= Solution
Write $u=\pi x^2/l_0^2$ and assume $V_0>0$. The derivative of the potential is $V'(x)=-2\pi V_0x\sin u/l_0^2$. Its equilibrium positions are zero and $x=\pm l_0\sqrt n$ for positive integers $n$. At a nonzero equilibrium,
$$
V''=-\frac{4\pi^2V_0n}{l_0^2}(-1)^n.
$$
Thus the stable equilibria and their small-oscillation <normal mode frequencies> are
$$
\boxed{x_n=\pm l_0\sqrt n,\quad\omega_n=\frac{2\pi}{l_0}\sqrt{\frac{nV_0}{m}},\quad n=1,3,5,\ldots.}
$$
Even positive $n$ give unstable maxima. At zero the quadratic term vanishes but $V(x)=V_0-V_0\pi^2x^4/(2l_0^4)+\cdots$, so the equilibrium is still unstable; a zero linearized frequency does not prove stability. If the unspecified sign of $V_0$ were chosen negative, even-$n$ positions would be stable instead, together with a degenerate quartic minimum at zero whose frequency tends to zero with oscillation amplitude.
\Image[/past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2015-s7-quadratic-phase-potential.png]
{title=Stable minima of the cosine-of-square potential and their small-oscillation frequencies}
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