A trumpet creates a sound intensity level of 1.22 102 dB at a distance of 1.10 m

Question

A trumpet creates a sound intensity level of 1.22 102 dB at a distance of 1.10 m. (a) What is the sound intensity of a trumpet at this distance? W/m2 (b) What is the sound intensity of five trumpets at this distance? W/m2 (c) Find the sound intensity of five trumpets at the location of the first row of an audience, 7.40 m away, assuming, for simplicity, the sound energy propagates uniformly in all directions. W/m2 (d) Calculate the decibel level of the five trumpets in the first row. dB (e) If the trumpets are being played in an outdoor auditorium, how far away, in theory, can their combined sound be heard? m (f) In practice such a sound could not be heard once the listener was 2-3 km away. Why can't the sound be heard at the distance found in part (e)? Hint: In a very quiet room the ambient sound intensity level is about 30 dB.

Explanation / Answer

a) I = (10^(1.22*10^2/10))*10^-12 = 1.58 W/m^2

b) 5 trumpets would produce 5*1.584W/m^2 = 7.924 W/m^2

c) Since sound is an inverse square law we have 7.924*1.10^2/7.40^2 = 0.175W/m^2

d) = 10db*log(0.175/10^-12) = 112.43 dB

e) The threshold of hearing has an intensity of 10^-12W/m^2

so 7.924*7.4^2/r^2 = 10^-12

so r = sqrt(7.924*7.4^2/10^-12) = 1.08x10^7m

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