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Sound Level

  1. Jan 13, 2008 #1
    [SOLVED] Sound Level

    1. The problem statement, all variables and given/known data

    If sound intensity level is reduced from 90dB to 75 dB, what is the ratio of the new sound level to the old?

    2. Relevant equations

    Sound Intensity Level = 10 log (I/Io)


    3. The attempt at a solution

    10 log 90 = 1.954
    10 log 75 = 1.875

    1.875/1.954 = .956
    This probably goes against every physical law that exists, but it's all I've come up with so far. Help, please!
     
  2. jcsd
  3. Jan 13, 2008 #2
    Its easy!

    Its 1/10^2.5

    Sorry that was for me!!

    Its 1/10^1.5 for you

    90--->75

    Mine was 90-65
     
    Last edited: Jan 13, 2008
  4. Jan 13, 2008 #3
    Could you please explain why?
     
  5. Jan 13, 2008 #4
    This 15 db difference corresponds to a 1.5-Bel difference. This difference is equivalent to a sound which is 10^1.5 more intense. Always raise 10 to a power which is equivalent to the difference in "Bels.

    Or less intense in this case thus dividing by 1.
     
  6. Jan 13, 2008 #5
    Thank you!
     
  7. Jan 13, 2008 #6
    Now i sent you a pm!! lol
     
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