Question

A particle free to move along the x-axis has potential energy given by $U\left(x\right)=k(1-e^{-x^2})$ For $-\infty <x<\infty$, 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 U, 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

$U\left(x\right)=k(1-e^{-x^2})$

U versus x graph will be as shown.

From the graph it is clear that at origin, potential energy U is minimum (therefore, kinetic energy will be maximum) and force acting on the particle is also zero because

Therefore, origin is the stable equilibrium position, Hence, particle will oscillate simple harmonically about x = 0 for small displacements. Therefore, correct option is (d).

(a), (b) and (c) options are wrong due to following reasons :

(a) At equilibrium position $F=-\frac{dU}{dx}$ = c

i.e., slope U-x graph should be zero and from the graph we can see that slope is zero at x = 0 and $x=\pm \infty$ Therefore, option (a) is wrong.

(b) For any finite non-zero value of x, force is directed towards the origin because origin is in stable equilibrium position.

(c) kinetic energy will be maximum since potential energy U is minimum at origin.