The Force Law for Simple Harmonic Motion — Question 7

Question 7

The slope of a force curve determines local stability

A one-dimensional force is $F(x)=\alpha x-\beta x^3$, with $\alpha,\beta>0$. A particle of mass $m$ is free to move along $x$.

1. Find all equilibria and classify their linear stability from the force slope.

2. Find the small-oscillation angular frequency about each stable equilibrium. Does this imply exact SHM for arbitrary amplitude?

Question 7 - Solution

The slope of a force curve determines local stability

(a) Derivation and calculation

$$F=x(\alpha-\beta x^2)=0\Rightarrow x_e=0,\quad x_e=\pm\sqrt{\alpha/\beta}.$$$$F^{\prime}(x)=\alpha-3\beta x^2.$$$$F^{\prime}(0)=+\alpha>0\quad(\text{unstable}),\qquad F^{\prime}(x_e)=-2\alpha<0\quad(\text{stable outer points}).$$

(b) Result and interpretation

Expand about either stable point, $x=x_e+q$:$$F(x_e+q)=-2\alpha q-3\beta x_e q^2-\beta q^3.$$$$m\ddot q\simeq-2\alpha q\Rightarrow\boxed{\omega=\sqrt{2\alpha/m}}.$$The higher-order terms remain; the linear approximation requires excursions small compared with $|x_e|=\sqrt{\alpha/\beta}$. It is not global finite-amplitude SHM.

Physical check

The two stable equilibria have the same linear curvature and frequency, although their quadratic corrections have opposite signs.