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 points 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

`U(x)=K[l-e^(-x^(2))] therefore F=-(dU)/(dx)=-2kxe^(-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=2kx//e^(x^(2))` for x to be small `F prop x `