What is the relationship between sound intensity level and power?

AI Thread Summary
To lower the sound intensity level from 90 dB to 70 dB, a significant reduction in acoustic power is required. The calculations indicate that the new power level would be 1% of the original power. This result seems surprising, as a 20 dB decrease corresponds to a dramatic drop in power. The relationship between sound intensity level and power is governed by the formula B = 10 * log(I/I0), where I0 is the reference intensity. Understanding this logarithmic relationship clarifies why such a large reduction in power is necessary for a relatively small change in dB.
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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