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Can you hear me now outside?

  1. Nov 29, 2006 #1
    A recording engineer works in a soundproofed room that is 45.0 dB quieter than the outside. If the sound intensity in the room is 1.30e-10 W/(m^2), what is the intensity outside?

    I know this intensity formula ...

    ... I = (P/A) OR I = ( P / [(4)(pi)(r^2)] )

    ... and for dB ...

    ... beta = (10 dB)(log (I/Io)


    ... I can't make sense of the 2 equations to know what to do next ... any ideas?
     
  2. jcsd
  3. Nov 29, 2006 #2
    Do I substitute 45 dB for the 10 in the dB equation?
     
  4. Nov 29, 2006 #3

    OlderDan

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    The outside sound is the more intense sound. The intensity inside is a fraction of the intensity of the outside sound such that the ratio gives a 45 dB difference. What is the value of beta, including its sign, if you take the outside sound intensity as the reference? What is it if you take the inside intensity as the reference?
     
  5. Nov 29, 2006 #4
    I think I have to revisit working with logs ... it's been a while ... LOL - I'll get back to you though.
     
  6. Nov 29, 2006 #5
    I came up with this ...

    ... 45 = 10 log ( I / 1e-10 ) = 10 log ( 1.3e-12 / 1e-10 ) ...

    ... does that sound right? LOL
     
  7. Nov 29, 2006 #6

    OlderDan

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    No it doesn't. The 45 is in the right place, but where did 1e-10 come from? With a +45 on the left, the sound intensity in the room should be the reference I_o and the intensity outside the I. Waht is the inverse function of the log functiuon?
     
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