Solution to 1996 Problem 92

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:
Therefore, answer (D) is correct.

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