Help understanding how to find the frequency of small oscillations I'm having trouble understanding how to calculate the frequency of small oscillations. 
I know that if a particle gently oscillates about the minimum of a potential, then we can approximate its motion as simple harmonic motion (SHM).
Let's say that I have a potential $U(x) = \beta (x^2-\alpha^2)^2$. The minima are at $x=\pm \alpha$. 
Because we can approximate motion about these minima as SHM, the potential about these points can be approximated to be $U(x)=\frac{1}{2}m\omega^2x^2$, where $\omega = $ frequency $=\sqrt{\frac{k}{m}}$.
Therefore, 
$$\beta (x^2-\alpha^2)^2 = \frac{1}{2}m\omega^2x^2$$
Solving for $\omega$:
$$\to \omega = \frac{ (x^2 - \alpha^2)}{x}\sqrt{\frac{2 \beta }{m}}$$
Have I understood this concept correctly?
 A: No. The potential $U(x)=\frac12 m\omega^2 x^2$ has a minimum at $x=0$ whereas your potential has minima at $\pm\alpha$. By equating them you don't get anything useful. (By the way, in your solution for $\omega$, what is $x$?) What you want to do to find the frequency of vibration about say, $x=\alpha$, is to translate your model potential to $x=\alpha$: $\frac12 m\omega^2(x-\alpha)^2$. Now find $\omega$ such that $\beta(x^2-\alpha^2)^2\approx \frac12 m\omega^2(x-\alpha)^2$ for $x\approx\alpha$. Hint: for $x\approx\alpha$,
$$\beta(x^2-\alpha^2)^2 = \beta(x+\alpha)^2(x-\alpha)^2\approx\beta(2\alpha)^2(x-\alpha)^2.$$
What you're really doing is expanding your potential into a second-order Taylor polynomial based at the minimum and looking at the coefficient on the square term. This is the generic approach that will work for all types of potentials.
A: I like to think of SHM as a "ball rolling around in a bowl representing a potential minimum", and the curvature of the bowl tells me something about the frequency.
For the simple harmonic oscillator in a parabolic well (and for small displacements, a minimum will almost always look like a parabolic well unless you deliberately do some mathematical skullduggery to make it not so) of the form $E=\frac12 k x^2$, an object with mass $m$ will have a frequency $\omega = \sqrt{\frac{\alpha}{m}}$
Looking at your potential, there are two minima, at $x=±\alpha$.
The curvature of the potential well will be given by the curvature, which is the second spatial derivative. The first derivative (from the chain rule) is:
$$\frac{dE}{dx} = 2\beta(x^2-\alpha^2)(2x)$$
and the second derivative becomes
$$\begin{align}\frac{dE^2}{d^2x} &= 2\beta(2x)(2x)+2\beta(x^2-\alpha^2)\cdot 2\\
&= 4\beta(2 x^2 + x^2 - \alpha^2)\\
&= 4\beta(3x^2-\alpha^2)\end{align}$$
Evaluating this at $x=±\alpha$ we conveniently get the same value at both points, $k = 8\beta\alpha^2$. You of course expected this because of the symmetry of the equation (the potential is symmetrical about x=0, so any curvature at a point +x must be the same as the curvature at point -x.)
The general procedure, then, is to take the second derivative and evaluate it at the minimum (or minima). This second derivative is $k$ in the general expression for SHM: $\omega = \sqrt{\frac{k}{m}}$
A: In fact, expanding potential to second order is a implicit method although it can also works well. An alternative way is to write down the motion equation, which can investigate the problem in details. Firstly, write down the field$$\vec{E}=-\nabla U(x)$$
Secondly, write the force(I assume the problem involves electron potential and electron)
$$\vec{F}=q\vec{E}$$
Thirdly, write down the motion eqution
$$m\ddot{x}=F$$
or
$$m\ddot{x}+q\frac{dU(x)}{dx}=m\ddot{x}+4q\beta(x^3-\alpha^2x)=0$$
Clearly, there is a perturbation $(4q\beta x^3)$ over the SHM, so the $exact$ frequency is of amplitude dependence. For small amplitude or say SHM, the perturbation term can be canceled. Finally,$$m\ddot{x}-4q\beta\alpha^2x=0$$
Clearly, a oscillation solution exists only in the case of $q<0$ or $\beta<0$, otherwise the amplitude will increase or decrease exponently. So for SHM, the frequency is $\omega=\sqrt{\frac{4|q\beta|\alpha^2}{m}}$.
Another way to this problem is the Hamiltonian treament.
