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Closed ended tube resonance problem

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data

    In class a 3 m long aluminum bar was made to resonate by clamping it at a node in the centre of the bar. The frequency heard was 1200Hz.


    2. Relevant equations

    [itex] L = \frac{n}{4}\lambda [/itex]

    [itex] v = f\lambda [/itex]

    3. The attempt at a solution

    Knowing that L is 3, we can substitute this into the formula and solve accordingly in order to find wavelength:

    [itex] 3 = \frac{1}{4}\lambda [/itex]
    [itex] (4)(3) = \lambda [/itex]

    Which gives us 12m. We can then use the wave equation, substituting 12 and 1200

    [itex] v = f\lambda [/itex]
    [itex] v =(1200)(12) [/itex]

    to arrive at [itex] 14400 m s^-1 [/itex]

    However, I am lead to believe that I am possibly wrong due to the relatively high value of velocity that I found. Am I wrong? If so, where is my mistake?

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 17, 2013 #2

    TSny

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    What must be at each end of the rod - a node or an antinode? Rethink the expression for ##L## in terms of ##\lambda##.
     
  4. Apr 17, 2013 #3
    I believe an antinode at each end, making it an open ended tube.

    [itex] L = \frac{n}{2}\lambda [/itex]

    L is therefore 6 m, and velocity is therefore 7200 m s-1

    Is this correct?

    Thanks,
     
  5. Apr 17, 2013 #4

    TSny

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    Homework Helper
    Gold Member

    Yes, that looks right to me. (According to a quick check on the internet, 7200 m/s is somewhat high for aluminum. But that's what your data gives.)
     
  6. Apr 17, 2013 #5
    Thanks
     
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