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A particle free to move along the (x - a...

A particle free to move along the (x - axis) hsd potential energy given by `U(x)= k[1 - exp(-x^2)] for -o o le x le + o o`, 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

Text Solution

Verified by Experts

The correct Answer is:
D
`U(x)= k (1-e^(-x^(2))), - oo le x le oo`
`F= (-dU(x))/(dx)= - [-k(e^(-x^(2)))(-2x)]`
`= -2k x e^(-x^(2))`
`F= -2k xe^(-x^(2)) = -2k x [1- x^(2) + (x^(4))/(2!)---]`
For small displacement- `F= -2kx rArr F prop -x`
Motion is SHM.
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