A particle starts from mean position and performs S.H.M. with period 4 second. At what time its kinetic energy is 50 % of total energy?

Step-by-step solution

The total energy is given by T.E = 1 2 kA ^2 Kinetic energy is given by K.E = 1 2 k (A^2-x^2 ) array ll & K.E = 1 2 P . E & 1 2 k (A^2-x^2 )= 1 2 ( 1 2 kA ^2 ) & ( A ^2- x ^2 )= 1 2 ~A ^2 & ( A ^2- x ^2 )= 1 2 ~A ^2 & x = A 2 array The equation of displacement in SHM is array ll & x = A 2 t T & A 2 = A 2 t 4 1 2 = t 2 & .( T =4 sec ) t 2 & = ⁻¹ 1 2 & t 2 = 4 & t =0.5 ~s array