Here's how I got the expressions that I plotted in post #5.
The solution ##y(x, t)## of the wave equation in terms of initial conditions at ##t = 0## is given by
d'Alembert's formula : $$y(x, t) = \frac 1 2 \left(f(x+ct) + f(x-ct)\right) + \frac 1 {2c} \int_{x-ct}^{x+ct}g(\bar x) d\bar x$$ Here, ##c## is the wave speed, ##f(x)## is the initial displacement of the string and ##g(x)## is the initial velocity profile of the string.
In our case, the string is semi-infinite along the positive x-axis and fixed at ##x = 0##. For ##t < 0##, we assume that the string is vibrating in the m
th harmonic standing wave in the region between ##x = 0## and ##x = 1## with amplitude 1. We'll also take ##c = 1##. So, $$y_{sw}(x, t) = \sin(m\pi x) \sin(m\pi t) \,\,\, \text{ for } 0 \le x \le 1 \text {, 0 otherwise.} $$
At ##t = 0## when the node at ##x = 1## is released,
##\,\,\,f(x) = 0## for ##x \ge 0## and
##\,\,\,g(x) = m\pi \sin(m\pi x)## for ##0 \le x \le 1##, ##g(x) = 0## for ##x > 1##.
It can be shown that the node at ##x = 0## is accounted for in d'Alembert's formula by extending ##f(x)## and ##g(x)## to all negative values of ##x## as odd functions about ##x = 0##. Thus, in d’Alembert’s formula we use ##f(x) = 0## for all ##x## and
$$g(x) =
\left\{
\begin{array}{lr}
m\pi \sin(m\pi x), & \text{if }|x|\le 1\\
\,\,\,\,\,\,0, & \text{if }|x| >1
\end{array}
\right\}
$$
So, with ##c=1##, the solution is $$y(x, t) = \frac 1 2 \int_{x-t}^{x+t}g(\bar x) d\bar x$$
Let ##G(x)## be an antiderivative of ##g(x)##. Then we can write the solution as $$y(x, t) = \frac 1 2 \left[G(x+t) – G(x-t)\right].$$
An antiderivative of ##g(x)## is $$G(x) = \int_{-\infty}^x g(\bar x)d \bar x.$$ Evaluating this for the extended ##g(x)## yields
$$
G(x) =
\left\{
\begin{array}{lr}
(-1)^m - \cos\left(m\pi x\right), & \text{if }|x| \le 1\\
\,\,\,\,\,\,0, & \text{if }|x| >1
\end{array}
\right\}
$$
Using the
Heaviside Pi function ##\Pi(x)##, ##G(x)## can be written as
$$
G(x) =
\left\{
\begin{array}{lr}
2\sin^2\large\left(\frac{m \pi x}{2}\right) \Pi(\frac x 2), & \text{if }m \text{ is even}\\
-2\cos^2\large\left(\frac{m \pi x}{2}\right) \Pi(\frac x 2), & \text{if }m \text{ is odd}
\end{array}
\right\}
$$
Using this in ##y(x, t) = \frac 1 2 \left[G(x+t) – G(x-t)\right]## gives the solution for any positive ##x## and positive ##t##.
Aside: The above solution for even ##m## will give “negative bumps” for the graphs rather than positive bumps. This is due to the fact that our expression for the standing wave ##y_{sw} (x, t) = \sin(m\pi x) \sin(m\pi t)## has the property that points of the string just to the left of ##x = 1## are moving downward at ##t = 0## for even ##m##. So, when the right node is released at ##t = 0##, the string is pulled downward as the wave propagates toward the right beyond ##x = 1##. This produces negative bumps for even harmonics. If you want the bumps to always be positive, you can change the overall sign of ##y_{sw}(x, t)## for the even harmonics. This is what I did for the m = 2 plot in post #5.
EDIT: The expression for ##G(x)## can be tidied up as a single expression for both even and odd values of ##m## as $$G(x) = -2\sin^2\left[ \frac{m \pi}{2}(x+1)\right] \Pi\left(\frac x 2\right)$$
.
