When a particle of mass m moves on the X-axis in a potential of the form V(x) = kx^(2), it performs simple harmonic motion. The corresponding time period is proportional to sqrt((m)/(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 kx^(2) 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)=a x^(2) (a gt 0) for |x| near the origin and becomes a constant equal to V_(0) for |x| ge X_(0) The acceleration of this particle for |X|>X_(0) is

When a particle of mass m moves on the X-axis in a potential of the fo

1.	When a particle of mass m moves on the X-axis in a potential of the form V(x) = kx^(2), it performs simple harmonic motion. The corresponding time period is proportional to sqrt((m)/(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 kx^(2) 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)=a x^(2) (a gt 0) for \|x\| near the origin and becomes a constant equal to V_(0) for \|x\| ge X_(0) The acceleration of this particle for \|X\|>X_(0) is
Answer» proportional to `(v_(0)/(mX_(0)))` propotinal to `sqrt(V_(0))/(mX_(0))` proportional to `V_(0)` Zero Solution :In this region `\|x\| GT X_(0)` POTENTIAL energy becomes CONSTANT HENCE `F = -(dU)/(dx) = 0` Hence is absence of force acceleration of the particle REMAINS zero.