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

  1. Mar 23, 2010 #1
    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.
    Answer in units of m.

    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. jcsd
  3. Mar 23, 2010 #2

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

    I think you mean 3.6 million meters. :wink:
     
  4. Mar 23, 2010 #3
    Re: Sound waves, finding distance to reach threshold of hearing, I'm getting wrong an

    Oops, typo... question still stands though :)
     
  5. Mar 23, 2010 #4
    Re: Sound waves, finding distance to reach threshold of hearing, I'm getting wrong an

    Could you help with my question though please?
     
  6. Mar 23, 2010 #5

    collinsmark

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    Gold Member

    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. :cool: 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,

    [tex] 0 \ \textbox{dB} = 10 \ \textbox{log} \left( \frac{I_0}{I_0} \right) [/tex]

    Also take note that

    [tex] I \propto \frac{1}{r^2} [/tex]

    So construct the problem as you have already done,

    [tex] 116 \ dB = 10 \ \textbox{log} \left( \frac{I_1}{I_0} \right) [/tex]

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

    So we can say,

    [tex] I_0 \propto \frac{1}{(r_2)^2} [/tex]

    and

    [tex] I_1 \propto \frac{1}{(r_1)^2} = \frac{1}{(5.77 m)^2}[/tex]

    So make your appropriate substitutions, and note that

    [tex] 10 \ log \left( x^2 \right) = 20 \ log \left( x \right) [/tex]

    Solve for r2.
     
    Last edited: Mar 23, 2010
  7. Mar 23, 2010 #6
    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. :(
     
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