A particle of mass 0.5 kg is executing simple harmonic motion. Its kinetic energy K varies with displacement x from the mean position as K = 36 - 4x^2 , where K is in joules and x is in meters. The magnitude of the maximum acceleration of the particle is ______ m/s ^2 .

Step-by-step solution

The kinetic energy of a particle is given by K = 1 2 mv^2 . Given m = 0.5 kg , we have: K = 1 2 (0.5)v^2 = v^2 4 Equating this to the given expression for kinetic energy: v^2 4 = 36 - 4x^2 v^2 = 144 - 16x^2 v^2 = 16(9 - x^2) Comparing this with the standard equation of velocity in SHM, v^2 = ^2(A^2 - x^2) , we get: ^2 = 16 = 4 rad/s A^2 = 9 A = 3 m The magnitude of the maximum acceleration in SHM is given by: a_ = ^2 A a_ = 16 3 = 48 m/s ^2 Answer: 48