Answer to Question #147269 in Differential Equations for Kiram

Question

The taut string of length 50 cm fastened at both ends, is distributed from its position of equilibrium by imparting to each of its points an initial velocity of magnitude kx for 0<x<50. Find the displacement function y(x, t).

Answer 1

$$Solution$$

The displacement of $y(x,t)$ any point $'x'$ of the sting at any time $t$ is given by

$$\frac{\delta^2 y}{\delta t}=a^2\frac{\delta^2 y}{\delta x}----------(1)$$

We have to solve the equation $(1)$ satisfying the following boundary condition

$y(0,t)=0$, for $t \geq 0------------------(2)$

$y(50,t)=0$, for $t \geq 0------------------(3)$

$y(x,0)=0$, for $0 \leq x \leq 50---------------(4)$

$\frac{\delta y}{\delta t}(x,0)=kx$, for $0 \leq x \leq 50--------------(5)$

The suitable solution of equation $(1)$, consistent with the vibration of the string, is,

$y(x,y)=(A\cos \rho x+B\sin \rho x)(C\cos \rho at+D\sin \rho at)---(6)$

Using boundary conditions $(1)$ and $(2)$ we have

$$B\ sin\ 50 \rho(C\ cos \rho at+D\ sin\ \rho at)=0\ \forall t\geq0$$

Either $B=0$ or $sin\ 50 \rho=0$

If we assume that $B=0$ , we get the trivial solution

$$sin\ 50\rho=0\\ 50 \rho=n \pi\\ \rho=\frac{n \pi}{50}$$

Where $n=0,1,2,---,\infty$

Using boundary conditions $(4)$ in $(6)$, we have

$B\ sin\ \rho x,\ C=0$, for $0 \leq x \leq 50$

As $B \neq 0$, we get $C=0$

Using these values of $A,\ \rho\ C$ in , the equation $(6)$, the solution reduces to

$$y(x,t)=k\ sin \frac{n \pi x}{50}sin \frac{n \pi at}{50}---------(7)$$

Where $n=0,1,2,---,\infty$

The most general solution of Equation $(1)$ is

$$y(x,t)= \sum_{x=1}^{\infty} \lambda_s\ sin \frac{n \pi}{50}cos \frac{n \pi at}{50}------(8)$$

Differentiating both sides of $(8)$ partially with respect to $t$, we have

$$\frac{\delta y}{\delta t}(x,t)=\sum_{x=1}^{\infty}(\frac{n \pi a}{50} \lambda_s)sin(\frac{n \pi x}{50})cos(\frac{n \pi at}{50})---(9)$$

Using boundary conditions $(5)$ in $(9)$, we have

$\sum_{x=1}^{\infty}(\frac{n \pi a}{50} \lambda_s)sin(\frac{n \pi x}{50})=kx$ for $0 \leq x \leq 50$

$=\sum_{x=1}^{\infty}b\ sin \frac{n \pi x}{50}$

Which is the Fourier half-range sine series of $kx$ in $(0,50)$

Comparing like terms, we get

$\frac{n\pi a}{50}\lambda_s=b_s=\frac{2}{50}\intop^{50}_{0} f(x) sin \frac{n \pi x}{25}\delta x$, by Euler's Formula

$=\frac{2}{50}\intop^{50}_{0}kx\ sin \frac{n \pi x}{25}\delta x$

$\therefore \begin{cases} 0 & if\ n\ is\ even \\ \frac{200k}{n^2\pi^2a} & if\ n\ is\ odd \end{cases}$

for Using this value as in , the required solution is

$$y(x,t)=\frac{200k}{n^2\pi^2a}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^4}sin\frac{(2n-1)\pi x}{50}cos\frac{(2n-1)\pi at}{50}$$