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We know that when a wave on a string is reflected from a hard boundary, the phase change is $\pi$, and from a soft boundary, the change is 0. My question is: this two conditions (hard and soft boundary) seem to be maxima and minima, thus, if the boundary is not absolute hard/soft, can the phase change be in somewhere 0 and $\pi$, or only 0 and $\pi$ are the possible values. For me, it would be a bit strange, because it would suggest something like a two-state universe...

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  • $\begingroup$ Yes, other phases are quite possible and common. What do you mean by "two-state universe"? What is strange about a wave reflecting with a phase that isn't 0 or $\pi$? $\endgroup$ – DanielSank Jan 20 '15 at 8:35
  • $\begingroup$ I was wrong, I wanted to write, if only 0 and $\pi$ are allowed, it would be a two-state thing, which would be strange for me $\endgroup$ – Amateur Physics Jan 20 '15 at 8:44
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Suppose the incident wave has amplitude $A$ and phase $0$. We think the reflected rays change the phase in mathematical context, but it actually changes the sign of the wave in the case of hard boundary, whereas in short boundary, it remains the same. It depends on our way of thinking.

In your case of neither soft or hard boundaries, the value of $A$ is damped, and the sign is reversed when it is relatively more hard.

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  • $\begingroup$ Oh, I see, so basically, the direction changes, which in this case can be only negative or positive, thanks :) $\endgroup$ – Amateur Physics Jan 20 '15 at 9:01
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It helps to realize that for the string, "hard boundary" means the displacement dy/dt=0, whereas for a soft boundary it means the force F=0.

In the non-extreme cases, neither dy/dt nor F is zero. Their ratio will determine the phase.

It's fairly clear that you can't have both F = 0 and dy/dt = 0 at one end, and there aren't any other relevant variables which can be zero all the time. This means there are exactly two extremes to consider.

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