Concordia PHYS 245 tutoring

The force is $F(x)=-U'(x)$. Equilibria satisfy $U'(x_e)=0$, where $x_e$ denotes an equilibrium position. A positive second derivative gives a strict local potential minimum. Linearize the force using the displacement $\xi=x-x_e$.

Since $U'(x)=\alpha x^3-\beta x=x(\alpha x^2-\beta)$, the equilibria are $x_e=0$ and $x_e=\pm\sqrt{\beta/\alpha}=\pm1\,\mathrm m$. Also $U''(x)=3\alpha x^2-\beta$. At zero it is $-4\,\mathrm{J/m^2}$: a local maximum, so the equilibrium is unstable. At either outer equilibrium it is $8\,\mathrm{J/m^2}$: a strict local minimum and a stable equilibrium.

For small $\xi$, retain the linear force term:

Dots mean time derivatives and $\omega$ is angular frequency. The ratio has units $\mathrm{s^{-2}}$, since $\mathrm{J/m^2}=\mathrm{kg/s^2}$. This is a local small-amplitude approximation; the full quartic potential does not give exactly sinusoidal motion at arbitrary amplitude. Stability is not settling: without damping, a nearby oscillation does not decay.

Setting $U=0$ would locate an arbitrary energy level, not equilibria. If $U''=0$, this second-derivative test would need further analysis.

Now $U'(x)=x(\alpha x^2+2\,\mathrm{J/m^2})$. Only $x_e=0$ is real. At that point $U''=2\,\mathrm{J/m^2}>0$, so it is stable and $\omega=\sqrt{2/2}=1\,\mathrm{rad/s}$. The pair of outer equilibria from the original model no longer exists.