why does the phase change when wave reflects from rigid boundary [duplicate]

This question already has an answer here:

when the wave is reflected from open end there is no phase change but when it reflects from the rigid surface its phase changes.

in open surface its phase change is 0 and in rigid surface its phase change is pi. How is it possible. Is is possible to reflect from open end.

marked as duplicate by user10851, Dilaton, Waffle's Crazy Peanut, Dan, Qmechanic♦Jul 23 '13 at 3:17

In one dimension, reflection of an incident wave from a fixed end of the string represents the simplest type of the wave-boundary interaction and is governed by the following boundary condition $w(0,t)=0$ where $w(x,t)$ is the transversal displacement.
Let us use the method of the images to solve the problem. Consider a displacement pulse $f(x-ct)$ propagating in the positive $x$−direction towards the fixation point as it is depicted in the figure below. Imagine now that the boundary at $x = 0$ is removed and the string is extended towards positive infinity. Let us consider this infinite string and in addition to the incident pulse $f(x-ct)$, introduce an “image” pulse so that the superposition of these two pulses would satisfy the above boundary condition. It is easy to understand that to reach this goal, the “image” pulse should be introduced symmetrically (with respect to $x=0$) to the incident pulse, be opposite in sign to the incident pulse and propagate in the negative $x$−direction with the wave speed $c$ as shown in the figure. 