# Application of One-Dimensional Wave Equation

• May 7th 2008, 11:26 PM
Chris L T521
Application of One-Dimensional Wave Equation
I'm working on my final project in this independent study I'm in. I worked out this problem, but I want to verify that I'm right.

A string of length L with both endpoints fixed, such as the string of a piano, is struck at its midportion, the impact thus imparting to the string an initial velocity $\psi(x)$ which is defined by:

$0, 0<= x<\frac{L-a}{2}$
$1, \frac{L-a}{2}
$0, \frac{L+a}{2}

Find the motion of the string.

I set up the following boundary value problem:

$y_{tt}=a^2y_{xx}$

$y(0,t)=0$

$y(L,t)=0$

$y(x,0)=0$

$y_{t}(x,0)=\psi(x)$

and I got the following Fourier Series Solution:

$y(x,t)=\frac{4L}{\pi^2a}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}sin{\left[\frac{(2n-1)\pi a}{2L}\right]}sin{\left[\frac{n\pi a}{L}t\right]}sin{\left[\frac{n\pi}{L}x\right]}$

Does this look right? I would appreciate any help!!!
• May 8th 2008, 07:33 AM
ThePerfectHacker
Quote:

Originally Posted by Chris L T521
$0, 0<= x<\frac{L-a}{2}$
$1, \frac{L-a}{2}
$0, \frac{L+a}{2}

The solution is given by,
$y(x,t) = \sum_{n=1}^{\infty} \left( A_n \cos \frac{\pi n at}{L} + B_n \sin \frac{\pi n at}{L} \right) \sin \frac{\pi n x}{L}$.

Where, $A_n = \frac{2}{L} \int_0^L 0\cdot \frac{\pi n x}{L} dx = 0$ because the string is streched.

And, $B_n = \frac{2}{\pi n a}\int_0^L \phi (x) \frac{\pi n x}{L} dx$.

All you need to do now is compute $B_n$ coefficient and you got your answer.
• May 8th 2008, 08:26 AM
Chris L T521
I understand how the Coefficients are found, but would you briefly explain how you got that series?
• May 8th 2008, 07:18 PM
ThePerfectHacker
Quote:

Originally Posted by Chris L T521
I understand how the Coefficients are found, but would you briefly explain how you got that series?

Are you familar with how to work with Boundary Value Problems? I would post out all the steps but it will take far to much typing. It would be better if you write your work and I tell you if I agree with it or disagree. Try reading this, see if it helps you.
• May 8th 2008, 07:29 PM
Chris L T521
I was thinking it over today, and I realized where that series came from (Because I had derived it a couple weeks back...but I was tired, and didn't catch that.) :)

There is one question I still have: Are you sure that your $B_n$ is correct? I got this for $B_n$

$B_n=\frac{2}{n\pi a}\int_{0}^{L}\psi(x)sin{\left(\frac{n\pi}{L}x\rig ht)}\,dx$.

When I simplify the integral, this is what I'm left with:

$B_n=\frac{2}{n\pi a}\int_{\frac{L-a}{2}}^{\frac{L+a}{2}}\psi(x)sin{\left(\frac{n\pi} {L}x\right)}\,dx$.

Does this seem right?
• May 8th 2008, 07:36 PM
ThePerfectHacker
Quote:

Originally Posted by Chris L T521
Are you sure that your $B_n$ is correct?

It is possible I got the $B_n$ wrong. Because maybed I mis-copied this solution directly out of my book because I am not going to rederive it each time. The link I gave you goes into more detail. There is a full solution there.

This is Mine 96:):)th Post!!!