Question

A particle located in a one - dimentional potential field has its potential energy functions as U(x) = $\frac{a}{x^4}-\frac{b}{x^2}$, where a and b are positive constants. The positionof equilibrium x corresponds to

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The position of equilibrium corresponds to F(x) = 0

Since $F\left(x\right)=\frac{-dU\left(x\right)}{dx}$

so $F\left(x\right)=\frac{d}{dx}(\frac{a}{x^4}-\frac{b}{x^2})orF\left(x\right)=\frac{4a}{x^5}-\frac{2b}{x^3}$

For equilibrium m, F(x) = 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 $x=\pm \sqrt{\frac{2a}{b}}$ gives $\frac{-d^{2}U\left(x\right)}{d{x}^2}$ as negative

So U is maximum. Hence, it is position of unstable equilibrium.