# Finding the equilibrium points of a potential system and determining their stability

I am having trouble solving a particular question

The question is: A particle of mass m moves on the x-axis under a force.

a) Let V(x) = $3x-x^3$. Sketch the potential and find the equilibrium points of the system and determine their stability.

b) Assuming that the particle is fired from the origin with speed $\dot{x}$ = 1, determine $m_{max}$, the maximum value of m, below which the particle’s motion must be oscillatory.

I have already sketched the potential and found the equilibrium points by doing $V(x)' = 0$ to be $\pm 1$. I don't know how to determine their stability.

For part b), I know the Energy $E = {1\over{2}}m\dot{x}^2 + V(x)$ and that the particle is fired from the origin so $x = 0$ and $\dot{x}$ = 1 so plugging in gives me $E = {1\over2} m$. Im stuck at this point and don't know how to get further. Any help will be appreciated thanks.

The equation of motion $$\frac{d^{2}x}{dt^{2}}=-\frac{d}{dx}U(x)=3x^{2}-3$$ Let $x(t)=\pm1+\delta{x}(t)$, where $|\delta(x)(t)|<<1$ is a small perturbation to the system when the former is in the equailibrium. Substituting into the equation $$\ddot{\delta{x}}=3(\delta{x})^{2}\pm6\delta{x}$$ $3(\delta{x})^{2}+6\delta{x}>0$, hence the $x=+1$ equilibrium is unstable, as the acceleration is positive, $3(\delta{x})^{2}-6\delta{x}<0$, hence the $x=-1$ equilibrium is stable as the acceleration is negative, i.e. the system wants to return back to equilibrium. For the part b) There is a theorem (look in any text on classical mechanics) that states that any bound 1D motion is periodic. For the particle to JUST escape from the bounding potential, the maximum potential energy in must have is $V(1)=2$ with zero velocity, that means his energy must be $E=V(1)=2$. The initial speed is $1$ and the particle is fiered from the origin, so $$2=\frac{1}{2}m_{max}$$ So $m_{max}=4$