Question

The potential energy of a particle as a function of its position is given by $U=x^3-6x^2$ (in SI units). The nature of equilibrium at $x=4$ is

• A. Stable
• B. Unstable
• C. Neutral
• D. Insufficient Data

Answer 1

The correct option is A Stable

$U=x^3-6x^2$
We know that,
$F=-\frac{\partial U}{\partial x}=-(3x^2-12x)$
$F=12x-3x^2$
The nature of equilibrium is stable if $F=0$ & $\frac{{\partial }^{2}U}{\partial x^2}$ is positive
So, when $F=0$
$\Rightarrow 3x(4-x)=0$
$\Rightarrow x=0\&x=4$
Now,
$\frac{{\partial }^{2}U}{\partial x^2}=\frac{\partial }{\partial x}\frac{\partial U}{\partial x}=\frac{\partial }{\partial x}(3x^2-12x)$
= $6x-12{|}_{x=4}$
= $24-12=12>0$
Therefore condition of stable equilibrium.