Chapter 12: Problem 16
We saw that the displacement of the plucked string is, on the one hand, $$ u(x, t)=\frac{4 L}{\pi^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2 n-1)^{2}} \cos \frac{(2 n-1) \pi a t}{L} \sin \frac{(2 n-1) \pi x}{L}, 0 \leq x \leq L, t \geq 0 $$ and, on the other hand, $$ u(x, \tau)=\left\\{\begin{aligned} x, & 0 \leq x \leq \frac{L}{2}-a \tau, \\ \frac{L}{2}-a \tau, & \frac{L}{2}-a \tau \leq x \leq \frac{L}{2}+a \tau, \\ L-x, & \frac{L}{2}-a \tau \leq x \leq L. \end{aligned}\right. $$ if \(0 \leq \tau \leq L / 2 a\). The first objective of this exercise is to show that (B) can be used to compute \(u(x, t)\) for \(0 \leq x \leq L\) and all \(t>0\) (a) Show that if \(t>0,\) there's a nonnegative integer \(m\) such that either (i) \(t=\frac{m L}{a}+\tau\) or (ii) \(t=\frac{(m+1) L}{a}-\tau\), where \(0 \leq \tau \leq L / 2 a\) (b) Use (A) to show that \(u(x, t)=(-1)^{m} u(x, \tau)\) if (i) holds, while \(u(x, t)=(-1)^{m+1} u(x, \tau)\) if (ii) holds. (c) Perform the following experiment for specific values of \(L\) and \(a\) and various values of \(m\) and \(k:\) Let $$ t_{j}=\frac{L j}{2 k a}, \quad j=0,1, \ldots k $$ thus, \(t_{0}, t_{1}, \ldots, t_{k}\) are equally spaced points in \([0, L / 2 a] .\) For each \(j=0,1,2, \ldots, k,\) graph the \(m\) th partial sum of (A) and \(u\left(x, t_{j}\right)\) computed from (B) on the same axis. Create an animation, as described in the remarks on using technology at the end of the section.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.