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 27 watts. A small microphone has an area of 0.76 cm2 and is located 167 meters from the point SOURCE

1) What is the sound intensity at the microphonE IN W/M2.

2)What is the power intercepted by the microphone?

3) How many minutes will it take for the microphone to recieve 0.2 Joules from this sound?

4) What is the sound intensity level at the microphone from this point source?

5)What would be the sound intensity level at the microphone if the point source doubled its power output?

Explanation / Answer

givendata

f = 2*10^3 hz

power = 27 W

A = 0.76 cm^2=0.76*10^-4 m^2

distance = 167 m

1) What is the sound intensity at the microphonE IN W/M2.

Sound intensity I = P/A = 27W/(4r^2) = 34W/(4*167^2) = 9.7*10^-5 W/m^2

2)What is the power intercepted by the microphone?

Power intercepted = I*A (microphone) = 9.7*10^-5*0.76*10^-4m^2 = 7.372*10^-9W

3) How many minutes will it take for the microphone to recieve 0.2 Joules from this sound?

E = P*t

t = E/P = 0.2/(7.372*10^-9) = 2.7*10^7s = 10^5min

4) What is the sound intensity level at the microphone from this point source?

Sound intensity I = P/A = 27W/(4r^2) = 34W/(4*167^2) = 9.7*10^-5 W/m^2

5)What would be the sound intensity level at the microphone if the point source doubled its power output?

If P doubled then I would double = 9.7*10^-5 W/m^2*2 = 19.4*10^-4W/m^2

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