Transverse displacement profile of the string using Fourier series

Question

A string of length $L$ fixed at $x=0$ and $x=L$ and released at time time $t=0$, the transverse displacement at a position $x$ along the string is given by: $y(x,0)=Ax(L-x)$

Assuming that the string is released from rest, write down an expression for $y(x, t)$, the transverse displacement profile of the string at an arbitrary later time.

My solution:

The initial displacement $y(x, 0)$ is can be represented as a sum of harmonic $y(x,0)=∑_{(n=1)}^∞a_n y_n (x) $ where $y_n(x)=\sin⁡(nπx/L)$ and $a_n=\frac{4AL^2 (1-(-1)^n )}{(π^3 n^3 )}$.

This is the part i am struggling with:

$y(x,t)=∑_{(n=1)}^∞(\frac{4AL^2 (1-(-1)^n )}{(π^3 n^3 )})(\sin⁡(nπx/L)) \cos(πvnt/L)$

Answer 1

I'm not sure what the problem is, exactly, so let me know if I've got it wrong. I have below a plot for the case $L=1$.

1. The blue line represents the function $y(x) = x(1-x)$,
2. The orange line represents the function $$y_1(x) = 4 \left(\frac{1- (-1)^1}{\pi^3 1^3}\right) \sin{(\pi x)},$$
3. The green line represent the function $$y_3(x) = \sum_{n=1}^3 4 \left(\frac{1- (-1)^n}{\pi^3 n^3}\right) \sin{(n \pi x)}$$

(I've leave it to you to figure out why I didn't plot $y_2(x)$.)

If you can't see the difference between the green and blue lines, it's because this Fourier Series converges extremely fast to give a very good approximation of the function!

However, if you are looking for a closed form solution to the series, I'm afraid you're going to be disappointed. Mathematica tells me there is one using special functions, but it's too horrid to write down here, and frankly doesn't give you much more information than this series does.