First we find the minimum of the potential.
$\begin{align*}\frac{d V}{d x} = -2 a x + 4 b x^3 = 0 \Rightarrow x = 0 \mbox{ or } x = \sqrt{\frac{a}{2b}}\end{align*}$
The second derivative of the potential is
$\begin{align*}\frac{d^2 V}{d x^2} = -2 a + 12 b x^2\end{align*}$
Plugging in $x = 0$ and $x = \sqrt{a}{2b}$ shows that $x = 0$ is a local maximum and $x = \sqrt{\frac{a}{2b}}$ is a local minimum. So, we consider small oscillations about $x = \sqrt{\frac{a}{2b}}$ . Define $x' = x - \sqrt{\frac{a}{2b}}$ . We would like to know the force on the particle as a function of $x'$ . This can be easily found as follows:
$\begin{align*}F = \frac{-d V}{d x} = 2 a x - 4 b x^3 = 2 a \left(x' + \sqrt{\frac{a}{2b}}\right) - 4 b \left(x' + \sqrt{\frac...$
Since we are told to consider only small oscillations about the minimum, we can ignore that $O(x'^2)$ terms. The angular frequency of oscillations in the case where we have a spring with spring constant $k$ is $\sqrt{k/m}$ , so, by analogy, the angular frequency of oscillations here is:
$\begin{align*}\boxed{\sqrt{\frac{4a}{m}}}\end{align*}$
Therefore, answer (D) is correct.

Solution to 1996 Problem 92