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Tuning fork and resonances

  1. Aug 18, 2010 #1
    1. The problem statement, all variables and given/known data

    A tuning fork is held over the top of a graduated cylinder that is slowly filled with water. Resonances are noted when the water level is 56.5cm and 17.5cm below the rim. Note that there is an end correction , meaning that the effective length of the air column is longer than the observed length by a fixed constant amount. If the speed of sound is 343m/s, then find the frequency of the tuning fork.

    2. Relevant equations

    f = v/lambda

    1st resonance = lambda/4

    2nd resonance = 3lambda/4

    3. The attempt at a solution

    I drew the cylinders. One cylinder is 56.5 cm below the rim and the other is 17.5cm below the rim. The first cylinder is at the second resonance of 3lambda/4 and the second is lamda/4. Now i don't know what to do. Can someone guide me?

    thanks
     
  2. jcsd
  3. Aug 19, 2010 #2

    diazona

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    Consider the resonances one at a time. For example, the first one, which is 56.5 cm below the rim, and is at 3λ/4. What is the relationship between those two distances?

    What about the other resonance?
     
  4. Aug 19, 2010 #3
    for the first one, 56.5cm = lambda/4

    the other resonance, 17.5 = 3lambda/4

    is that right?
     
  5. Aug 19, 2010 #4

    diazona

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    I don't believe you've taken this piece of information into account.
     
  6. Aug 19, 2010 #5
    oh right...

    first resonance: 56.5 + D = lamda/4

    second resonance: 17.5 + D = 3lamda/4
     
  7. Aug 19, 2010 #6
    k. I solved these equations, but I get a negative wavelength and therefore a negative frequency.

    How can that be?:confused:
     
  8. Aug 19, 2010 #7

    diazona

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    How do you know which resonance corresponds to which distance?
     
  9. Aug 19, 2010 #8
    i don't know. I'm just guessing the the first would be the second and the second distance is the first.
     
  10. Aug 19, 2010 #9

    diazona

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    Guesses are wrong sometimes, y'know :wink:
     
  11. Aug 19, 2010 #10
    k. let me think about this one. We are dealing with a tube that is open on one end and closed in the other. Equations i know to use for this situation: lambda = 4L , f = n(v/4L) when n= 1,3,5,etc.

    In the question, the tube is slowly filled with water and the resonances are noted at the distances above..the first resonance happens at 56.5cm.

    am i going in the right direction?
     
  12. Aug 19, 2010 #11
    K. I tried drawing a picture so i can visualize this. The second resonance will happen at 56.5cm and the first resonance will happen at 17.5cm.

    I got the answer.
     
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