A particle free to move along the x-axis has potential energy given by…

Q:  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)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

Ans: (d)

Sol: At origin U = 0 . Where Potential energy is minimum Kinetic energy is maximum .

As F = -dU/dx = 0 . Therefore origin is the stable equilibrium position & particle will oscillates simple harmonically about x = 0 for small displacement .