A particle of mass ' m ' is executing S.H.M. about the origin on x-axis with frequency Ka m , where K is a constant and a is the amplitude of S.H.M. If ' x ' is the displacement of a particle at time ' t ', the potential energy of a particle will be

Step-by-step solution

The frequency of the particle executing S.H.M. is given by f = Ka m The angular frequency is = 2 f = 2 Ka m Squaring both sides, we get ^2 = 4 ^2 ( Ka m ) = 4 Ka m The potential energy of a particle in S.H.M. at a displacement x is given by U = 1 2 m ^2 x^2 Substituting the value of ^2 , we get U = 1 2 m ( 4 Ka m ) x^2 U = 2 K a x^2 Answer: 2 Kax^2