A particle is executing simple harmonic motion with an amplitude A . At a displacement of A 2 from the mean position, the potential energy of the particle is 12 J . The maximum kinetic energy of the particle during its motion is

Step-by-step solution

The potential energy of a particle executing simple harmonic motion at a displacement x is given by U = 1 2 m ^2 x^2 . Given that at x = A 2 , the potential energy is 12 J : U = 1 2 m ^2 ( A 2 )^2 = 1 4 ( 1 2 m ^2 A^2 ) Let the total mechanical energy (which is equal to the maximum kinetic energy) be E = 1 2 m ^2 A^2 . Therefore, U = E 4 . Given U = 12 J , we have: E 4 = 12 E = 48 J In simple harmonic motion, the maximum kinetic energy is equal to the total mechanical energy. Thus, the maximum kinetic energy is 48