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Temperature's affect on frequency

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    A marching band member tunes her flute indoors in a 26 oC room. She then goes outside to play on a chilly 4 oC day in January. What percent change does this cause in the frequency played by her flute?)

    2. Relevant equations

    v = f(Lambda)

    3. The attempt at a solution

    I know the wavelength will not change with temperature. The speed of the sound does so I need to use an equation that uses wavelength, freq, and wave speed.

    I'm struggling with this. I calculated the speeds at 347.6 for 26 degrees C, and 337.4 for 4 degrees C. I chose the frequency at random as a 'C' she was tuning the flute to, the freq of this is 16.35. That's as far as I've gotten.
     
  2. jcsd
  3. Sep 16, 2009 #2
    I'm new to this forum so if this is posted in the wrong place or I've formatted it incorrectly, I apologize and if you could put it in the right direction that was would great!

    Thanks for any help.
     
  4. Sep 16, 2009 #3

    rl.bhat

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    Homework Helper

    You have done it correctly.

    As you have said, the wavelength remains constant. So
    v2 = f2*λ
    v1 = f1*λ
    take the ratio
    v2/v1 = f2/f1
    1 - v2/v1 = 1 - f2/f1
    Now find the percenatge change in the frequency.
     
  5. Sep 16, 2009 #4
    But how do I find V2=F2*λ if I don't know λ? Should I just select a wavelength at random? Like 1 meter?

    And how does V2/V1 = F2/F1 if the frequencies are the same?

    Strugging to understand how I can find the percentage change in the frequency.

    Thanks so much for helping me so far. I've been out of school for 8 years and am so very confused.
     
  6. Sep 16, 2009 #5

    rl.bhat

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    When the temperature changes, the length of the flute does not change. So the wave length does not change, because it depends on the length of the flute. So the frequency changes.
     
  7. Sep 16, 2009 #6
    Okay, still not understanding it I guess. How do I find the change in frequency? I have so far that

    347.6/16.35 = 21.260 = λ1
    337.4/16.35 = 20.640 = λ2

    So the equation V=fλ comes out to

    347.6=16.35*21.260
    337.4=16.35*20.640

    These work out obviously. I just don't understand how to find the change in frequency or how temperature effects it.
     
  8. Sep 16, 2009 #7

    rl.bhat

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    337.4/347.6 = f2/f1
    1- 337.4/347.6 = 1 - f2/f1
    (10.2/347.6)*100 = ( 1 - f2/f1)*100
    Find the percent change in the frequency.
     
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