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Frequency of wire?

  1. Jan 28, 2007 #1
    1. The problem statement, all variables and given/known data
    Two students perform an experiment to measure the frequency of vibration of a stretched wire. In the experiment a steel wire was stretched over two fixed wooden bridges. A small paper rider was placed on the wire. When student A placed a vibrating tuning fork, F1 on one of the bridges, no effect was noted. However, when student B placed another vibrating tuning fork, F2 on the bridge, the paper rider immediately jumped off. How do you explain what happend?


    2. Relevant equations
    none


    3. The attempt at a solution
    After F1 was on the bridge, the wire should be oscillating at the frequency of F1. After F2 was installed, the wire is forced to oscillate a second time at a different frequency. Would it be correct to say the wire is oscillating at a frequency equal to the difference between F1 and F2? Given that the paper rider fell off, it was the case that the beat frequency equals the resonance frequency of the steel wire.

    The other choice is to say that the wire is oscillating at a frequency equal to the combined frequency of F1+F2. But that wouldn’t work because imagine you had two vibrators both oscillating at the same frequency f. The wire would still be oscillating at f no matter how many vibrators there are although the intensity is proportional to the number of vibrators.
     
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  3. Jan 29, 2007 #2

    andrevdh

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    I get the impression that the tuning forks were placed separately on one of the bridges and not simultaneously in the second case.

    The fact that the rider was jolted off in the second case implies that a standing wave formed in the wire and that the rider was positioned at an antinode.

    In the first case one cannot say wether a standing wave formed or not since the rider could have been at a nodal position if it did form.
     
  4. Jan 30, 2007 #3
    I intepret the question as a vibrator is placed first and paper dosen't fall off. Then another vibrator is placed and the paper fell off (while the first vibrator was still vibrating). So is it the case that the frequency difference between the two vibrators happened to be a reasonant frequency and such that an antinode is present at the position of the paper? In the first case, the frequency could also be a reasonant frequency but the position of the paper was an a node (although low probability). It's more likely that the frequency was a non reasonant frequency and that is why the paper didn't fall off the first time.
     
  5. Jan 30, 2007 #4

    andrevdh

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    If both tuning forks were applied in the second case a standing wave will not form in general. Beats form when the frequencies differ slightly from each other. The beats will also "run along" the length of the wire like a wave with a low frequency and most likely not kick the rider off.
     
  6. Jan 30, 2007 #5
    Since it is not a sound wave but a physics wire, the reasonant frequency might be very low. The speed of the pulse could be tens of metre/second which with a reasonably long wire will allow for low frequencies. So is the beat frequency not ruled out as the resonating frequency?
     
  7. Jan 31, 2007 #6

    andrevdh

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    Well , I am would not call it "resonant frequency" since it is not a standing wave that forms in such a case, but beats could account for the rider being kicked off under such conditions. I am not sure why you are pushing towards a low frequency anyway in the case of beats forming since the propagation velocity in the wire will not determine the frequency of the beats.
     
  8. Feb 1, 2007 #7
    Actually in thinking about this problem again, the problem stated '...an experiment to measure the frequency of vibration of a stretched wire.' So the wire would be shaking. And the frequency refers to the shaking of the wire from side to side? There is a frequency which maximises the shaking of the wire (i.e the amplitude of it travelling from side to side). This frequency as you pointed out could well be a result of the beat frequency.

    Propagation of the wire i.e physically shaking it and create a propagating wave is meaningless in this question?
     
  9. Feb 2, 2007 #8

    andrevdh

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    I do have my doubts that beats will form under these conditions.

    Placing a tuning fork on the bridge will cause it to vibrate up and down in phase with the tuning fork (a tuning fork forces the bottom piece up and down as the prongs vibrate to and fro - this is how a soundbox works). This vibration is then transferred to the wire. The disturbance propagates in the string and is reflected at the endpoints. We have a situation where the vibration is supported in the wire if the frequency of the tuning fork is such that a standing wave forms in it.
     
  10. Feb 3, 2007 #9
    So not vibrate sideways? But transversly. But the fork is vibrating from side to side?

    Are you saying either the first or second fork was vibrating at a frequency such that a standing wave was formed hence oscillating at a harmonic frequency? So the forks were placed one after the other, independently? i.e the first was taken away when the second was placed on the bridge?
     
  11. Feb 5, 2007 #10

    andrevdh

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    When transferring the vibrations of the tuning fork to a surface its bottom is pressed up against the surface. The prongs will still vibrate sideways, but these movements of the prongs pushes and pulls the bottom part up and down as the prongs vibrate. This is then transferred to the surface, generating a disturbance in its surface - a two dimensional wave is formed on the surface of the object.
     
    Last edited: Feb 5, 2007
  12. Feb 5, 2007 #11
    So up and down vibration is more due to gravity.
     
  13. Feb 6, 2007 #12

    andrevdh

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    Not really. The metal structure reacts that way to deformations. When the prongs bends outwards the bottom part is pulled upwards and when they bend inwards it is pushed downwards a bit due to the deformations of the metal.
     
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