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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?
Wiki User
∙ 13y ago
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.
Wiki User
∙ 13y ago
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