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
Equilibrium: $F = -\dfrac{dU}{dx} = -(3x^2 - 12) = 0 \Rightarrow x = \pm 2$ m.
Stable equilibrium is a minimum of $U$: $\dfrac{d^2U}{dx^2} = 6x > 0$ at $x = 2$. (At $x = -2$ it is a maximum, which is unstable.)
Solution by Sreeraj P, M.Sc Physics