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Phase difference in standing waves

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data
    We have the standard standing wave equation, [tex]y=2Acos(2\pi \frac{x}{\lambda})sin(2\pi \frac{t}{T}).[/tex] We must prove that if two x-positions on the wave have an even number of nodes between them, they have a phase difference of 0, whereas in the opposite condition, they have a phase difference of [tex]\pi[/tex].


    2. Relevant equations



    3. The attempt at a solution
    My teacher already gave us a graphical approach of the proof. but I would like to know if there is another one.
     
  2. jcsd
  3. Sep 22, 2009 #2

    tiny-tim

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    Hi petermer! :smile:

    (have a pi: π :wink:)

    I'm not sure what your question means, but anyway a node is a value of x for which y = 0 for all t.

    So nodes are at t/T = n + π/2.

    Can you take it from there? :smile:
     
  4. Sep 23, 2009 #3
    Well, I'm just saying that if two x's have an even number of nodes between them, the phase inside the sine is [tex]2\pi \frac{t}{T}[/tex] if their y's are positive, or [tex]2\pi \frac{t}{T} + \pi[/tex] if their y's are negative. So they have a phase difference of 0. On the other hand, when the two x's have an odd number of nodes between them, it occurs that they have a phase difference of [tex]\pi[/tex] (with the same thought applied before). Therefore, I'm asking for a non-graphical proof of this.

    P.S.: I'm using the equation of the standing wave, where x=0 is an anti-node. That is, [tex]y=2Acos(2\pi \frac{x}{\lambda})sin(2\pi \frac{t}{T})[/tex], x being the position on the x-axis, y the position on the y-axis, λ the wavelength, t the time in sec, and T the period.
     
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