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Stationary waves and Resonance

  1. Mar 19, 2013 #1
    I don't really understand the relationship between the wavelength of a stationary wave and the length of the air column. I also don't know what happens when the wavelength changes.
    I would appreciate it if you could help.
     
  2. jcsd
  3. Mar 19, 2013 #2

    Drakkith

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    Well, what is a stationary wave? What makes it a stationary wave and not another type of wave?
     
  4. Mar 19, 2013 #3
    All I know is that a stationary wave is not really a wave but it's the pattern formed when two progressive waves of the same frequency moving in opposite directions interfere.
    I also know that the name stationary comes from the fact that the nodes and antinodes are always in the same place (as if they're standing still). But I don't really know how the wavelength relates to the nodes and antinodes.
     
  5. Mar 19, 2013 #4

    Drakkith

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    Take a standing wave with 10 nodes. What would happen if the wavelength of each wave were doubled? Would the position of the nodes change? Would the number of nodes change?
     
  6. Mar 19, 2013 #5

    sophiecentaur

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    You can get a standing wave without a resonance. A standing wave will occur where there is a reflection at a boundary (say at one end of a string or at a wall) and if the reflected wave can dissipate itself. The standing wave will be there, irrespective of the wavelength involved, with a series of nodes and antinodes.
    To get a resonance on a string, both ends need to be clamped (and there's an equivalent for pipes and Electromagnetic transmission lines). Then, the resonance will only occur when the reflections from both ends are in step and energy builds up on the string. This will happen when the length of the vibrating string (or whatever) is a whole number of half wavelengths - or in some resonators, it can be quarter wavelengths. Change the excitation frequency and the condition for the waves being in step doesn't apply and there is no resonance.
    Losses and energy escaping will prevent the energy level from building up to an infinite level at resonance.
     
  7. Mar 19, 2013 #6

    davenn

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    Here's a link to a video showing info on standing waves, reflections etc

    hope it enlightens you :)

    Dave
     
  8. Mar 20, 2013 #7
    I think since the wavelengths were doubled the frequency must have halved. I've also seen in a video before that by increasing the frequency the number of nodes and antinodes increases as well. So I guess the number of nodes in our case will decrease. Maybe 5 nodes instead of 10? I guess the positions of the nodes won't change. Only some nodes will disappear and the rest will remain in the same positions.
     
  9. Mar 20, 2013 #8
    So the length of the vibrating string (or the air column in case of a sound wave in a tube)must be a whole number of quarter or half wavelengths for a stationary wave or resonance to occur? Is that why the frequency must be adjusted to give the required wavelength that will produce the stationary wave?
     
  10. Mar 20, 2013 #9
    davenn

    Thanks a lot that's a very clear demonstration of wave behaviour :)
     
  11. Mar 20, 2013 #10

    sophiecentaur

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    There can be 3 node resonances and 5 node resonances, for example and none of the nodes would be in the same places. It's down to common factors in ratios.
     
  12. Mar 20, 2013 #11
    Could you please explain more ?
     
  13. Mar 20, 2013 #12

    sophiecentaur

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    For an ideal string (perfectly well defined ends. It will resonate at any frequency where there is a whole number of half wavelengths - hence you can have :
    no nodes
    1 node (one half wavelength)
    2 nodes (two half wavelengths)
    3 nodes (etc.)
    4 nodes
    5 nodes
    etc.
    as you increase the frequency of excitation f, 2f, 3f, 4f, 5f,... nf ....
    When n is non prime, there will be two (or more) lower frequency resonances, mf and lf, where lm=n that will share some antinode positions but for n1 and n2 where they are both prime, they cannot share node positions. I'm not stating anything very deep here - it just makes sense.

    Real strings and particularly real air columns and other resonators don't resonate at exact harmonics of a fundamental. The pukka word for the resonances of real resonators is 'Overtones'. ( I moan a lot about the misuse of the word 'harmonic' in this respect but no one takes any notice. Sob sob)
     
  14. Mar 20, 2013 #13
    sophiecentaur
    Thank you so much. That really helped a lot. :smile:
    ( and now that i know the difference I won't confuse overtones with harmonics )
     
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