During takeoff, the sound intensity of a jet engine is 140 dB at a distance of 1

Question

During takeoff, the sound intensity of a jet engine is 140 dB at a distance of 15 m.

a) What is the sound intensity in W/m2?

W/m2

b) If you are standing at this location for 1 minute, how much energy will be transferred to your eardrum? (Assume that your eardrum is 6.0 mm in diameter.)

J

c) What is the sound intensity in W/m2 if you are standing 1 km from the jet?

W/m2

d) What is the new decibel rating at this distance?

dB

e) How far from the jet would you need to be standing to hear a sound with a decibel rating of 70 dB?

m

Explanation / Answer

(a)
dB = 10 * log( I / (1.0 * 10^-12 W/m2))
Now,
140 = 10 * log( I / (1.0 * 10^-12 W/m2))
14 =  log( I / (1.0 * 10^-12 W/m2))
Taking antilog both sides,
10^14 =  I / (1.0 * 10^-12 W/m2))
I = 1.0 * 10^2 W/m2
I = 100 W/m^2

(b)
Area of ear drum, = *d^2/4 = (3.14 * (6.0*10^-3))/4 = 2.826 * 10^-5 m^2

Total Power, = 100 W/m^2 * 2.826 * 10^-5 m^2
P = 2.83 * 10^-3 W

Energy = Power * time
Energy = 2.83 * 10^-3 * 60 s
Energy = 0.169 J

(c)
The total power is distributed over the area of a sphere whose radius is the distance from the airplane.
Total Area = 4**r^2
A = 4**(1*10^3)^2 = 1.257*10^7 m^2
Now,
I = P/Area
I = (2.83 * 10^-3)/  1.257*10^7 W/m^2
I = 2.25 * 10^-10 W/m^2

(d)
dB = 10 * log( I / (1.0 * 10^-12 W/m2))
dB = 10 * log( (2.25 * 10^-10) / (1.0 * 10^-12))
dB = 23.52 dB

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