Explain the reflection of a wave at a rigid support.

(N/A) When a wave pulse traveling in the $+x$-direction hits a rigid (fixed) support,it reflects back in the $-x$-direction.
If we assume that no energy is absorbed at the boundary,the shape of the reflected pulse remains the same as the incident pulse,but its phase changes by $180^{\circ}$ or $\pi$ radians.
This occurs because the boundary is fixed,meaning the displacement at that point must always be zero.
Let the displacement of the incident progressive wave at time $t$ be $y_{i}(x, t) = a \sin(kx - \omega t)$.
Let the displacement of the reflected wave be $y_{r}(x, t)$.
According to the principle of superposition,the resultant displacement $y(x, t)$ is given by:
$y(x, t) = y_{i}(x, t) + y_{r}(x, t)$
Since the support is rigid,the displacement at the fixed end is always zero,so $y(x, t) = 0$.
Therefore,$0 = y_{i}(x, t) + y_{r}(x, t)$
This implies $y_{r}(x, t) = -y_{i}(x, t)$
Substituting the expression for the incident wave,we get:
$y_{r}(x, t) = -a \sin(kx - \omega t) = a \sin(kx - \omega t + \pi)$