Question

A particle located in a one-dimensional potential field has its potential energy function as $U\left(x\right)\frac{a}{x^4}-\frac{b}{x^2}$where $a$ and $b$ are positive constants. The position of equilibrium $x$ corresponds to

• A. $\frac{b}{2a}$
• B. $\sqrt{\frac{2a}{b}}$
• C. $\sqrt{\frac{2a}{a}}$
• D. $\frac{a}{2a}$

Answer 1

Answer: (B)

The position of equilibrium corresponds to $F\left(x\right)=0$
Since $F\left(x\right)=\frac{-dU\left(x\right)}{dx}$
so $F\left(x\right)=-\frac{d}{dx}\left(\frac{a}{x^4}-\frac{b}{x^2}\right)$ or$F\left(x\right)=\frac{4a}{x^5}-\frac{2b}{x^3}$
For equilibrium, $F\left(x\right)=0$, therefore
$\frac{4a}{x^5}-\frac{2b}{x^3}=0$
$\Rightarrow x=\pm \sqrt{\frac{2a}{b}}$
$\frac{d^{2}U\left(x\right)}{d{x}^2}=-\frac{20a}{x^6}+\frac{8b}{x^4}$
Putting $x=\pm \sqrt{\frac{2a}{b}}$ given $\frac{d^{2}U\left(x\right)}{d{x}^2}$ as negative
So $U$ is maximum. Hence, it is position of unstable equilibrium