# Why standing wave conditions don't include a phase difference of π

So in order to set up a standing wave on string I can set a travelling wave on it first then let this travelling wave get reflected from a fixed end and this reflected wave interferes with the incoming wave to produce a standing wave.

So while superimposing one right travelling wave with another left travelling reflected wave I set the phase difference of the reflected wave ( w.r.t to the incoming wave) as ∆, then when I solve for the boundary conditions ( that y=0 at x=0 for all time ) I get the phase difference ∆ to be zero but the reflected wave should be shifted by π because of reflection from fixed end. Where am I making the mistake?

Edit: I added a image of my calculation, can anyone say what's wrong in it and why (Kx+wt) doesn't give a phase difference of π but ( -kx-wt) does give phase difference of π?

• I don't know what you mean by "I get the phase difference...". Are you doing some calculation? Without showing more of the calculation, I'm afraid I don't understand the details. – BowlOfRed Apr 2 at 17:50
• I'll post a picture, since I can't do latex.. – Lucifer Apr 2 at 17:53

I think your logic is correct, you probably just overlooked something in the algebra.

Right traveling wave: $$y_R=\mathrm{e}^{i(kx-\omega t)}$$

Left traveling wave: $$y_L=\mathrm{e}^{i(-kx-\omega t +\Delta)}$$

The sum $$y=y_R+y_L$$ must be constant in time at position $$x=0$$: \begin{align} y(x=0) &= \mathrm{e}^{-i\omega t} + \mathrm{e}^{-i\omega t +i\Delta}\\ &= \mathrm{e}^{-i\omega t} [1+\mathrm{e}^{i\Delta}]. \end{align} If this is to be independent of time then the term in brackets must be $$0$$ and so $$\Delta=\pi$$. (You can also do all this with sines and cosines instead of complex exponentials, but you'll need to use some trig identities).

• Oh I see where it went wrong instead of (-kx-wt) for the left wave I took (Kx+wt).. – Lucifer Apr 2 at 18:34
• But why doesn't taking ( Kx+wt) work? it's an equivalent expression for left moving wave. – Lucifer Apr 2 at 18:49
• @Lucifer It does work, but you then got confused about how to find the phase, after you put the minus signs in "random" places instead of being consistent. $a \sin x$ and $-a \sin x$ have a phase difference of $\pi$. If you write $a \sin x$ and $a \sin -x$ instead, it's not so obvious what the phase difference is! – alephzero Apr 2 at 19:36
• You're doing it right. The issue is just the definition of what "phase difference" actually means. In this context it means that at the boundary the two waves are out of phase. Fix $x=0$ then one wave looks like $\sin(\omega t)$ and the other looks like $-\sin(\omega t)$, which is equal to $\sin(\omega t + \pi)$. At the boundary one wave is always the inverse of the other. – Alex Apr 3 at 20:42
• (continuing) This case is kind of obvious because we forced the total amplitude to be zero at the boundary. But in other situations you have an incoming wave, a reflected wave, and also a transmitted wave. Then the derivation is more complicated and you end up finding that the phase shift being $0$ or $\pi$ depends on the properties of the materials on either side of the boundary compared to the direction of the incoming wave. – Alex Apr 3 at 20:45