An electronic point source emits sound isotropically at a frequency of 2 × 103 H

Question

An electronic point source emits sound isotropically at a frequency of 2 × 103 Hz and a power of 30 watts. A small microphone has an area of 0.74 cm2 and is located 182 meters from the point source. 1) What is the sound intensity at the microphone ? I = W/m2 Submit 2) What is the power intercepted by the microphone? Pin = W Submit 3) How many minutes will it take for the microphone to recieve 0.1 Joules from this sound? T = min Submit 4) What is the sound intensity level at the microphone from this point source? = 10^-12 dB Submit 5) What would be the sound intensity level at the microphone if the point source doubled its power output? =

Explanation / Answer

1) I = power/Area = 30/(4*3.14*182^2) = 7.21*10^-5 W/m^2

2) Pin = I*A = 7.21*10^-5*0.74*10^-4 = 5.335*10^-9 watts

3) E = Pin*t

t = E/pin

= 1/5.335*10^-9

= 1.87427*10^8 s


4)
B = 10dB*log(I/Io)

= 10 dB*log(7.21*10^-5/10^-12)

= 78.6 dB

5)

B = 10dB*log(2*I/Io)

= 10 dB*log(2*7.21*10^-5/10^-12)

= 81.59 dB

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