JEE Advance - Physics (2010 - Paper 1 Offline - No. 23)

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).

When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).

The acceleration of this particle for $$|x| > {X_0}$$ is
proportional to V0.
proportional to V0/mX0.
proportional to $$\sqrt {{V_0}/m{X_0}} $$.
zero.

Explanation

For $$|x| > {X_0}$$, $$V = {V_0}$$ = constant

Force $$ = - {{dV} \over {dx}} = 0$$

Hence, acceleration of the particle is zero for $$|x| > {X_0}$$.

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