A particle of mass m is executing bounded motion along the x -axis. Its kinetic energy K varies with position x as K = x - x^2 , where and are positive constants. The time period of the motion is

Step-by-step solution

Given, K = x - x^2 We know that kinetic energy K = 1 2 mv^2 1 2 mv^2 = x - x^2 Differentiating both sides with respect to x : 1 2 m (2v dv dx ) = - 2 x Since v dv dx = a (acceleration), we get: ma = - 2 x a = - 2 m (x - 2 ) This equation is of the form a = - ^2 (x - x₀) , which represents Simple Harmonic Motion about the mean position x₀ = 2 . Comparing the terms, we get ^2 = 2 m = 2 m The time period of the motion is T = 2 = 2 m 2 . Answer: 2 m 2