Q. A particle free to move along the $x$-axis has potential energy given as $U(x) = k\left\lbrack 1 - \exp\left( - x^{2} \right) \right\rbrack\ \ \ \ \ \ \ \ \ \ (for - \infty \leq \ + \infty)$ Where $k$ is a positive constant of appropriate dimensions. Then
A) At points away from origin, the particle is in equilibrium
B) For any finite non-zero value of $x$, there is a force directed away from the origin
C) Its total mechanical energy is $k/2$ and it is equal to its kinetic energy at origin
D) At $x = 0$, the motion of the particle is simple harmonic
Explanation
(d) If a force acting on an object is a function of position only, it is said to be conservative force and it can be represented by a potential energy ($U$) function which for one dimensional case satisfies the derivative condition.
$F(x) = - \frac{dU}{dx}$
Given, $U_{x} = k\left\lbrack 1 - \exp\left( - x^{2} \right) \right\rbrack$
$\therefore$ $F = - \frac{dU}{dx} = - 2kx\exp{( - x^{2})}$
At $x = 0,\ F = 0$.
Hence, at equilibrium force exerted on particle is zero.
Also, potential energy of the particle is minimum at $x = 0\ and\ x = \pm \infty$ the potential energy is maximum. Hence, at $x = 0$, the motion of particle is simple harmonic.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.