Question

Directions: The following question has four choices, out of which one or more are correct. The potential energy function of a particle moving in one dimension is $U=k[x-\frac{x^3}{3a^2}],$ where $a$ and $k$ areconstants. Then,

• (A) equilibrium is stable at x = a/2
• (B) equilibrium is stable at x = -a
• (C) at stable equilibrium, PE = -2ka/3
• (D) equilibrium is stable at x = +a

Answer 1

The Correct Option is A, D

Explanation:
Force corresponding to potential energy is,$$F=-\frac{dU}{dx}=-k[1-\frac{x^2}{a^2}]$$For equilibrium, $F=0.$ Therefore,$$\begin{array}{rl}1-\frac{x^2}{a^2}& =0\\ x& =\pm a\end{array}$$For stable equilibrium,$$\frac{d^{2}U}{d{x}^2}>0$$$$\frac{d^{2}U}{d{x}^2}=-\frac{2x}{a^2}$$$$\begin{array}{l}{\frac{d^{2}U}{d{x}^2}|}_{\text{at }x=a}=-\frac{2}{a}<0,\text{ so unstable }\\ {\frac{d^{2}U}{d{x}^2}|}_{\text{at }x=-a}=\frac{2}{a}>0,\text{ so stable }\end{array}$$Potential energy at $x=-a$ is$$U=k[-a+\frac{a^3}{3a^2}]=-\frac{2ka}{3}$$Hence, this is the required solution.