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

Options

A. at points away from the origin, the particle is in unstable equilibrium.
B. For any finite nonzero 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

Let us plot the graph of the mathematical equation
`U(x) = K[1 - e^(-x 2)], F = -(d U)/(d x)= 2k x e^(-x^2)`
(##JMA_CHMO_C10_003_S01##).
From the graph 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 from (x = 0) then for displacements the particle tends to regain the position (x = 0) with a force `F = (2k x)/(e^(x2)`. Therefore for small values of (x) we have `F prop x`.