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Homework Statement
A non-linear oscillator consisting of a mass on a spring has a potential energy of the form \frac{1}{2}kx^2 - \frac{1}{3}\alpha x^3, where k and \alpha are positive constants, and x is displacement. Using conservation of energy, show that the motion is oscillatory if the initial position x_0 satisfies 0 < x_0 < \frac{k}{\alpha} and the initial velocity satisfies v_0 < \frac{k}{\alpha}\sqrt{\frac{k}{m}}.
Homework Equations
E = T + U = \mathrm{constant}
F = -\frac{dU}{dx}
The Attempt at a Solution
By conservation of energy, the quantity E = \frac{1}{2}mv^2 + U(x) 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 E = U(x_1) = U(x_2). Since E 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 x values must lie between the two critical points of U. By setting dU/dx = 0, we get x = 0 and x = \frac{k}{\alpha} as the two critical points, and so we must have 0 < x < \frac{k}{\alpha}, 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 \frac{k}{\alpha}\sqrt{\frac{k}{m}}, 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 \frac{1}{2}mv^2 somehow. Could anyone help with this part?