1. A sound wave with a sound level (Beta) of 80.0 dB is incident on an eardrum o
ID: 1300342 • Letter: 1
Question
1.
A sound wave with a sound level (Beta) of 80.0 dB is incident on an eardrum of area 0.600 x10 ^-4 m^2. how much ebergy is absorbed by the eardrum in 3 minutes?
2.
At point A, 3,00 m from a small source of sound that is emitting uniformly in all directions, the sound level is 53.0 dB.
a) what is the intensity of the sound at point A?
b) How far from the source would the intensity of the sound be 1/4 the intensity at point A?
c) How far from the source would the sound level be 1/4 of the sound level at point A?
Explanation / Answer
1) sound intensity level, beta = 10*log(I/Io)
80 = 10*log(I/10^-12)
8 = log(I/10^-12)
10^8 = I/10^-12
I = 10^-4 W/m^2
Intensity = power/Area
power = Intensity*Area
Energy = power*time
= Intentisty*Area*time
= 10^-4*0.6*10^-4*3*60
= 1.08*10^-6 J <<<<<<<<<<<----Answer
2)
a)sound intensity level, beta = 10*log(I/Io)
53 = 10*log(I/10^-12)
5.3 = log(I/10^-12)
10^5.3 = I/10^-12
I = 1.9953*10^-7 W/m^2 <<<<<<<<<<<----Answer
b) I1/I2 = (r2/r1)^2
==> r2 = r1*sqrt(I1/I2)
= r1sqrt(4)
= 2*r1
= 2*3
= 6 m <<<<<<<<<<<----Answer
c) sound intensity level, beta = 10*log(I/Io)
53/4 = 10*log(I/10^-12)
1.325 = log(I/10^-12)
10^1.325 = I/10^-12
I = 2.1135*10^-11 W/m^2
I2/I1 = (r2/r1)^2
r2 = r1*(sqrt(I2/I1)
= 3*sqrt(1.9953*10^-7/2.1135*10^-11)
= 291.5 m <<<<<<<<<<<----Answer
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