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Resonance in tube filled with air

  1. May 9, 2008 #1
    1. The problem statement, all variables and given/known data

    A long tube contains air at a pressure of 1.00 atm and a temperature of 77.0 degrees celsius. The tube is open at one end and closed at the other by a movable piston. A tuning fork near the open end is vibrating with a frequency of 500 Hz. Resonance is produced when the piston is at distances 18.0, 55.5, and 93.0 cm from the open end.

    These data show that a displacement antinode is slightly outside of the open end of the tube. How far outside is it?

    2. Relevant equations

    f1 = v/4L
    f1 = v/2L

    3. The attempt at a solution

    I have tried this a couple times, and got .008m. I use mastering physics, and don't have many attempts left, because apparently my answer is wrong. I can't figure out what it is I am doing incorrectly.
  2. jcsd
  3. May 9, 2008 #2


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    Hi bharp24! :smile:

    Show us what you did, so that we can see where you've gone wrong and help you. :smile:
  4. May 9, 2008 #3
    my TA helped me in class, and even he said he is not sure if this was the right way to go about it, but this is what we did:

    antinode occurs at wavelength/2
    we assigned a variable to the different distances the piston is
    from knowing the frequency we found the velocity by f/wavelength (this is 375 m/s)
    and then, he told me to set up another equation--> v = Sqrt((gamma*R*T)/(M)) and solve for gamma

    all of this has gotten me nowhere and the problem is due soon. I have no idea how to produce the correct answer....I don't think my TA has a clue either. If you could help, I would greatly appreciate it!!
  5. May 10, 2008 #4


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    Hi bharp24,

    I don't think you can find the velocity from the wavelength; you will not know the wavelength until you know the speed of sound. (You can find the speed of sound because you know the air temperature.)

    For example, you have in your equation that the wavelength equals 4L (for a specific frequency); in another you have 2L. That is true, but what is L? L is the distance from one node to the next antinode. In many problems (most?) you would say that L is the length of the tube, but here they are being more realistic: the antinode is slightly outside the tube, so L = d + x, where d is the length of the tube and x is the distance you are trying to find.
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