At 25 m from a sound source the intensity level is 55 dB. What At 25 m from a so

Question

At 25 m from a sound source the intensity level is 55 dB. What

At 25 m from a sound source the intensity level is 55 dB. What's the level at distances of (a) 50 m, (b) 250 m? Write the equation for the loudness in decibels, twice. Write it once for the first distance d1, using I = P/(4?d2) and leaving Io in the equation (it will eventually divide out). Write it again for distance d2. Subtract the two loudness's (?1 - ?2). Rearrange the subtraction of the log portions into a division as shown in the review above. You should see that the power P and the 4? divide out.

Explanation / Answer

equation for intensity I = power/area = power/(4pi r1^2)

intnesity level B = 10 dB log (I/Io) where I0 = 10^-12 W/m^2

so given B = 55dB

so 10 dB log(I/I0) = 55

log (I/Io ) = 5.5

I/Io = 244.69

I = 2.44*10^-10 W/m^2

power = IA = 2.44*10^-10 * 4pi * 25*25

Power = 1.525 *10^-7

intensity art 50 m = I2 = 1.525*10^-7 /(4pi*50*50)

I2 = 0.00485

intensity I3 at 250 m is I3 = 1.525*10^-7 /(4pi*250*250)

I3 = 0.194

so B1-B2 = log (I2/I3)

B1-B2 = log(0.194/0.00485)

B1-B2 = 40

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