# Homework Help: Wave Equation

1. May 25, 2010

### bon

1. The problem statement, all variables and given/known data

I'm given that the motion of an infinite string is described by the wave equation:

(let D be partial d)

D^2 y /Dx^2 - p/T D^2/Dt^2 = 0

I'm asked for what value of c is Ae^[-(x-ct)^2] a solution (where A is constant)

Then im asked to show that the potential and KE of the wave packet are equal..

2. Relevant equations

3. The attempt at a solution

So im guessing the value of c is root(T/p)?since the solution is a function of (x-ct) so this corresponds to D'Alembert..But then PE and KE dont seem equal...

KE = integral from -infinity to + infinity of 1/2 p A^2 e^[4c^2(x-ct)] while PE = integral from - inf to + inf of 1/2 p A^2 c^2 e^[-(4x-ct)]..and these dont seem equal..

any help?

thanks!

2. May 29, 2010

### bon

Any ideas on this?
To me it doesn't even seem to obey the wave equation - though it is of the form (x-ct) which corresponds to the d'alembert solution.

yxx = Ae^-2 whereas ytt = Ae^(-2c^2)

Any ideas? Thanks

3. May 29, 2010

### eys_physics

Hi
As you say one solution is $$c=\sqrt{T/p}$$ but the kinetic energy is given by
$$E_k=p\frac{1}{2}\int_{-\infty}^{\infty}(\frac{dy}{dt})^2 dx$$ and the potential energy by $$E_p=T\frac{1}{2}\int_{-\infty}^{\infty}(\frac{dy}{dx})^2 dx$$.

I hope this helps.

4. May 29, 2010

### bon

But if you look at my post above..it doesn't seem that c = root T/p will satisfy the equation..

Also i cant get hte KE and PE to be equal...

5. May 30, 2010

### eys_physics

Okay, you have done something wrong in the calculations of $$\frac{d^2 y}{dx^2}$$ and $$\frac{d^2 y}{dt^2}$$. I have
$$\frac{dy}{dx}=-2A(x-ct)e^{-(x-ct)^2}$$,
$$\frac{d^2}{dx^2}=-2A(1-2(x-ct)^2)e^{-(x-ct)^2}$$,
$$\frac{dy}{dt}=2Ac(x-ct)e^{-(x-ct)^2}$$
and
$$\frac{d^2}{dt^2}=-2Ac^2(1-2(x-ct)^2)e^{-(x-ct)^2}$$.
Thus for $$c=\pm \sqrt{T/p}$$ is
$$\frac{d^2}{dx^2}-\frac{p}{T}\frac{d^2 y}{dt^2}=0$$
satisfied. With the expressions above it's also easily seen that $$E_k=E_p$$.

6. May 30, 2010

### bon

Oh i see where I went wrong.. Thank you so much for your help! :)