A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as $F (x) = - k x + a x^{3}$ . Here k and a are positive constants. For $x \geq 0$ , the functional form of the potential energy U(x) of the particle is

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Solution

The correct option is D

The potential energy of the particle is given by
$U = - \int F d x = - \int (- k x + a x^{3}) d x$
$\Rightarrow U = k \frac{x^{2}}{2} - a \frac{x^{4}}{4} = \frac{x^{2}}{4} (2 k - a x^{2}) . . . . . . . (1)$
From Eq. (1) it follows that U = 0 at two values of x which are x = 0 and $x = \sqrt{\frac{2 k}{a}}$ . Hence, graphs (b) and (c) are not possible.
Also, U is maximum or minimum at a value of x given by $\frac{d U}{d x} = 0$
$\Rightarrow \frac{d}{d x} (\frac{k x^{2}}{2} - \frac{a x^{4}}{4}) = 0 \Rightarrow k x - a x^{3} = 0$
$\Rightarrow x = \sqrt{\frac{k}{a}}$
At this value of x, U is maximum if $\frac{d^{2} U}{d x^{2}} < 0$ .
Now, $\frac{d^{2} U}{d x^{2}} = \frac{d}{d x} (k x - a x^{3}) = k - 3 a x^{2}$
At $x = \sqrt{\frac{k}{a}}$ , $\frac{d^{2} U}{d x^{2}} = k - 3 a \frac{k}{a}$
$= k - 3 k = - 2 k$ , which is negative.
Hence U is maximum at $x = \sqrt{\frac{k}{a}}$ .
Hence, graph (a) is also not possible.
Moreover, U is negative for $x > \sqrt{\frac{2 k}{a}}$ .
Therefore, the correct graph is (d).

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