# Time varying potential

1. Oct 15, 2011

### Piyu

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

A time dependent potential energy is given by
V(r,t) = $\frac{M}{2}$f(t)$\omega^{2}(x^{2}+y^{2}-2z^{2}$)

where f(t) = 1 for 0<t<$\frac{T}{2}$ and f(t)= -1 for $\frac{T}{2}$<t<T.

and f(t+T) = f(t)

Find r(T) and v(T) in terms of r(0) and v(0)
2. Relevant equations

F=-$\nabla$V

3. The attempt at a solution

So far i have tried resolving the forces to each of the cartesian coordinate seperately and finding out x,y,z in terms of t. I solve the 2nd order differential equation for t<T/2 and express x(T/2) in terms of x(0). THen i move on to solve the differential equation for T/2<t<T and substituting x(T/2) as initial conditions to solve for constant and before finding X(T). This works pretty fine for the x and y coordinates but the z coordinate part becomes hell as the equation become absurdly long.

2. Oct 15, 2011

### vela

Staff Emeritus
Show us your calculations for the z-components. It sounds like your approach would be fine.

3. Oct 16, 2011

### Piyu

Ok here goes then,

For 0<t<T/2

2m$\omega^{2}$z=m$\ddot{z}$

general solution for such a 2nd order differential is

z= A$e^{\sqrt{2}\omega t}$+B$e^{-\sqrt{2}\omega t}$

z(0) = A+B
$\frac{\dot{z(0)}}{\sqrt{2}\omega}$=A-B

so A = $\frac{1}{2}$(z(0)+$\frac{\dot{z(0)}}{\sqrt{2}\omega}$)
and B = $\frac{1}{2}$(z(0)-$\frac{\dot{z(0)}}{\sqrt{2}\omega}$)

therefore Z($\frac{T}{2}$)= $\frac{1}{2}$(z(0)+$\frac{\dot{z(0)}}{\sqrt{2}\omega}$)$e^{\sqrt{2}\omega \frac{T}{2}}$ + $\frac{1}{2}$(z(0)-$\frac{\dot{z(0)}}{\sqrt{2}\omega}$)$e^{-\sqrt{2}\omega \frac{T}{2}}$

Moving on to $\frac{T}{2}$<t<T

$\ddot{z}$ = -2$\omega^{2}$z

So general Solution is z = C cos($\sqrt{2}\omega t$) + D sin($\sqrt{2}\omega t$)
$\frac{\dot{z}}{\sqrt{2}\omega}$ = -C sin($\sqrt{2}\omega t$) +Dcos($\sqrt{2}\omega t$)

Heres wheres the maths gets really long and in fact pretty much too long for such an exercise. I substitute z($\frac{T}{2}$) as the intial conditions to to find constants C and D and i end up with a multitudes of trigonometrics and hyperbolics. Am i on the right track?

4. Oct 16, 2011

### vela

Staff Emeritus
Try this trick. Write your solution for t>T/2 as $z(t)=C\cos[\sqrt{2}\omega(t-T/2)]+D\sin[\sqrt{2}\omega(t-T/2)]$.

I'd also try writing the solution for t<T/2 in terms of sinh and cosh, just to make it less unwieldy.

5. Oct 16, 2011

### Piyu

Wow didnt think of that trick thanks!

I've applied it and gotten a much shorter term for z(T) and $\dot{z}$(T)

However, to express r(T) in terms of r(0) ill need to merge all the terms together. but i cannot combine the terms together where r(T) = x(T) i + y(T) j + z(T) k. this is because the z(0) has different coefficients from the y(0) and x(0). in particular the sine and cosine terms have an extra $\sqrt{2}$ inside. i used eulers equation to pull out an e$^{\sqrt{2}}$ from both the trigo and hyperbolic trigo in the z term and end up with more z(0) than x(0) and y(0) i dont know how now to express r(T) in terms of r(0) since r requires equal amounts of x y and z.

EDIT: Working out what i said i realised the mistake in the above paragraph i cant remove the $\sqrt{2}$ from the exponential haha what a stupid maths error. Anyways now im totally stuck as to how to express generally r(T) in terms of r(0). i expand out and the x(0) y(0) and z(0) have different coefficients and hence i cannot pull out the r(0) term.

Last edited: Oct 16, 2011
6. Oct 16, 2011

### vela

Staff Emeritus
I don't think that's necessary or possible. I'm sure the problem simply wants you to write r(T) in terms of the initial conditions.

7. Oct 16, 2011

### Piyu

Hmmm but the problem explicitly states express r(T) and v(T) in terms of r(0) and v(0). because for the next part we must find the value of ωT that shows a trapped particle.

I could equate each of the coordinate at t=T with the one at t=0 but then ill still have to solve them simultaneously for times which all 3 coordinates x(T) y(T) and z(T) are equal to their initial counterparts.