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Time varying potential

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data

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


    where f(t) = 1 for 0<t<[itex]\frac{T}{2}[/itex] and f(t)= -1 for [itex]\frac{T}{2}[/itex]<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=-[itex]\nabla[/itex]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. jcsd
  3. Oct 15, 2011 #2

    vela

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    Show us your calculations for the z-components. It sounds like your approach would be fine.
     
  4. Oct 16, 2011 #3
    Ok here goes then,

    For 0<t<T/2

    2m[itex]\omega^{2}[/itex]z=m[itex]\ddot{z}[/itex]

    general solution for such a 2nd order differential is

    z= A[itex]e^{\sqrt{2}\omega t}[/itex]+B[itex]e^{-\sqrt{2}\omega t}[/itex]

    z(0) = A+B
    [itex]\frac{\dot{z(0)}}{\sqrt{2}\omega}[/itex]=A-B

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

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

    Moving on to [itex]\frac{T}{2}[/itex]<t<T

    [itex]\ddot{z}[/itex] = -2[itex]\omega^{2}[/itex]z

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

    Heres wheres the maths gets really long and in fact pretty much too long for such an exercise. I substitute z([itex]\frac{T}{2}[/itex]) 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?
     
  5. Oct 16, 2011 #4

    vela

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

    I'd also try writing the solution for t<T/2 in terms of sinh and cosh, just to make it less unwieldy.
     
  6. Oct 16, 2011 #5
    Wow didnt think of that trick thanks!

    I've applied it and gotten a much shorter term for z(T) and [itex]\dot{z}[/itex](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 [itex]\sqrt{2}[/itex] inside. i used eulers equation to pull out an e[itex]^{\sqrt{2}}[/itex] 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 [itex]\sqrt{2}[/itex] 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
  7. Oct 16, 2011 #6

    vela

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    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.
     
  8. Oct 16, 2011 #7
    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.
     
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