A particle free to move along the x-axis has potential energy given by $U (x) = k [1 - \exp (- x^{2})]$ for , $- \infty \leq x \leq + \infty$ where k is a positive constant of appropriate dimensions. Then

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For any finite non-zero value of x, there is a force directed away from the origin

For small displacements from $x = 0$ , the motion is simple harmonic

At points away from the origin, the particle is in unstable equilibrium

If its total mechanical energy is $\frac{k}{2}$ , it has its minimum kinetic energy harmonic

answer is D.

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Detailed Solution

Let us plot the graph of the mathematical equation

$U (x) = K [1 - e^{- x^{2}}] ∴ F = - \frac{d U}{d x} = - 2 k x e^{- x^{2}}$

It is clear that the potential energy is minimum at $x = 0$ . Therefore, $x = 0$ is the state of stable equilibrium. Now if we displace the particle from $x = 0$ , then for small displacements the particle tends to regain the position $x = 0$ with a force $F = \frac{2 k x}{e^{x^{2}}}$ for x to be small $F \propto x .$