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Question about a kx force

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m is subject to a force F(x) = kx, with k > 0. What is the most general form of x(t)? If the particle starts out at x0, what is the one special value of the initial velocity for which the particle doesn't eventually get far away from the origin?


    2. Relevant equations
    [itex]m\ddot{x} = kx[/itex]


    3. The attempt at a solution
    So the general solution is [itex]x(t) = Ae^{\omega t} + Be^{-\omega t}[/itex], with [itex]\omega = \sqrt{\frac{k}{m}}[/itex] of course. My issue is with the second part of the question; as far as I understand it, it wants me to figure out for what value of v0 makes it so that the first term of the general solution doesn't dominate (since that term would see x increase to infinity, whereas the second term would see x approaching the x-axis). If that's what the question is asking, I'm not quite sure how I'd go about that! Any hints and help would be much appreciated. :smile:
     
  2. jcsd
  3. Sep 5, 2011 #2

    HallsofIvy

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    Yes, x will go indefinitely away from [itex]x_0[/itex] as long as A is not 0.

    You have the initial values [itex]x(0)= x_0[/itex] and [itex]x'(0)= v_0[/itex]. What value of [itex]v_0[/itex] gives A= 0?
     
  4. Sep 5, 2011 #3
    Okay, so if A is going to be 0, I get
    [itex] v_{0} = -B\omega [/itex].
    Owing to the fact that [itex] x(0) = x_{0} = A + B [/itex], and seeing as how A = 0, this gives me a value of
    [itex] v_{0} = -x_{0}\omega [/itex].
    Does that sound right?
     
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