1. A train sounds its horn as it approaches an intersection. The horn can just b

Question

1. A train sounds its horn as it approaches an intersection. The horn can just be heard at a level of 48 dB by an observer 10 km away. (a) What is the average power generated by the horn? (Treat the horn as a point source and neglect any absorption of sound by the air.) (b) If someone is waiting at an intersection 41 m from the train, what intensity level the horn's sound does he hear? (c) What is the pressure amplitude of the wave at this point? (Bair = GammaP = 1.4 x 10^5 N/m^2, Pair=1.3 kg/m^3)

Explanation / Answer

First find the intensity of a 48dB sound using ? = 10dB*log(I/10^-12)

so I = 10^(48/10)*10^-12 = 6.3*10^-8 W/m^2

Now P = I*A = 6.3x10^-8*4?*(10000m)^2 = 79.16W

b) at 41m the intensity = P/A = 79.16 W/(4?*41^2) = 0.00374W/m2

This corresponds to a level of 10dB*log(0.00374/10^-12) = 95.7dB

c). pressure amplitude of a sound wave with an intensity level of 95.7 dB

if P is amplitude of pressure, then sound intensity (I) is proportional to P^2
I/Io = (P/Po)^2
dB = sound intensity level = 10 log[I/Io]
95.7 = 10 log[I/Io] = 10 log[P/Po]^2 = 20 log(P/Po)
log(P/Po) = 95.7/20 = 4.785
P = Po [10]^4.785
Po = reference pressure amplitude for human ear = 2*10^-5 N/m^2
P = 2*10^-5*[10]^4.785
P = 1.219 N/m^2

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