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How to calculate sound intensity when given decibels

  1. Jan 13, 2009 #1
    1. The problem statement, all variables and given/known data

    When a person wears a hearing aid, the sound intensity level increases by 30.0 dB. By what factor does the sound intensity increase?



    2. Relevant equations

    B = (10 dB) log (Ia/Ib)

    Ia = sound intensity
    Ib = threshold of human hearing (1.0 * 10^-12 W/m^2)


    3. The attempt at a solution


    I used the equation above, substituting into it:

    30 dB = (10 dB) log (Ia/1.0 * 10^-12 W/m^2)

    I divided both sides by 10 dB to get rid of the 10 dB on the right side of the equation.

    3 = log (Ia/1.0 * 10^-12 W/m^2)

    Then I converted this into regular exponential form without logs:

    10^3 = (I/1.0 * 10^-12 W/m^2)

    I get 1.0 * 10^-9 for I, but my answer key says the answer is 1000. This is 10^3, which I seem to have, but I still have an unknown - I - in the equation.

    :confused:
     
  2. jcsd
  3. Jan 13, 2009 #2

    Delphi51

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    Homework Helper

    Your answer of I = 1.0 * 10^-9 is correct.
    But the question doesn't ask for I, it asks by what factor has I increased - presumably from 0 db which is the threshold of hearing. So divide your answer for I by the value of Ib.
     
  4. Jan 13, 2009 #3

    rock.freak667

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    Homework Helper

    Right there is your answer. The wanted the factor, I/I0, not the intensity to give 30dB
     
  5. Jan 13, 2009 #4
    So I got the right answer all along...it was just that I had to realize that the question asked for what (Ia/Ib) was equal to, not just Ia.

    Thanks!
     
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