Non-SHM oscillatory motion 
How to solve these kind of questions , where $|F| \propto x^2$? 
How to find time period and velocity type related things to the oscillatory motion?
$$m\dfrac{d^2x}{dt^2}=F=-\dfrac{dU}{dx}=-3kx|x|.$$
But after this $$m\dfrac{d^2x}{dt^2}=-3kx|x|.$$
What is general solution of this ODE? I think it would give the $x$ in terms of $t$ and I would be able to get time period from it.
 A: The dimensional analysis in zkf's answer completely solves the exercise. 
Nevertheless, it is possible to give a closed formula for the period
$$ T~=~ 4 ~\sqrt{\frac{m}{2k}} \int_0^a\! \frac{dx}{\sqrt{a^3-x^3}}
~\stackrel{x=au}{=}~ 4 ~\sqrt{\frac{m}{2ka}} \int_0^1\! \frac{du}{\sqrt{1-u^3}}. $$
Can you see why? Unsurprisingly, this just confirms zkf's answer.
A: zkf gives you enough to answer this question but I would like to make a few extra points:


*

*The absolute value operation in the potential makes this a nonlinear problem, which are generally pretty difficult to deal with. I got impatient waiting for Mathematica to come up with a closed form solution for $x(t)$, so there probably isn't one. This is the typical situation in physics and in this case you have to resort to a numerical calculation.

*Since this is a one dimensional problem you can still solve it using the conservation of energy (you only need one conserved quantity to solve a single particle in 1D problem). If you write the conservation of energy for this system:
$$ \frac{1}{2} m \dot{x}^2 + k |x|^3 = E = \text{constant}, $$
you can rearrange this a bit into
$$ \mathrm{d}t = \frac{\mathrm{d}x}{\sqrt{\text{stuff involving x}}}. $$
Integrating up gives you the answer QMechanic posted just now. :)
A: These kinds of proportionality questions are often best answered with dimensional analysis. You want to know a form a quantity with the units of time in terms of what you have.
You have a quantity $k$ with units $\frac{\text{Energy}}{\text{Distance}^3} = \frac{\text{Mass}}{\text{Distance} \times \text{Time}^2}$. You also have the mass $m$ (units of Mass) and amplitude $a$ (units of Distance). The only way you could possibly combine these quantities to get an answer with the units of time is if the expression is some pure number times $\sqrt{\frac{m}{ak}}$. So this tells you how the period must scale with respect to the dimensionful constants.
