A particle of mass 0.5 kg executes simple harmonic motion. Its acceleration a varies with displacement x as a = -16 ^2 x . At t=0 , the particle is at the mean position moving in the positive x -direction with a kinetic energy of 4 J . The kinetic energy of the particle at t = 1 12 s is ______ J .

Step-by-step solution

The acceleration of a particle in SHM is given by a = - ^2 x . Comparing this with a = -16 ^2 x , we get: ^2 = 16 ^2 = 4 rad/s At t=0 , the particle is at the mean position ( x=0 ), so its velocity is maximum. The maximum kinetic energy is given as 4 J . 1 2 m v_ max ^2 = 4 1 2 0.5 v_ max ^2 = 4 v_ max ^2 = 16 v_ max = 4 m/s Since the particle starts from the mean position and moves in the positive direction, its velocity as a function of time is: v(t) = v_ max ( t) v(t) = 4 (4 t) At t = 1 12 s , the velocity is: v