  • Understanding of sound intensity levels and decibels (dB)
  • Familiarity with logarithmic functions, specifically base 10 logarithms
  • Knowledge of acoustic power and its relationship to area
  • Basic principles of physics related to sound propagation
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LeakyFrog
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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
 
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You can just solve B = 10 * log (I/I_0) for I with B=90 and B is 70

remember to use the base 10 logarithm and not the natural logarithm
 
willem2 said:
You can just solve B = 10 * log (I/I_0) 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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