Question

The potential energy of a particle is given by relation $U=x^3-6x^2$ (In SI units). The nature of equilibrium at $x=0$ is

• A. Stable
• B. Unstable
• C. Neutral
• D. Undefined

Answer 1

The correct option is B Unstable

It is given that, $U=x^3-6x^2$

We also know that $F=-\frac{dU}{dx}=-3x^2+12x$
$\therefore \frac{dU}{dx}=3x^2-12x$

For the condition of equilibrium, force $F=0\Rightarrow F=-3x^2+12x=0$

$\Rightarrow F=0\Rightarrow 3x(4-x)=0$
$\therefore x=0,x=4$

$\frac{d^{2}U}{d{x}^2}=\frac{d}{dx}(3x^2-12x)=6x-12$

At $(x=0)$
$\therefore \frac{d^{2}U}{d{x}^2}=-12<0$
$\because F=0$ & $\frac{d^{2}U}{d{x}^2}<0$ is the condition for unstable equilibrium

$\Rightarrow$ Unstable Equillibrium