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DB difference, intensity

  1. Apr 30, 2009 #1

    Oijl

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    1. The problem statement, all variables and given/known data
    Two sounds differ in sound level by 2.80 dB. By what factor is the one intensity greater than the other?


    2. Relevant equations



    3. The attempt at a solution

    Is the equation

    [tex]B[/tex] = 10dB log(I/Io)

    relevant?
     
    Oijl, Apr 30, 2009
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  3. Apr 30, 2009 #2

    rl.bhat

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    Homework Helper

    The equation is
    Bdb = 10log(I/Io)
    Now let B1 = 10log(I1/Io) and
    B2 = 10log(I2/Io)
    Take the difference and find the ratio of I1/I2.
     
    rl.bhat, Apr 30, 2009
  4. Apr 30, 2009 #3

    danago

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    Gold Member

    Lets say sound 1 has an intensity of I1 and intensity level B, and sound 2 has an intensity of I2 with intensity level B+2.8

    B = 10 log (I1 / I0)
    B+2.8 = 10 log (I2 / I0)

    See if you can use those

    EDIT: rl.bhat beat me to it :P
     
    danago, Apr 30, 2009
  5. Apr 30, 2009 #4

    Oijl

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    I'm sorry, but the algebra is confusing me. Could you show the first few steps, please?
     
    Oijl, Apr 30, 2009
  6. Apr 30, 2009 #5

    rl.bhat

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    Homework Helper

    B1 = 10log(I1/Io)
    B2 = 10log(I2/Io)
    B1 -B2 = 10[log(I1/Io) - log(I2/Io)]
    Use laws of logarithm to further simplification.
     
    rl.bhat, Apr 30, 2009
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