<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\ud}{\underline}
\newcommand{\lin}{\int_0^\infty}
\parindent=0pt
\begin{document}


{\bf Question}

The transverse displacement $y(x,t)$ of a vibrating string of
length $l,$ fixed at its endpoints, satisfies the equation
 $$ {{\partial^2y}\over
{\partial x^2}}-{1 \over {c^2}} {{\partial^2t}\over {\partial
t^2}}=0. $$
\begin{description}
\item[(a)]
If the string is initially at rest with displacement
$y(x,0)=f(x),$ where $$f(x)=\left\{\begin{array}{ll}x,&amp;0\le
x&lt;l/4\\ l/4,&amp;l/4\le x&lt;3l/4\\l-x,&amp;3l/4\le x\le l
\end{array}\right.$$
show that the displacement at later times is

$$y(x,t)=\sum_{n=1}^{\infty}Q_n\sin\left(\frac{n\pi
x}{l}\right)\cos\left(\frac{n\pi ct}{l}\right)$$ with

$$Q_n=\frac{4l}{\pi^2n^2}\sin\left(\frac{n\pi
}{2}\right)\cos\left(\frac{n\pi}{4}\right)$$

\item[(b)]
Find the displacement if instead the string is initially in its
equilibrium position, so that $y(x,0)=0,$ but has non-zero initial
velocity $y_t(x,0)=g(x)$ given by

$$g(x)=\left\{\begin{array}{ll}x,&amp;0\le x\le l/2\\ l-x,&amp;l/1&lt;x\le
l\end{array}\right.$$



\end{description}

\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]
Let $\ds y(x,t) = X(x)T(t),$ so

$$ \frac{X''}{X} = \frac{1}{c^2}\frac{T''}{T} = \lambda$$
$$\Rightarrow X'' - \lambda X = 0$$

Using $X(0) = 0$ and $\ds X(l) = 0 \Rightarrow \lambda =
-\frac{n^2\pi^2}{l^2}$

Thus \begin{eqnarray*} X_n(x) &amp; = &amp; A_n \sin \left(\frac{n\pi
x}{l} \right) \\ T_n(t) &amp; = &amp; B_n \sin \left( \frac{n\pi c t}{l}
\right) +  C_n \cos \left( \frac{n\pi c t}{l}
\right)\end{eqnarray*}

Using $T'(0) = 0 \Rightarrow B_n = 0$

So \begin{eqnarray*} y_n(x,t) &amp; = &amp; Q_n \sin \left(\frac{n\pi
x}{l} \right)\cos \left(\frac{n\pi c t}{l} \right) \\ y(x,t) &amp; = &amp;
\sum Q_n \sin \left(\frac{n\pi x}{l} \right)\cos \left(\frac{n\pi
c t}{l} \right)\end{eqnarray*}

and $$y(x,0) = f(x) - \sum_{n=1}^\infty Q_n \sin \left(\frac{n\pi
x}{l} \right)$$ $$Q_n = \frac{2}{l} \int_0^l f(x) \sin
\left(\frac{n\pi x}{l} \right)\, dx$$
\begin{eqnarray*} \frac{Q_n}{2} &amp; = &amp; \int_0^\frac{l}{4}x  \sin
\left(\frac{n\pi x}{l} \right) \, dx + \frac{l}{4}
\int_\frac{l}{4}^\frac{3l}{4} \sin \left(\frac{n\pi x}{l} \right)
\, dx  \\  &amp; &amp; + \int_\frac{3l}{4}^l (l - x)  \sin
\left(\frac{n\pi x}{l} \right) \, dx \\ &amp; &amp; \frac{}{}\\ &amp; = &amp;
\left[ -\frac{lx}{n\pi} \cos \left( \frac{n \pi x}{l} \right)
\right]_0^\frac{l}{4} + \int_0^\frac{l}{4} \frac{l}{n\pi} \cos
\left( \frac{n \pi x}{l} \right) \, dx  \\ &amp; &amp; + \left[
-\frac{l^2}{4n\pi} \cos \left( \frac{n \pi x}{l} \right)
\right]_\frac{l}{4}^\frac{3l}{4}
 + \left[ \frac{-(l-x)}{n\pi} \cos \left( \frac{n \pi x}{l}
\right) \right]_\frac{3l}{4}^l  \\ &amp; &amp; - \int_\frac{3l}{4}^l
\frac{l}{n \pi} \cos \left( \frac{n \pi x}{l} \right) dx \\ &amp; = &amp;
-\frac{l^2}{4n \pi} \cos \left( \frac{n\pi}{4} \right) + \left[
\frac{l^2}{n^2\pi^2} \sin \left( \frac{n\pi x}{l} \right)
\right]_0^\frac{l}{4}  \\ &amp; &amp; -\frac{l^2}{4n\pi} \cos \left(
\frac{3n \pi}{4} \right)  + \frac{l^2}{4n\pi} \cos \left(
\frac{n\pi}{4} \right) \\ &amp; &amp; + \frac{l^2}{4n\pi} \cos \left(
\frac{3n\pi}{4} \right) - \left[
\frac{l^2}{n^2\pi^2}\sin\left(\frac{n\pi
x}{l}\right)\right]_\frac{3l}{4}^l\\ &amp; = &amp;
\frac{l^2}{n^2\pi^2}\left\{ \sin\left( \frac{n\pi}{4} \right)
 + \sin \left( \frac{3n\pi}{4} \right) \right\} \\  \\
 \frac{lQ_n}{2} &amp; = &amp; \frac{2l^2}{n^2\pi^2} \sin\left( \frac{n\pi}{4} \right)
 \cos \left( \frac{3n\pi}{4} \right)
\end{eqnarray*}

