# Phase difference in standing waves

1. Sep 22, 2009

### petermer

1. The problem statement, all variables and given/known data
We have the standard standing wave equation, $$y=2Acos(2\pi \frac{x}{\lambda})sin(2\pi \frac{t}{T}).$$ 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 $$\pi$$.

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. Sep 22, 2009

### tiny-tim

Hi petermer!

(have a pi: π )

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?

3. Sep 23, 2009

### petermer

Well, I'm just saying that if two x's have an even number of nodes between them, the phase inside the sine is $$2\pi \frac{t}{T}$$ if their y's are positive, or $$2\pi \frac{t}{T} + \pi$$ 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 $$\pi$$ (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, $$y=2Acos(2\pi \frac{x}{\lambda})sin(2\pi \frac{t}{T})$$, 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.