Q 11-05-045JEE MainMedium
The potential energy of a particle moving along the $x$-axis is $U(x) = x^3 - 12x$ (SI units). The particle is in stable equilibrium at
Answer: (D) $x = 2$ m
Concept. Equilibrium where $\dfrac{dU}{dx} = 0$; stable where $\dfrac{d^2U}{dx^2} > 0$.
Step 1. Equilibria. $3x^2 - 12 = 0 \Rightarrow x = \pm2$.
Step 2. Stability. $6x > 0$ at $x = 2$.
Answer. $x = 2$ m (Option D).
Solution by Sreeraj P, M.Sc Physics