Harmonic Response
Consider an elastic string stretched between two fixed points a distance $L$ apart. The harmonic response is a standing wave with speed $c=\sqrt{E/\rho}$
$$ \begin{align} y(x,t) &=\sum_{i=1}^{\infty}\sin\left(\frac{i\pi x}{L}\right)\left(A_{i}\sin\left(\frac{i\pi ct}{L}\right)+B_{i}\cos\left(\frac{i\pi ct}{L}\right)\right) \\
\dot{y}(x,t) &=\sum_{i=1}^{\infty}\sin\left(\frac{i\pi x}{L}\right)\frac{i\pi c}{L}\left(A_{i}\cos\left(\frac{i\pi ct}{L}\right)-B_{i}\sin\left(\frac{i\pi ct}{L}\right)\right) \end{align}$$
with unknown coefficients $A_{i}$ and $B_{i}$ depending on the initial conditions. This is a result of the equation of motion
$$T\,\frac{\partial^{2}y(x,t)}{\partial x^{2}}-\rho\,S\,\frac{\partial^{2}y(x,t)}{\partial t^{2}}=0$$
where $T$ is the elastic tension in $\rm{[N]}$, $E$ the elastic modulus in $[{\rm N/m^2}]$, $S$ is the string area in $[{\rm m^2}]$ and $\rho$ the mass density in $[\rm{kg}/{m^{3}}]$.
Initial Conditions
Now consider at $t=0$ the shape and speed of the string to be $y(x,0)=U(x)$ and $\dot{y}(x,o)=V(x)$.
For example, for a slow pluck at point $x_p$ the initial speed is zero $V(x)=0$ and the shape is $$U(x)=\begin{cases}
U_{0}\left(\frac{x}{x_{p}}\right) & 0\geq x\geq x_{p}\\
U_{0}\left(1-\frac{x-x_{p}}{L-x_{p}}\right) & x_{p}>x\geq L
\end{cases}$$
where $U_0$ is the amplitude of the pluck
Fourier Analysis
By expanding out the following frequency decomposition we get the coefficients $A_i$ and $B_i$.
$$\left. \begin{aligned} \int\limits _{0}^{L}\sin\left(\frac{j\pi x}{L}\right)U(x)\,{\rm d}x&=\frac{L}{2}\sum_{i=1}^{\infty}\left[B_{i}\delta_{ij}\right]=\frac{L}{2}B_{j} \\
\int\limits _{0}^{L}\sin\left(\frac{j\pi x}{L}\right)V(x)\,{\rm d}x&=\frac{L}{2}\sum_{i=1}^{\infty}\left[\frac{i\pi c}{L}A_{i}\delta_{ij}\right]=\frac{i\pi c}{2}A_{j} \end{aligned} \right\} \boxed{ \begin{aligned} A_{i}&=\frac{2}{i\pi c}\int\limits _{0}^{L}\sin\left(\frac{i\pi x}{L}\right)V(x)\,{\rm d}x \\ B_{i}&=\frac{2}{L}\int\limits _{0}^{L}\sin\left(\frac{i\pi x}{L}\right)U(x)\,{\rm d}x \end{aligned} } $$
Again for the slow plucked string we get $$\begin{aligned} A_{i}&=0 \\ B_{i}&=U_{0}\frac{2L^{2}\sin\left(\frac{\pi ix_{p}}{L}\right)}{\pi^{2}i^{2}x_{p}\left(L-x_{p}\right)} \end{aligned}$$
Example
A slow plucked string in the middle $x_p = \frac{1}{2} L$ has solution
$$y(x,t) = \sum_{i=1}^{\infty} \frac{8 U_0}{i^2 \pi^2} \sin\left( \frac{i \pi x}{L} \right) \sin\left( \frac{i \pi}{2} \right) \cos\left( \frac{\pi c i t}{L} \right) $$
$$ y(x,t) \approx \frac{8 U_0}{\pi^2} \left[ \cos(\theta) \sin\left( \frac{\pi x}{L} \right) - \frac{1}{9} \cos(3\theta) \sin\left( \frac{3 \pi x}{L} \right) +\frac{1}{25} \cos(5\theta) \sin\left( \frac{5 \pi x}{L} \right) + \ldots \right]$$
where $\theta = \frac{\pi c t}{L}$ is a new independent variable replacing time.
