A particle free to move along the `x-`axis has potential energy given by `U(x) = k[1 - e^(-x^(2))]` for `-oo le x le + oo`, where `k` is a positive constant of appropriate dimensions. Then select the incorrect option

Options

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

Answer

D

Explanation

Potential energy of the particle `U = k(1-e^(-x^(2)))`
Force on particle `F = (-dU)/(dx) = -k[-e^(x^(2)) xx(-2x)]`
`F = - 2kx e^(-x^(2)) = -2k x[1-x^(2) + (x^(4))/(2!)-…..]`
For small displacement `F = -2kx`
`rArr F prop -x` i.e., motion is simple harmonic motion.