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Sound wave rock band problem

  • Thread starter JinM
  • Start date
66
0
1. Homework Statement
A rock group is playing in a club. Sound emerging
outdoors from an open door spreads uniformly in
all directions. If the decibel level is 70 dB at a dis-
tance of1.0 m from the door,at what distance is the
music just barely audible to a person with a normal
threshold ofhearing? Disregard absorption.

Given:
dB1 = 70 dB
r1 = 1.0 m

Unknowns:
I1 = ?
r2 = ?
Io = 10^-12 W/m^2

dB1 is the decibel level at a distance of 1m from the door. r1 is distance from the door. I1 is intensity at a distance 1m (r1) from the door. Io is threshold of hearing, at a distance of r2 (unknown) from the door.
2. Homework Equations
dB = 10 log(I/Io)
I = P/A = P/4πr^2


3. The Attempt at a Solution
db = 10 log(I/Io)
70 = 10 log(I/Io)
7log10 = log(I/Io)
10^7 = I/Io
10^-12 * 10^7 = I

I1 = 10^-5 W/m^2 -> Intensity 1

I1 = P/A
10^-5 = P/(4π)
P = 1.3 x 10^-4 W

Now I used the same equation to find r of threshold of hearing:

Io = P/A
10^-12 = (1.3 x 10^-4)/(4πr^2)

r = sqrt([1.3x10^-4]/[10^-12x4π]) = 3216 m

This is what I got for r, but it just doesn't seem reasonable. Is this the correct answer? Have I done something wrong?

Thank you, and please do tell if my work is incomprehensible.
 

Answers and Replies

nrqed
Science Advisor
Homework Helper
Gold Member
3,540
181
1. Homework Statement
A rock group is playing in a club. Sound emerging
outdoors from an open door spreads uniformly in
all directions. If the decibel level is 70 dB at a dis-
tance of1.0 m from the door,at what distance is the
music just barely audible to a person with a normal
threshold ofhearing? Disregard absorption.

Given:
dB1 = 70 dB
r1 = 1.0 m

Unknowns:
I1 = ?
r2 = ?
Io = 10^-12 W/m^2

dB1 is the decibel level at a distance of 1m from the door. r1 is distance from the door. I1 is intensity at a distance 1m (r1) from the door. Io is threshold of hearing, at a distance of r2 (unknown) from the door.
2. Homework Equations
dB = 10 log(I/Io)
I = P/A = P/4πr^2


3. The Attempt at a Solution
db = 10 log(I/Io)
70 = 10 log(I/Io)
7log10 = log(I/Io)
10^7 = I/Io
10^-12 * 10^7 = I

I1 = 10^-5 W/m^2 -> Intensity 1

I1 = P/A
10^-5 = P/(4π)
P = 1.3 x 10^-4 W

Now I used the same equation to find r of threshold of hearing:

Io = P/A
10^-12 = (1.3 x 10^-4)/(4πr^2)

r = sqrt([1.3x10^-4]/[10^-12x4π]) = 3216 m

This is what I got for r, but it just doesn't seem reasonable. Is this the correct answer? Have I done something wrong?

Thank you, and please do tell if my work is incomprehensible.
I did not check every numerical value but your reasoning is completely correct. Sounds good.

It's not surprising that calculations like this give distances that seem to be way too large compared to what we expect in real life. The reason is that th ecalculation neglect all absorption of energy by the air which is always present in real life (the air gets heated up a little, etc). It's one of those calculation where neglecting air friction is a very bad approximation (like when we calculate trajectories of projectiles and we neglect air friction..we get results that are completely different from real life observations).
 
66
0
Oh, I see. Thanks nrged. :-)
 

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