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) For periodic motion os small amplitude A . The time periodic T of this particle is proportional to

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) For periodic motion os small amplitude A . The time periodic T of this particle is proportional to
Answer» `Asqrt(m)/(ALPHA)` `(1)/(A)sqrt(m)/(alpha)` `Asqrt(alpha)/(m)` `(1)/(A)sqrt(alpha)/(m)` SOLUTION :Here dimensions of m I s[M], A is [L] It is given that V(x) = `alphax^(4)` hence dimensions of `alpha` is `[ML^(-2)T^(-2)]` From dimensional analysis we can see that time PERIOD is proportional to `(1)/(A) sqrt((m)/(alpha))`, hence option (b) is correct.