An electronic point source emits sound isotropically at a frequency of 4 time 10
ID: 1631865 • Letter: A
Question
An electronic point source emits sound isotropically at a frequency of 4 time 10^3 Hz and a power of 29 watts. A small microphone has an area of 0.71 cm^2 and is located 229 meters from the point source. What is the sound intensity at the microphone ? l = W/m^2 What is the power intercepted by the microphone? P_in = W How many minutes will it take for the microphone to recieve 0.1 Joules from this sound? DeltaT = min What is the sound intensity level at the microphone from this point source? B = dB What would be the sound intensity level at the microphone if the point source doubled its power output? B = dBExplanation / Answer
(1)
sound intensity at the microphone is
I = P/A = P/ 4pir^2 = 29/ 4pi ( 229)^2 = 4.4* 10^-5 W/m^2
(2)
P_in = IA = 4.4* 10^-5 W/m^2(0.71* 10^-4)= 3.124 * 10^-9 W
(3)
t = U/P = 0.1 J/3.124 * 10^-9 W= 32010243.27 s = 533504 min
(4)
beta = 10log ( I / I_0 )
= 10 log ( 4.4*10^-5 W/m^2 / 10^-12 )
= 76.43 dB
(5)
I = 2*4.4* 10^-5 = 8.8 * 10^-5 W/m^2
beta = 10 log ( 8.8*10^-5 W/m^2 / 10^-12 )
= 89.44 dB
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