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Velocity & Frequency Wave on String

  1. Aug 8, 2012 #1
    Please help me, I'm so bewildered it's not even funny.

    Does the frequency only depend on the source? This all seems so paradoxical to me...for example:

    An 100 Hz oscillator produces a wave on a rope with a wavelength of .1 meters. If I increase the tension of the rope by sticking a weight at the end of it, the velocity increases. So does the wavelength, but the frequency stays he same.

    But...if I have a standing wave on a string and I increase the tension of the string, the frequency of the wave definitely increases. This frequency given as f=nv/2L (because the velocity increased the frequency does too).

    Is this all because in the first example the oscillator produces a constant frequency? If the oscillator had not been there, wouldn't the frequency increase?

    Also, if I take the first example and turn the oscillator to 200 Hz, wouldn't the wavelength drop correspondingly but the velocity of the wave would stay constant (velocity only determined by (Tension/linear mass density)^1/2.
  2. jcsd
  3. Aug 8, 2012 #2


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    If you do not have reflections at the other end, you get an oscillation with a frequency equal to the oscillator frequency. With reflections, it can get tricky to define "frequency" at all. You can always write the whole system as superposition of resonance frequencies, but then you do not have "one frequency", but many.

    I do not understand which setup you have in mind here.

  4. Aug 8, 2012 #3
    Thank you mfb.

    I guess I'm talking about something similar to what they have set up here:


    Now, the only way to increase the speed of that wave would be to put a bigger mass on the other end, right? But what happens to the frequency of the wave if I do that. Wouldn't it stay the same because the oscillator is directing the frequency? Normally as the velocity of the wave increases the frequency does too (freq: nv/2L), but does the oscillator take prescience?
    So the only things that should happen would be the freq of that wave increases and correspondingly the wavelength (but the frequency remains at the frequency set by the oscillator).
  5. Aug 8, 2012 #4


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    We just had a long thread about this same basic thing. (Here)
    Ignoring the transitory effect on an oscillating system when an excitation is initially switched on (in which some of the natural modes can be excited briefly by a step function, but then die out), the only vibrations that can exist can be at the excitation frequency. Depending upon how close the exciting frequency is to a natural mode of vibration and the Q factor of the oscillator, the standing wave will be at a high or low level.
    There is no way that such a linear system can produce a change of frequency because you need to have phase continuity with the energy supply and a change in frequency would violate this.
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