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We're not sure of the solution, so we'll post the work here, and let others

tear it apart:

Sound intensity is proportional to the inverse square of the distance from the source.

So the new sound intensity is ( 1/0.67 )2 times the original value.

db = 10 log(P2/P1) = 10 log (1/0.67)2 = 20 log (1/0.67) = -20 log (0.67) =

+3.479 dB

Check:

(all together now . . . everybody sing, dance, and wave your hands in the air)

That sounds about right . . . we know that 0.707 is 1/2 sqrt(2),

which would produce a change of 3 dB.

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13y ago
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1mo ago

The change in decibels experienced by the observer is approximately -6 dB. This is because the change in distance is inversely proportional to the change in intensity, which is logarithmically related to decibels. So, if the distance is decreased by 67%, the intensity (and therefore the decibel level) will decrease by about -6 dB.

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Q: What approximate change in decibels does an observer experience from a sound source if the observer moves to a new location that is 67 percent as far from the source?
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