Conservation of energy in a non-linear oscillator

Question

I have a homework question about a "non-linear oscillator". I actually have an answer to this question, but the answer I get is stronger than what is needed according to the question. The question says ". . . the potential energy of a non-linear oscillator is given by $U(x) = \frac{1}{2}kx^2 - \frac{1}{3} \alpha x^3$. 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 $v_0 < \frac{k}{\alpha}\sqrt{\frac{k}{m}}$".

If the motion is oscillatory, then the velocity will be zero at exactly two different positions. So the potential energy must be confined to the region where it reaches its local max/min value twice. In other words, the initial position $x_0$ must lie between the critical points of $U(x)$. Setting $dU/dx = 0$, I get that the critical points are $x = 0$ and $x = \frac{k}{\alpha}$, so that the initial position must satisfy $0 < x_0 < \frac{k}{\alpha}$ as required.

For the initial velocity, I reasoned that the initial kinetic energy must be less than the local max of the potential energy. The local max occurs at $x = \frac{k}{\alpha}$ and is equal to $U(\frac{k}{\alpha}) = \frac{1}{6}\frac{k^3}{\alpha^2}$. So the initial kinetic energy cannot exceed this value; i.e. $$\frac{1}{2}mv_0^2 < \frac{1}{6}\frac{k^3}{\alpha^2}.$$ Rearranging, this gives $$v_0 < \frac{1}{\sqrt{3}}\frac{k}{\alpha}\sqrt{\frac{k}{m}}.$$ This condition is stronger than the one in the question: in other words, if my condition holds, then so does the one in the question. So in that sense I technically have answered the question, but I'm unsure if my reasoning is correct. Is what I have done correct, or should I actually be getting the weaker inequality given in the question?

Answer 1

The problem is asking you to show that if $x_0$ and $v_0$ satisfy certain conditions, the motion will be oscillatory. So in order to answer it (technically or otherwise), you need to demonstrate that for all possible $x_0$ and $v_0$ satisfying those conditions, the motion is oscillatory. You haven't done that. So yes, you should be getting the inequality specified in the problem.

That being said, the inequality specified is actually wrong! I can easily choose a value of $x_0$ such that $0 < x_0 < \frac{k}{\alpha}$ and a value of $v_0 < \frac{k}{\alpha}\sqrt{\frac{k}{m}}$ such that the particle will not oscillate. There is an additional condition relating $v_0$ and $x_0$ that must be satisfied for you to have an oscillator. You can consider it an extension of the homework problem to find that condition ;-)

For the initial velocity, I reasoned that the initial kinetic energy must be less than the local max of the potential energy.

Remember that the law of energy conservation always compares total energy at one position/time to total energy at another position/time. Think about this: in order for the motion to be oscillatory (i.e. bounded), what must be true of the total energy at the local maximum of the potential energy curve $x = \frac{k}{a}$? (What's the potential energy at the point? What's the kinetic energy at that point?)

Once you figure that out, you can set it equal to the total (not just kinetic) initial energy.

Answer 2

The proper condition for bound motion is that

$-r{\alpha\over k} < x < {\alpha\over k}$

Where $r$ is the unique positive solution to the cubic equation $2r^3 + 3r^2 - 1 =0 $. This condition is that the potential energy at the beginning of the motion is less than ${k^3\over 6\alpha2}$, which is the maximum bound energy, as you calculated.

The kinetic energy constraint is just that

$${m v_0^2\over 2} + V(x_0) < {k^3\over 6\alpha^2}$$

That is, that the kinetic energy plust the potential energy is less than the potential at the top of the hill. The book didn't formulate the problem correctly, and further, it screwed up this simple calculation. Throw it away.