The sound intensity level of a certain sound source is measured by two listeners

Question

The sound intensity level of a certain sound source is measured by two listeners located at different positions along a line from the source. The listeners are located on the same side of the source and are separated by 155.2 m. The listener that is closest to the source hears the sound with a sound intensity level of 65.5 dB. The sound intensity level of the sound heard by the more distant listener is 50.9 dB. 1) What is the distance from the source to the closest listener? 2) What is the power output of the source?

Explanation / Answer

let distance between source and closest observer = d m therefore distance between other observer and source = ( d + 155.2 ) m L1 = 65.5 dB , L2 = 50.9 dB since , L = 10* log ( I1 / I0 ) where I0 = 1*10^-12 W/m^2 on substituting and solving I1 = 3.55*10^-6 W/m^2 I2 = 0.123*10^-6 W/m^2 since I1 / I2 = ( d + 155.2)^2 / d^2 therefore d = 35.5 m a) distance from the source to the closest listener i.e. d = 35.5 m b) since P / 4*pi*r^2 = I taking r = d , and I = I1 or r = ( d+152.2 ) , I = I2 we get , P = 0.056 W

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