Question

A particle free to move along the x-axis has potential energy given by $U\left(x\right)=k[1-exp(-x{)}^2]$ for $-\infty \le x\le +\infty$, where k is a positive constant of appropriate dimensions. Then

• 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 1

The correct option is D

For small displacements from x = 0, the motion is simple harmonic

Potential energy of the particle $U=k(1-e^{-x^2})$

Force on particle $F=\frac{-dU}{dx}=-k[-e^{x^2}\times (-2x\left)\right]$

$F=-2kx{e}^{-x^2}=-2kx[1-x^2+\frac{x^4}{2!}-........]$

For small displacement F = -2kx

$\Rightarrow F\infty -xi.e.$ motion is simple harmonic motion.