First slide

Concept of potential energy

Question

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

Moderate

A

$\frac{b}{2 a}$
B

$\sqrt{\frac{2 a}{b}}$
C

$\sqrt{\frac{2 b}{a}}$
D

$\frac{a}{b}$

Solution

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

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