Intensity and Power An electronic point source emits sound isotropically at a fr

Question

Intensity and Power
An electronic point source emits sound isotropically at a frequency of 3000 Hz and a power of 27 watts. A small microphone has an area of 0.7 cm2 and is located 199 meters from the point source. a) What is the sound intensity at the microphone ?

I = W/m2 *
5.43*10^-5 OK
b) What is the power intercepted by the microphone?

Pin = W
4.28*10^-9 NO
c) How many minutes will it take for the microphone to receive 0.3 Joules from this sound?

?T = min
d) What is the sound intensity level at the microphone from this point source?

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

? = dB
10.86*10^-5 NO

thanks!

Explanation / Answer

An electronic point source emits sound at a frequency is f = 3000 Hz An electronic point source at a power is P = 27 W Area of the microphone is A = 0.7 cm^2 a ) Sound intensity of the microphone is I = P / A                                                                   = P / 4r^2                                                                   = 27 W / 4(199m)^2                                                                   = 5.42*10^-5 W/m^2 b ) Power intercepted by the microphone is P = I A                                                                         = 5.42*10^-5 * 0.7 *10^-4 m^2                                                                         = 3.79*10^-9 W c ) time will take by the microphone is t = E / P                                                                 = 0.3 J / 3.79*10^-9 W                                                                 = 1.31*10^6 min d ) Sound intensity level is      I = P / A                                                    = P / 4r^2                                                    = 27 W / 4(199m)^2                                                      = 5.42*10^-5 W/m^2                      = 10log ( I / I_0 )                           = 10 log ( 5.42*10^-5 W/m^2 / 10^-12 )                           = 77.3 dB e ) If Power is doubled the intensity is also doubled then I =   5.42*10^-5 W/m^2 * 2                                                                                           = 10.84*10^-5 W /m^2                      = 10log ( I / I_0 )                           = 10 log ( 10.84*10^-5 W/m^2 / 10^-12 )                           = 80.35 dB                                                    = P / 4r^2                                                    = 27 W / 4(199m)^2                                                      = 5.42*10^-5 W/m^2                      = 10log ( I / I_0 )                           = 10 log ( 5.42*10^-5 W/m^2 / 10^-12 )                           = 77.3 dB e ) If Power is doubled the intensity is also doubled then I =   5.42*10^-5 W/m^2 * 2                                                                                           = 10.84*10^-5 W /m^2                      = 10log ( I / I_0 )                           = 10 log ( 10.84*10^-5 W/m^2 / 10^-12 )                           = 80.35 dB                      = 10log ( I / I_0 )                           = 10 log ( 10.84*10^-5 W/m^2 / 10^-12 )                           = 80.35 dB

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