A particle of mass 0.5 kg executes simple harmonic motion. Its maximum kinetic energy is 9 ^2 J and the maximum restoring force acting on it is 6 ^2 N. If the particle starts its motion from the positive extreme position at t=0 , what is its position at t = 1 3 s?

Step-by-step solution

The maximum kinetic energy is given by K_ max = 1 2 m v_ max ^2 = 1 2 m (A )^2 . Substituting the given values: 1 2 0.5 (A )^2 = 9 ^2 (A )^2 = 36 ^2 A = 6 m/s. The maximum restoring force is given by F_ max = m a_ max = m (A ^2) . Substituting the given values: 0.5 (A ^2) = 6 ^2 A ^2 = 12 ^2 m/s ^2 . Dividing the two results to find angular frequency : = A ^2 A = 12 ^2 6 = 2 rad/s. Substituting back to find amplitude A : A(2 ) = 6 A = 3 m. Since the particle starts from the positive extreme position at t=0 , its eq