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Wavelength - distance between a node and adjacent antinode

  1. Dec 14, 2008 #1
    1. The problem statement, all variables and given/known data
    A tube of air is open at only one end and has a length of 1.7 m. This tube sustains a standing wave at its third harmonic. What is the distance between one node and the adjacent antinode?


    2. Relevant equations
    v = f (lambda)


    3. The attempt at a solution
    If it's in the third harmonic, that means the length of the tube is equal to 1.25 wavelengths. So, 1.7 / 1.25 = 1.36. The distance between an antinode and the adjacent node is .25 lambda, so 1.36 * .25 = .34m. So .34m is the answer.
     
  2. jcsd
  3. Dec 14, 2008 #2

    rl.bhat

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    Re: Wavelength

    If it's in the third harmonic, who many nodes and anti nodes are formed in the tube?
     
  4. Dec 14, 2008 #3
    Last edited: Dec 14, 2008
  5. Dec 14, 2008 #4
    Re: Wavelength

    Can someone tell me what I did wrong?
     
  6. Dec 14, 2008 #5

    rl.bhat

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    Re: Wavelength

    In the tube closed at one end, third harmonic is the first overtone. So L = 3*lambda/4
     
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