A very noisy chain saw operated by a tree surgeon emits a total acoustic power o
ID: 2010808 • Letter: A
Question
A very noisy chain saw operated by a tree surgeon emits a total acoustic power of 22.0 W uniformly in all directions.
A: At what distance from the source is the sound level equal to 107 dB?
R1 = ? m
B: At what distance from the source is the sound level equal to 50.0 dB ?
R2 = ? km
Also, A trumpet player is tuning his instrument by playing an A note simultaneously with the first-chair trumpeter, who has perfect pitch. The first-chair player's note is exactly 440 Hz, and 2.00 beats per second are heard.
What are the two possible frequencies of the other player's note?
f = ? Hz
Help with how to work these problems would really be appreciated! Thanks!
Explanation / Answer
Given that Total acoustic power P = 22.0 W ------------------------------------------------------------ We have the formula for the intensity level is = (10 dB)log [I/Io] A) Given that = 107 dB (107 dB) = (10 dB)log [I/Io] I = Io 10^10.7 = (10 ^-12 W/m2)(10 ^10.7) = 0.050 W/m2 But I = P/(4R^2) R = [P/(4I)] = [(22.0 W)/(4(0.050 W/m2))] = 6.0 m B) Given that = 50.0 dB (50.0 dB) = (10 dB)log [I/Io] I = Io 10^5 = (10 ^-12 W/m2)(10 ^5) = 10 ^-7 W/m2 But I = P/(4R^2) R = [P/(4I)] = [(22.0 W)/(4(10 ^-7 W/m2))] = 1323.14 m B) Given that = 50.0 dB (50.0 dB) = (10 dB)log [I/Io] I = Io 10^5 = (10 ^-12 W/m2)(10 ^5) = 10 ^-7 W/m2 But I = P/(4R^2) R = [P/(4I)] = [(22.0 W)/(4(10 ^-7 W/m2))] = 1323.14 m (50.0 dB) = (10 dB)log [I/Io] I = Io 10^5 = (10 ^-12 W/m2)(10 ^5) = 10 ^-7 W/m2 But I = P/(4R^2) R = [P/(4I)] = [(22.0 W)/(4(10 ^-7 W/m2))] = 1323.14 mRelated Questions
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