Time Varying Potential Homework Solution

[Piyu]

Homework Statement

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)

Homework Equations

F=-\nablaV

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.

 

[vela]

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

 

[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= Ae^{\sqrt{2}\omega t}+Be^{-\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?

 

[vela]

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.

 

[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 don't 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 realized the mistake in the above paragraph i can't remove the \sqrt{2} from the exponential haha what a stupid maths error. Anyways now I am 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.

 

[vela]

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.

 

[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.