Work Energy And Power Question 68
Question: A particle free to move along the x-axis has potential energy given by $ U(x)=k[1-\exp {{(-x)}^{2}}] $ for $ -\infty \le x\le +\infty $ , where k is a positive constant of appropriate dimensions. Then [IIT-JEE 1999; UPSEAT 2003]
Options:
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
Show Answer
Answer:
Correct Answer: D
Solution:
Potential energy of the particle $ U=k(1-{e^{-x^{2}}}) $
Force on particle $ F=\frac{-dU}{dx}=-k[-{e^{-x^{2}}}\times (-2x)] $ F $ =-2kx{e^{-x^{2}}} $
$ =-2kx[ 1-x^{2}+\frac{x^{4}}{2!}-…… ] $
For small displacement $ F=-2kx $
therefore $ F(x)\propto -x $ i.e. motion is simple harmonic motion.
