A particle of unit mass is moving along the x-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column I (a and U0 are constants). Match the potential energies in column I to the corresponding statement (s) in column II.Column IColumn II(A) (P) The force acting on the particle is zero at x = a.(B) (Q) The force acting on the particle is zero at x = 0.(C) (R) The force acting on the particle is zero at x = – a.(D) (S) The particle experiences an attractive force towards x = 0 in the region | x | < a.(T) The particle with total energy can oscillate about the point x =

Engineering

Physics

Potential Energy Conservative Forces and Non Conservative Forces

Question

A particle of unit mass is moving along the x-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column I (a and U₀ are constants). Match the potential energies in column I to the corresponding statement (s) in column II.

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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(A) $U_{1} = \frac{U_{0}}{2} {[1 - \frac{x^{2}}{a^{2}}]}^{2}$

$F = - \frac{dU}{dx} = + \frac{U_{0}}{0} \times 2 (1 - \frac{x^{2}}{a^{2}}) (1 + \frac{2 x}{a^{2}})$

at x = a, F = 0

at x = 0, F = 0

at x = –a, F = 0

at x = –a position of stable equilibrium

x = +a position of stable equilibrium

(B) $U_{2} (x) = \frac{U_{0}}{2} (\frac{x^{2}}{a})$

$F_{2} = - \frac{dU}{dx} = - \frac{2 U_{0} x}{2 a} = \frac{- U_{0}}{a} x$

$F_{3} (x) = \frac{- dU}{dx} = - \frac{U_{0}}{2} \frac{2 x}{a} e^{- x^{2} / a^{2}} + \frac{U_{0}}{2} \frac{x^{2}}{a} e^{- x^{2} / a^{2}} (\frac{2 x}{a^{2}})$

$= - \frac{U_{0}}{2} \frac{2 x}{a} e^{- x^{2} / a^{2}} {1 - \frac{x^{2}}{a^{2}}}$

x = –a, a

will be positions of unstable equilibrium.

(D) $U_{4} (x) \frac{U_{0}}{2} {\frac{x}{a} - \frac{x^{3}}{3 a^{3}}}$

$F = \frac{- dU}{dx} = - \frac{U_{0}}{2 a} + \frac{U_{0}}{6 a^{3}} 3 x^{2} = - \frac{U_{0}}{2 a} (1 - \frac{x^{2}}{a^{2}})$

x = –a ⇒ position of stable equilibrium

x = +a ⇒ position of unstable equilibrium