8.4 Potential energy diagrams and stability  (Page 3/7)

You can just eyeball the graph to reach qualitative answers to the questions in this example. That, after all, is the value of potential energy diagrams. You can see that there are two allowed regions for the motion $\left(E>U\right)$ and three equilibrium points (slope $dU\text{/}dx=0),$ of which the central one is unstable $\left(d^{2}U\text{/}d{x}^2<0\right),$ and the other two are stable $\left(d^{2}U\text{/}d{x}^2>0\right).$

Solution

1. To find the allowed regions for x , we use the condition
   
   $K=E-U=-\frac{1}{4}-2\left(x^4-x^2\right)\ge 0.$
   
   If we complete the square in $x^2$ , this condition simplifies to $2{\left(x^2-\frac{1}{2}\right)}^2\le \frac{1}{4},$ which we can solve to obtain
   
   $\frac{1}{2}-\sqrt{\frac{1}{8}}\le x^2\le \frac{1}{2}+\sqrt{\frac{1}{8}}.$
   
   This represents two allowed regions, $x_p\le x\le x_R$ and $\text{−}{x}_R\le x\le -x_p,$ where $x_p=0.38$ and $x_R=0.92$ (in meters).
2. To find the equilibrium points, we solve the equation
   
   $dU\text{/}dx=8x^3-4x=0$
   
   and find $x=0$ and $x=\text{±}{x}_Q$ , where $x_Q=1\text{/}\sqrt{2}=0.707$ (meters). The second derivative
   
   $d^{2}U\text{/}d{x}^2=24x^2-4$
   
   is negative at $x=0$ , so that position is a relative maximum and the equilibrium there is unstable. The second derivative is positive at $x=\text{±}{x}_Q$ , so these positions are relative minima and represent stable equilibria.

Significance

Before ending this section, let’s practice applying the method based on the potential energy of a particle to find its position as a function of time, for the one-dimensional, mass-spring system considered earlier in this section.