A particle starts from mean position and performs S.H.M. with period 6 second. At what time its kinetic energy is 50 % of total energy? ( 45^ = 1/ 2 )

Step-by-step solution

The displacement of a particle starting from the mean position is given by x = A ( t) . The velocity of the particle is v = dx dt = A ( t) . The kinetic energy is K = 1 2 m v^2 = 1 2 m A^2 ^2 ^2( t) . The total energy of the particle in S.H.M. is E = 1 2 m A^2 ^2 . Given that the kinetic energy is 50 % of the total energy: K = E 2 1 2 m A^2 ^2 ^2( t) = 1 2 ( 1 2 m A^2 ^2 ) ^2( t) = 1 2 ( t) = 1 2 t = 4 Substituting = 2 T : 2 T t = 4 t = T 8 Given T = 6 s: t = 6 8 = 0.75 s. Answer: 0.75 s