Question

Potenntial energy of a particle moving along x-axis given by U =$\frac{x^3}{3}-\frac{9x^2}{2}+20x$. Find out position of the stable equilibrium state.

Answer 1

Potential energy of particle

$u=\frac{x^3}{3}-\frac{9x^2}{2}+20x$

Particle would be at stable equilibrium when 'u' is minimum

For a minima, we need, $\frac{du}{dx}=0$ and $\frac{d^{2}u}{d{x}^2}>0$

Thus,

$\frac{du}{dx}=x^2-9x+20$ and $\frac{d^{2}u}{d{x}^2}=2x-9$

$=(x-4)(x-5)$

Which means,

$\frac{du}{dx}=0$ given $(x-4)(x-5)=0\Rightarrow x=4$ or $x=5$.

also, for $x=4,\frac{d^{2}u}{d{x}^2}{|}_{x=4}<0$ but for $x=5,\frac{d^{2}u}{d{x}^2}{|}_{x=5}>0$

$\therefore$ particle is at state equilibrium at $x=5$