= Solution
Let $x_0$ be the equilibrium, so $V'(x_0)=0$. For $\xi=x-x_0$, the <Taylor expansion> of the potential gives
$$
V(x)=V(x_0)+\frac12V''(x_0)\xi^2+O(\xi^3),\qquad
m\ddot\xi=-V''(x_0)\xi+O(\xi^2).
$$
An ordinary small-amplitude <simple harmonic motion> therefore requires a nondegenerate local minimum, $V''(x_0)>0$, with the quadratic term dominating the higher terms over the excursion. Then
$$
\boxed{\omega_0^2=\frac{V''(x_0)}m,\qquad f_0=\frac1{2\pi}\sqrt{\frac{V''(x_0)}m}.}
$$
A local minimum with $V''(x_0)=0$, such as a quartic minimum, need not give small-amplitude harmonic motion with a nonzero limiting <angular frequency>. If $V''(x_0)<0$, the equilibrium is unstable rather than oscillatory.
Back to article page