\item[(b)]
As Before: $\ds y(x,t) = X(x)T(t)$

$$ \frac{X''}{X} = \frac{1}{c^2}\frac{T''}{T} = \lambda$$
$$\Rightarrow X'' - \lambda X = 0$$

Using $X(0) = 0$ and $X(l) = 0 \Rightarrow \lambda =
-\frac{n^2\pi^2}{l^2}$

Thus \begin{eqnarray*} X_n(x) &amp; = &amp; A_n \sin \left(\frac{n\pi
x}{l} \right) \\ T_n(t) &amp; = &amp; B_n \sin \left( \frac{n\pi c t}{l}
\right) +  C_n \cos \left( \frac{n\pi c t}{l}
\right)\end{eqnarray*}

Using $\ds y(x,0) = 0 \Rightarrow T(0) = 0 \Rightarrow C_n = 0$

So \begin{eqnarray*} y_n(x,t) &amp; = &amp; Q_n \sin \left(\frac{n\pi
x}{l} \right)\sin \left(\frac{n\pi c t}{l} \right) \\ y(x,t) &amp; = &amp;
\sum_{n = 1}^\infty Q_n \sin \left(\frac{n\pi x}{l} \right)\sin
\left(\frac{n\pi c t}{l} \right)\end{eqnarray*}

\begin{eqnarray*}\frac{\pl y}{\pl t}(x,t) &amp; = &amp; \sum_{n = 1}^\infty
Q_n \sin \left(\frac{n\pi x}{l} \right) - \frac{l}{n\pi c} \cos
\left( \frac{n \pi c t}{l} \right) \\ {\rm so \ \ } g(x) &amp; = &amp;
\sum_{n = 1}^\infty -\frac{l}{n \pi c} Q_n \sin \left( \frac{n \pi
x}{l} \right)\\ - \frac{l Q_n}{n \pi c}&amp; = &amp; \frac{2}{l} \int_0^l
g(x) \sin \left(\frac{n \pi x}{l} \right) dx \\ - \frac{l^2
Q_n}{2n \pi c}&amp; = &amp;  \int_0^l g(x) \sin \left(\frac{n \pi x}{l}
\right) dx \\ &amp; = &amp; \int_0^\frac{l}{2} x \sin \left(\frac{n \pi
x}{l} \right) dx + \int_\frac{l}{2}^l (l - x) \sin \left(\frac{n
\pi x}{l} \right) dx \\ &amp; = &amp; \left[-\frac{lx}{n\pi} \cos
\left(\frac{n\pi x}{l}\right) \right]_0^\frac{l}{2} +
\int_0^\frac{l}{2} \frac{l}{n\pi} \cos \left(\frac{n \pi x}{l}
\right) dx \\ &amp; &amp; + \left[ -\frac{(l - x)}{n \pi} \cos
\left(\frac{n \pi x}{l} \right)\right]_\frac{l}{2}^l -
\int_\frac{l}{2}^l \frac{l}{n\pi} \cos \left(\frac{n \pi x}{l}
\right) dx \\ &amp; = &amp; -\frac{l^2}{2n\pi} \cos \left(\frac{n \pi}{2}
\right) + \left[ \frac{l^2}{n^2\pi^2} \sin \left(\frac{n \pi x}{l}
\right)\right]_0^{\frac{l}{2}} \\ &amp; &amp;  + \frac{l^2}{2n\pi} \cos
\left(\frac{n \pi}{2} \right) - \left[ \frac{l^2}{n^2\pi^2} \sin
\left(\frac{n \pi x}{l} \right)\right]_{\frac{l}{2}}^l \\ &amp; = &amp;
\frac{l^2}{n^2\pi^2} \left( 2 \sin \left( \frac{n\pi}{2} \right)
\right) \\ &amp; =  &amp; \left\{ \begin{array}{ll} 0 &amp; n {\rm \ \ even}
\\  \ds\frac{2l^2}{(2k+1)^2\pi^2}\sin \left((2k+1)\frac{\pi}{2}
\right) &amp; n {\rm \ \ odd} \end{array}\right. \\  - \frac{l^2
Q_n}{2n \pi c}&amp; = &amp; \left\{ \begin{array}{ll} 0 &amp; n = 2k \\
\ds\frac{2(-1)^kl^2}{(2k+1)^2\pi^2}\sin \left((2k+1)\frac{\pi}{2}
\right) &amp; n = 2k + 1
\end{array}\right. \\ \Rightarrow Q_n &amp; = &amp; \left\{ \begin{array}{ll}
0 &amp; n = 2k \\  \ds\frac{4c(-1)^{k+1}}{(2k+1)\pi}\sin
\left((2k+1)\frac{\pi}{2} \right) &amp; n = 2k+1 \end{array}\right. \\
\Rightarrow y(x,t) &amp; = &amp; \sum_{k = 0}^\infty
\frac{4c(-1)^k}{(2k+1) \pi}\sin \left( \frac{(2k+1)\pi x}{l}
\right) \sin \left( \frac{(2k+1) \pi c t}{l} \right)
\end{eqnarray*}
\end{description}







\end{document}
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