Sound waves, finding distance to reach threshold of hearing, I'm getting wrong answer

1. Mar 23, 2010

stevenbhester

1. The problem statement, all variables and given/known data
A rock group is playing in a bar. Sound
emerging from the door spreads uniformly in
all directions. The intensity level of the music
is 116 dB at a distance of 5.77 m from the
door.
At what distance is the music just barely
audible to a person with a normal threshold
of hearing? Disregard absorption.

So,
Given-
I1(dB) (the intensity level 5.77 meters from the door)=116 dB
r1 (distance from door when intensity is 116 dB)= 5.77 m
Io (Intensity at threshold of hearing)= 1e-12

Unknown-
r2 (Radius at threshold of hearing)
P (power of sound source)
I1(w/m^2) (intensity 5.77 meters from door in watts/meters squared)

2. Relevant equations
dB=10log(I/Io)
P=4*I*π*r^2
r=√(P/4πI)

3. The attempt at a solution
First, I changed the given Intensity into W/m^2 instead of hertz.

dB=10log(I1/Io)
116=10log(I1/1e-12)
11.6=log(I1/1e-12)
10^11.6=I1/1e-12
(1e-12)(10^11.6)=I1

.3981071706=I1

So that's the Intensity at the spot from the door mentioned, so now I calculated the power source.

P=4*I1*π*r1^2
P=4*.3981071706*π*5.77^2

P=166.5564633

So, now that I had the power source, I calculated the radius needed to achieve threshold of hearing

r2=√(P/4πIo)
r2=√(166.5564633/4*π*1e-12)
r2=√(166.5564633/1.256637061e-11)
r2=√1.325414222e13

r2=3640623.878

Doesn't seem right... 3.6 million miles seems overkill.

Last edited: Mar 23, 2010
2. Mar 23, 2010

collinsmark

Re: Sound waves, finding distance to reach threshold of hearing, I'm getting wrong an

I think you mean 3.6 million meters.

3. Mar 23, 2010

stevenbhester

Re: Sound waves, finding distance to reach threshold of hearing, I'm getting wrong an

Oops, typo... question still stands though :)

4. Mar 23, 2010

stevenbhester

Re: Sound waves, finding distance to reach threshold of hearing, I'm getting wrong an

Could you help with my question though please?

5. Mar 23, 2010

collinsmark

Re: Sound waves, finding distance to reach threshold of hearing, I'm getting wrong an

It may seem like a lot, I agree. But 116 dB is pretty loud, and we are ignoring absorption and all that. I mean it's really only 3640 km. That's just peanuts compared to the size of, say, the galaxy.

Seriously though, I don't see any mistakes with your math. There is a much easier way to solve this problem, but the answer comes out the same as yours.

Edit:
Okay, I'll bite with the easier solution. After all, you did already get the answer.

Note that I'll use 0 dB use the human threshold of hearing. In other words,

$$0 \ \textbox{dB} = 10 \ \textbox{log} \left( \frac{I_0}{I_0} \right)$$

Also take note that

$$I \propto \frac{1}{r^2}$$

So construct the problem as you have already done,

$$116 \ dB = 10 \ \textbox{log} \left( \frac{I_1}{I_0} \right)$$

Now note that at some distance r2 we're going to end up with an intensity I0

So we can say,

$$I_0 \propto \frac{1}{(r_2)^2}$$

and

$$I_1 \propto \frac{1}{(r_1)^2} = \frac{1}{(5.77 m)^2}$$

So make your appropriate substitutions, and note that

$$10 \ log \left( x^2 \right) = 20 \ log \left( x \right)$$

Solve for r2.

Last edited: Mar 23, 2010
6. Mar 23, 2010

stevenbhester

Re: Sound waves, finding distance to reach threshold of hearing, I'm getting wrong an

Thanks, I guess the website is at fault. It maintains that I'm wrong. :(