Phase difference in standing waves

  • Thread starter petermer
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  • #1
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Homework Statement


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].


Homework Equations





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.
 

Answers and Replies

  • #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:
 
  • #3
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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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