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Question about standing wave

  1. Jul 3, 2015 #1

    kelvin490

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    By considering the superposition of two waves propagating through a string, one representing the original or incident wave and the other representing the wave reflected at the fixed end, if both ends of the string is fixed then the waves can reflected and travel back and forth. Standing wave can be formed if the length of the string is an integer numbers of half wavelength.

    I just wonder what will we get if the length of the string is NOT an integer numbers of half wavelength and both ends are fixed?
     
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  3. Jul 3, 2015 #2

    Orodruin

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    You will get a superposition of standing waves which do have a half integer as ratio between string length and wave length. It is possible to describe any wave on a finite interval like this (it follows directly from Fourier analysis).
     
  4. Jul 3, 2015 #3

    kelvin490

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    Thanks. Will the nodes and anti-nodes will be moving?

    In addition, I am also wondering whether those simple mathematical description like those in p. 493 of ( https://books.google.com.hk/books?id=7S1yAgAAQBAJ&pg=PA489&lpg=PA489&dq=string+is+now+stretched,+giving+increased+tension,+so+the+free+end+of+the+string+is&source=bl&ots=oSDs7tqt4M&sig=5zzzUMnsC8dG_qdObx9nJxQRPkg&hl=zh-TW&sa=X&ei=MT2VVYbREdP-8QWssoDgAw&ved=0CCQQ6AEwAQ#v=onepage&q&f=true [Broken] ) is only suitable for the length of string being an integer number of wavelength?

    I have this question because it seems like only in this situation the reflected wave can be described as -Acos(kx-wt), otherwise there should be some phase difference other than 180 degree out of phase at the same position.
     
    Last edited by a moderator: May 7, 2017
  5. Jul 3, 2015 #4

    sophiecentaur

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    The usual description of a standing wave (which you are working with) is too limited and doesn't bear close examination. A 'real' standing wave on a string, in the case where there is resonance, is what you get when there are a large (infinite) number of waves. The same wave train will have been that have reflected back and forth for a long (infinite) time. If the path length between the ends is an integer number of half wavelengths, the energy will build up in step to form the antinodes and you will get your familiar standing wave pattern.
    Unless you are prepared to have an infinite build up of energy in your system, there will be some loss. There has to be some loss, in fact, because, at the very least, the source of energy will have a finite 'source resistance' and there is always friction in one form or another. This limits the build up of energy until the rate of energy supplied equals the rate of energy dissipated. (The Q factor of the resonator) If the frequency of excitation is not equal to the resonant frequency, the waves will troll up and down the string without producing any stable interference pattern.
    The reason that a resonance actually allows the energy to be stored in the standing wave (it will carry on for some time after the source is switched of) is a clever one and it's because, at resonance, the Impedance presented to the source by the string is very different from the source impedance so energy takes a while to get into the resonator (build up) but also doesn't all "get out" until the amplitude of the oscillations has built up to its maximum.

    PS. You will get a standing wave on a string, tied at just one end and with a perfectly 'matched' excitation source, at all frequencies. In this case, the peak amplitude will be just twice the amplitude of the applied wave.
    PPS But strings are really not the best examples of easily analysed standing waves. You can get much more measurable results using an electrical transmission line which can be seen to follow the theory much closer. You just don't know what your hand is doing to the end of a string as you wave it up and down. :wink:
     
  6. Jul 3, 2015 #5

    sophiecentaur

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    By definition, they will not be identifiable antinodes if they are the result of waves adding up in different relative phases at different places and times. My previous point about the multitude of waves that actually form the coherent interference pattern, is relevant here.
     
  7. Jul 3, 2015 #6

    sophiecentaur

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    @kelvin irungu With a handle like that, I guess you start every question assuming that we know 'absolute zero' ? :biggrin:
     
  8. Jul 3, 2015 #7

    kelvin490

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    Can we say that in a musical instrument like guitar when we pluck a string we produces waves of different wavelengths on the string. If a wave cannot have integer number of half-wavelength on the string, the wave attenuates very quickly and only the waves with suitable wavelengths can remain, so the corresponding frequencies are called "natural" frequencies.
     
  9. Jul 3, 2015 #8

    Orodruin

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    No, only waves of the appropriate wavelengths were ever there to begin with, a Fourier series expansion will tell you this. Any function you can come up with that satisfies your boundary conditions must have this property.
     
  10. Jul 4, 2015 #9

    sophiecentaur

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    . . . . . . . any shape that you may start off with the plucked string will only have spatial harmonics of the fundamental string length. Those modes are the only ones that plucking can excite. It's a different mater if you excite the string with a nearby loudspeaker, when what Kelvin suggests will apply.
    BTW, in real instruments, because of the end effects on strings and air columns etc. the high order modes will not correspond exactly to harmonics so it is strictly correct to refer to Overtones in this case. The non-harmonic relationship gives the characteristic timbre of the instrument.
     
  11. Jul 4, 2015 #10

    Orodruin

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    No. This is wrong. Any shape and behaviour of the string is describable in terms of the harmonics of the string. This follows directly from Sturm-Liouville's theorem. You may get time dependent coefficients, but the string shape and velocity at any time will be given by a Fourier series, as will the action of any external force.
     
  12. Jul 4, 2015 #11

    kelvin490

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    OK. In the case that the two ends are strictly fixed only integer multiples of first harmonic can exist. Is there any difference in the situation that we can continuously input waves into the string? e.g. Swing the string with some particular frequency so that the total length of the string is NOT a integer number of the half-wavelength of wave. i.e. the frequency is not integer multiples of the first harmonic. Then after a large enough number of such wave is transmitted and reflected back and forth we let both ends fixed. I guess the wave remains in this string is not a standing wave.

    BTW, if we keep on swinging at one end without stop, should this end be treated as fixed or not?

    Thank you.
     
  13. Jul 4, 2015 #12

    sophiecentaur

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    That's true as long as you make it clear that they are spatial harmonics and not harmonic frequencies. The spatial harmonics are the Modes I referred to. What I wrote is not "wrong" when you read it appropriately. The frequencies involved are also affected if there is dispersion on the string - not an ideal string of course but it can be relevant when the resonator is not a simple one dimensional string. This is why I get so picky when people use the term 'harmonic' when 'overtone' is more appropriate.
    I don't understand how a Fourier Analysis over a finite length of string (which assumes that shape is repeated ad infinitum) can contain anything other than harmonics of the string length. Perhaps we are talking at cross purposes, which could be resolved, perhaps by making it clear if we are talking space or time domain.
    Also, excitation of a string with an impressed signal cannot generate any frequencies other than are contained in the signal - if the medium is linear. But, of course, if the string has a finite Q, there may be a response at other frequencies than the natural fundamental or overtones.
     
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