A particle free to move along the x-axis has potential energy giv… - Vidyadip

MCQ

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

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

✓

Answer

Correct option: D.

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

d
(d)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\left[ {1 - {x^2} + \frac{{{x^4}}}{{2\,!}} - ......} \right]$
For small displacement $F = - 2kx$
$⇒$ $F(x) \propto - x$ i.e. motion is simple harmonic motion.

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