Answer to Question #279760 in Differential Equations for Khan

Question

A string of length l has it's ends x=0 and x=l fixed. It is released from its rest in position y=[4 lamda x(l-x)]/l². Find an expression of the displacement of the string at any subsequent time.

Answer 1

wave equation:

$y_{tt}=\lambda^2 y_{xx}$

solution:

$y(x,t)=\sum sin(n\pi x/l)(a_ncos(\lambda \pi nt/l)+b_n sin(\lambda \pi nt/l))$

where

$a_n=\frac{2}{l}\int^l_0 f(x) sin(n\pi x/l)dx$

$b_n=\frac{2}{\lambda n \pi}\int^l_0 g(x) sin(n\pi x/l)dx$

we have:

$f(x)=y(x,0)=4\lambda x(l-x)/l^2$

$g(x)=y_t(x,0)=0$

$y(0,t)=y(l,t)=y_t(0,t)=y_t(l,t)=0$

then:

$b_n=0$

$a_n=\frac{2}{l}\int^l_0(4\lambda x(l-x)/l^2) sin(n\pi x/l)dx=-\frac{8\lambda}{l^3}\frac{l^3(\pi nsin(\pi n)+2cos(\pi n)-2)}{\pi^3n^3}=$

$=-\frac{8\lambda(\pi nsin(\pi n)+2cos(\pi n)-2)}{\pi^3n^3}$

$y(x,t)=-\sum\frac{8\lambda(\pi nsin(\pi n)+2cos(\pi n)-2)}{\pi^3n^3}sin(n\pi x/l)cos(\lambda \pi nt/l)$