Q. A particle free to move along the $x$-axis has potential energy given by $U(x) = k\lbrack 1 - \exp\left( - x^{2} \right)\rbrack$ for $- \infty \leq x \leq + \infty$, where $k$ is a positive constant of appropriate dimensions. Then
A) At point away from the origin, the particle is in unstable equilibrium
B) For any finite non-zero value of $x$, there is a force directed away from the origin
C) If its total mechanical energy is $k/2$, it has its minimum kinetic energy at the origin
D) For small displacements from $x = 0$, the motion is simple harmonic
Explanation
(d) Potential energy of the particle $U = k(1 - {e^{- x}}^{2})$
Force on particle $F = \frac{- dU}{dx} = - k\lbrack - {e^{- x}}^{2} \times ( - 2x)\rbrack$
$F = - 2kx{e^{- x}}^{2} = - 2kx\left\lbrack 1 - x^{2} + \frac{x^{4}}{2!} - \ldots \right\rbrack$
For small displacement $F = - 2kx$
$\Rightarrow F \propto - x$ $i.e.$ motion is simple harmonic motion
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.