Q: The potential energy of a particle varies with x according to the relation U(x) = x^{2} – 4 x. The point x = 2 is a point of :

(A) stable equilibrium

(B) unstable equilibrium

(C) neutral equilibrium

(D) none

Ans: (A)

Solution: U(x) = x^{2} – 4x

F = 0

$\frac{dU(x)}{dx} = 0 $

2 x – 4 = 0

x = 2

$\frac{d^2U(x)}{dx^2} = 2 > 0 $

i.e. U is minimum hence x = 2 is a point of stable equilibrium