Solving the Wave Equation for c and Examining Potential/Kinetic Energy Equality

[bon]

Homework Statement

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 I am asked to show that the potential and KE of the wave packet are equal..

Homework Equations

The Attempt at a Solution

So I am 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 don't 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 don't seem equal..

any help?

thanks!

 

[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

 

[eys_physics]

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

I hope this helps.

 

[bon]

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

Also i can't get hte KE and PE to be equal...

 

[eys_physics]

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

 

[bon]

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