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Middle ear amplifies

  1. Jun 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Hei!
    I have problem with this quest:
    Middle ear amplifies the sound waves entering cochlea by 30dB. This means that in case of total dysfunction of ossicles the sound energy transferred into cochlea is reduced:
    a)3 times b)30 times c)1000 times d)3000 times

    I don know how should I start with it..which formula?
     
  2. jcsd
  3. Jun 23, 2012 #2

    tiny-tim

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    welcome to pf!

    hei ilona! welcome to pf! :smile:

    total dysfunction means no amplification, so it's 30dB quieter

    sooo … what is the definition of dB ? :wink:
     
  4. Jun 23, 2012 #3
    logarithmic unit of ratio of intensity to standard treshold of hearing(10^-12)
    dB=log^10*I/Io
     
  5. Jun 23, 2012 #4

    ehild

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    That is almost the definition of loudness, but not quite.
    Amplification is defined in terms of output intensity and input intensity. And you meant base 10, not log on the 10th power... and it is decibel, that "deci" means something...
    See: http://en.wikipedia.org/wiki/Decibel#Definition

    ehild
     
  6. Jun 23, 2012 #5

    tiny-tim

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    (try using the X2 and X2 buttons just above the Reply box :wink:)
    isn't that B, not dB ? :confused:
     
  7. Jun 23, 2012 #6

    ehild

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    It is not anything. log without argument multiplied by I/Io...
     
  8. Jun 23, 2012 #7
    L(B)=log10*I/Io

    but i wonder if there is no amplification we have 1 dB ,haven't we?
    so why is not it just reduced by 30 times???
     
  9. Jun 23, 2012 #8

    ehild

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  10. Jun 23, 2012 #9

    tiny-tim

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    yes, ilona, if I/Io = 1, then log I/Io = … ? :smile:
     
  11. Jun 23, 2012 #10
    log1= 0
     
  12. Jun 23, 2012 #11

    tiny-tim

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    right, if the amplification ratio is 1, then the dB difference is 0

    and if the amplification ratio is 10, then the B difference is 1, and the dB difference (10 times as much) is 10

    so if the amplification ratio is 100, then the B difference is … ? :smile:
     
  13. Jun 23, 2012 #12
    ok , i think answer will be 1000times
     
  14. Jun 23, 2012 #13
    but once again , cause I don't think I understand it..
    I have this formula LB=log10p1/p0
    so I have 30 dB so it is 3 B
    and now why p1/po is 1?
     
  15. Jun 23, 2012 #14

    tiny-tim

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    yes :smile:
    forget the formula :rolleyes:

    (you'll never remember it in the exam, anyway)

    just remember ('cos it's easy) …

    10 dB = 10 times louder

    then you know that 1 B = 10 times louder, so 2 B = 100 times etc, and finally convert to dB by always multiplying by 10 (same as converting grams to decigrams, or metres to decimetres) :wink:
     
  16. Jun 23, 2012 #15
    ok, thank you very much =)
     
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