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Sound Intensity Please help

  1. Mar 3, 2008 #1
    [SOLVED] Sound Intensity Please help!!!

    1. The problem statement, all variables and given/known data
    The sound intensity level 5.0m from a point source is 95 dB. At what distance will it be 75 dB?


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


    3. The attempt at a solution
    I honestly don't know what to do with this problem. I started out by plugging 5.0m and 95 dB into the equation >>

    dB = 10log((P/4*pi*r^2)/Io)
    >> 95 = 10log((P/4*pi*25)/1*10^-12)
    >> 95 =10log(P/4*pi*25)-10log(1*10^-12)
    >> 95 + 10log(1*10^-12) = 10log(P/4*pi*25)
    >> -25 = 10log(P)-10log(4*pi*25)
    >> -25+10log(4*pi*25) = 10log(P)
    >> -.0307 = 10log(P)
    >> -.00307 = log(P)
    >> P = 10^-.00307

    After that, I used the P that I found in the equation >>

    75 = 10log((10^-.00307/4*pi*r^2)/1*10^-12)
    >> 75 + 10log(1*10^-12)=10log(10^-.00307/4*pi*r^2)
    >> -45 = 10log(10^-.00307)-10log(4*pi*r^2)
    >> -44.95 =-10log(4*pi)+10log(r^2)
    >> -33.96 = 10log(r^2)
    >> -3.39 = log(r^2)
    >> r^2 = 10^-3.39
    >> r = .0200

    This doesn't make any sense at all, but I don't know what I'm doing wrong. Our physics teacher didn't teach us the lesson on this. He just gave us a bunch of formulas and told us to figure it out. Am I using the wrong formulas, or am I making a stupid math error? Please help!!!
     
  2. jcsd
  3. Mar 3, 2008 #2
    deleted
     
  4. Mar 3, 2008 #3
    r=50m
    95=10log(I/Io)
    75=10log(I'/Io)
    20=10log(I/I') - log properties
    2=log(I/I')
    100=I/I'
    I=100I'
    I/I' = r^2/R^2
    100=r^2/25
    r= 50

    in fact 10dB = 10 times , 20dB = 10^2 times.
    =.= my unlucky day
    always got mistakes, haha~
     
  5. Mar 3, 2008 #4
    Thank you so much for your help!

    So... let me see if I understand this...
    Instead of solving for P, you get the difference between the decibels and solve for r using the first distance given (R=5). Is that right?
     
  6. Mar 3, 2008 #5
    Yes, that's.:wink:
    A better approach is to "delete" as much "constants" as possible,
    it makes the calculation easier..
     
  7. Mar 3, 2008 #6
    er... one more question...

    would I add 25 to 50? r=50, which equals the difference between the two distances, right?
     
  8. Mar 3, 2008 #7
    No

    at distance = 5m 95dB
    at distance = 50m 75dB
    difference between two distance =50-5 = 45m
     
  9. Mar 3, 2008 #8
    Oh! I get it now! Thank you!!! ^^
     
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