At the new location, because of the inverse square law, the sound intensity will be higher by a factor
I2/I1 = (1/0.67)^2 = 2.23
10 log(10) I2/I1 = 10*0.348 = 3.48 db
I agree with your answer. You did nothing wrong.
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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% as far from the source?
My answer choices are 23, -23, 40, 2.76, -2.76, 1.94, -1.94, 2.57, -2.57, and 0
I worked out this problem and came up with the following:
db = 10 log(P2/P1) = 10 log (1/0.67)2 = 20 log (1/0.67) = -20 log (0.67) =
+3.479 dB
This isn't one of my choices. Can you please help and advise me what I did wrong? Thank you!
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