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DB difference, intensity

  1. Apr 30, 2009 #1
    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)

  2. jcsd
  3. Apr 30, 2009 #2


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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.
  4. Apr 30, 2009 #3


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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
  5. Apr 30, 2009 #4
    I'm sorry, but the algebra is confusing me. Could you show the first few steps, please?
  6. Apr 30, 2009 #5


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    B1 = 10log(I1/Io)
    B2 = 10log(I2/Io)
    B1 -B2 = 10[log(I1/Io) - log(I2/Io)]
    Use laws of logarithm to further simplification.
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