# Conservation of energy in a non-linear oscillator

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?

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The book screwed up. I posted a complete answer, but it was deleted, and replaced with an inferior answer by Zaslavski. I would recommend throwing the book away. –  Ron Maimon Sep 4 '11 at 2:07
Well, David's answer is perfect for the purposes of solving this question ("solving it" in the sense of getting the answer that they want). He also points out that there is a mistake in the problem as stated. So I don't know what your objection to his answer is. (The problem isn't from a book by the way). –  saurs Sep 4 '11 at 4:12
I deleted Ron 's answer in accordance with the policy that giving away a complete answer to a homework question is not allowed. I did ask the other moderators to review this though. (By the way, I wouldn't say "replaced" since we posted our answers at almost the same time) –  David Z Sep 4 '11 at 5:36

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.

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Ah, right. So actually I should be using $\frac{1}{2}mv_0^2 + \frac{1}{2}kx_0^2 - \frac{1}{3}\alpha x_0^3 < \frac{k^3}{6 \alpha^2}$, and I can derive the inequality for $v_0$ from there. Thanks! (As for the condition relating $x_0$ and $v_0$, I'll have to think about that for a while. . .) –  saurs Sep 4 '11 at 1:17
(Uh, that less than sign should be an equals sign I guess) –  saurs Sep 4 '11 at 1:19
If you mean the one in $v_0 < \frac{k}{\alpha}\sqrt{\frac{k}{m}}$, then I did mean $<$, not $=$. I was just restating the condition from the problem as given. –  David Z Sep 4 '11 at 1:35
I was referring to the one in my comment, sorry. –  saurs Sep 4 '11 at 4:08

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.

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I wonder if you might be giving away too much for the solution to a homework question? –  David Z Sep 4 '11 at 0:48
I just want this question closed and removed from the list. I don't know any other way to do this. Further, the OP had the correct solution, it was the book that was faulty. –  Ron Maimon Sep 4 '11 at 0:52
This is a good question (at least as far as I'm concerned, though you could ask dmckee and mbq) and it is not getting deleted. When someone posts an answer to a homework question that gives away too much, standard practice is to delete the answer (it can be undeleted later if needed). And yes, the book (assuming it was a book) screwed up but the problem still has some value if you ignore the stated condition on $v_0$. –  David Z Sep 4 '11 at 0:58