Question

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

(a) The particle is in unstable equilibrium at points away from the origin

(b) There is a force directed away from the origin for any finite nonzero value of x

(c) It has its minimum kinetic energy at the origin if its total mechanical energy is k/2

(d) The motion is simple harmonic for small displacement from x=0

This question was posed to me in exam.

This key question is from Harmonic Motion topic in section Oscillations of Engineering Physics I

Answer 1

Right answer is (d) The motion is simple harmonic for small displacement from x=0

To explain: U=k(1-e^(-x^2))

F=-dU/dx=-2kxe(-x^2)=-2kx(1-x^2+⋯)

For small x, F ≅-2kx

This shows that the force is directed towards the origin and for smaller x, F∝x. Hence the motion is simple harmonic.