A firework rocket explodes at an altitude of 350 m, producing an average sound i

Question

A firework rocket explodes at an altitude of 350 m, producing an average sound intensity of 8.6 x 10-2 W/m2 for 0.20 sec at a point on the ground directly below it.

a) What is the average intensity of the sound at a distance 5.0 m from the rocket?

b) What is the sound level at a point just below the rocket (at ground level) compared to a point 5.0 m from the rocket?

c) What is the difference in sound level between Tom, standing (on the ground) directly below the explosion and Jane, standing 180 m from Tom? Is this a noticeable reduction in sound heard by Jane as compared to Tom? Where could Jane stand, with respect to Tom, for there to be a “noticeable” reduction in noise (i.e., 3.0 dB or more)?

Explanation / Answer

a)

intensity I = P/(4*pi*r^2)

since the power of source is same

I2/I1 = (r1/r2)^2

I2/(8.6*10^-2) = (350/5)^2

I2 = 421.4 W/m^2

==================

(b)

sound level = 10*log(I/I0)

I0 = threshold intensity = 10^-12 W/m^2

sound level = 10*log(421.4/(10^-12))

sound level = 146.2 dB

====================

for Jane

r2 = sqrt(350^2+180^2) = 393.6 m

I2 = I1*(r1/r2)^2 = 8.6*10^-2*(350/393.6)^2 = 0.068 W/m^2

sound level dB2 = 10*log(I2/Io) = 10*log(0.068/(10^-12)) = 108.32 dB

for Jane

sound level dB1 = 10*log(I1/Io) = 10*log((8.6*10^-2/(10^-12)) = 109.34

difference = 142.6 - 109.34 = 33.26 dB

for Jane intensitylevel = 109.34 - 3 = 106.34 dB

10*log*(I2/I0) = 106.34

10*log(I2/10^-12) = 106.34

I2 = 0.043

I2/I1 = (r1/r2)^2

0.043/(8.6*10^-2) = (350/r2)^2

r2 = 495 m

sqrt(350^2+d^2) = 495

d = 350 m

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