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Restoring Force

The potential...

Question

The potential energy of a particle varies with distance x as shown in the graph. The force acting on the particle is zero at:

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B and C

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A and D

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Solution

The correct option is C B and C
$F = \frac{- d U}{d x}$
F=0 where slope is 0 i.e. at maxima and minima
Hence, F=0 at B & C

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Column I	Column II
(A) $U_{1} (x) = \frac{U_{0}}{2} {[1 - {(\frac{x}{a})}^{2}]}^{2}$	(P) The force acting on the particle is zero at x = a.
(B) $U_{2} (x) = \frac{U_{0}}{2} {(\frac{x}{a})}^{2}$	(Q) The force acting on the particle is zero at x = 0.
(C) $U_{3} (x) = \frac{U_{0}}{2} {(\frac{x}{a})}^{2} exp [- {(\frac{x}{a})}^{2}]$	(R) The force acting on the particle is zero at x = – a.
(D) $U_{4} (x) = \frac{U_{0}}{2} [\frac{x}{a} - \frac{1}{3} {(\frac{x}{a})}^{3}]$	(S) The particle experiences an attractive force towards x = 0 in the region $\| x \| < a .$
	(T) The particle with total energy $\frac{U_{0}}{4}$ can oscillate about the point x = – a.

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