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Non-linear oscillator

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data
    A non-linear oscillator consisting of a mass on a spring has a potential energy of the form [itex]\frac{1}{2}kx^2 - \frac{1}{3}\alpha x^3[/itex], where [itex]k[/itex] and [itex]\alpha[/itex] are positive constants, and [itex]x[/itex] is displacement. Using conservation of energy, show that the motion is oscillatory if the initial position [itex]x_0[/itex] satisfies [itex]0 < x_0 < \frac{k}{\alpha}[/itex] and the initial velocity satisfies [itex]v_0 < \frac{k}{\alpha}\sqrt{\frac{k}{m}}[/itex].

    2. Relevant equations
    [itex]E = T + U = \mathrm{constant}[/itex]

    [itex]F = -\frac{dU}{dx}[/itex]

    3. The attempt at a solution
    By conservation of energy, the quantity [itex]E = \frac{1}{2}mv^2 + U(x)[/itex] must be constant. So if the motion is oscillatory then the velocity will be zero at two (and only two) different positions, i.e. we have [itex]E = U(x_1) = U(x_2)[/itex]. Since [itex]E[/itex] is the maximum value of the potential energy, this is equivalent to saying that the potential energy must reach the value of its local max/min at two (and only two) positions. In other words, the [itex]x[/itex] values must lie between the two critical points of [itex]U[/itex]. By setting [itex]dU/dx = 0[/itex], we get [itex]x = 0[/itex] and [itex]x = \frac{k}{\alpha}[/itex] as the two critical points, and so we must have [itex]0 < x < \frac{k}{\alpha}[/itex], as required.

    I think the above is OK, but feel free to correct me if you see a problem. What I don't understand is how the initial velocity comes into it. The question is saying that if the initial velocity is greater than [itex]\frac{k}{\alpha}\sqrt{\frac{k}{m}}[/itex], then the motion won't be oscillatory, but I don't know how to derive this. Presumably this has to do with the conservation of energy as well -- I guess I have to use the kinetic energy term [itex]\frac{1}{2}mv^2[/itex] somehow. Could anyone help with this part?
     
  2. jcsd
  3. Sep 3, 2011 #2
    Draw the potential, and with that draw the force. For small initial velocity the particle can be trapped by the potential.
     
  4. Sep 3, 2011 #3

    dynamicsolo

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    Homework Helper

    Consider also that a system is only bound for E < 0 . When the mass is at the extreme of displacement [itex]x = \frac{k}{\alpha}[/itex], what would the maximum permissible velocity be?

    (I think there is some margin in the given conditions: I am getting some dimensionless multipliers on the order of 1 for the limits on displacement and velocity.)
     
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