A particle starts oscillating simple harmonically from its equilibrium position with time period ' T '. At time t = T 12 , the ratio of its kinetic energy to potential energy is [ 3 6 = 3 2 , 6 = 3 = 1 2 ]

Step-by-step solution

Take equation of motion as x = a t , we have t = T 12 x ( T 12 )= ( 2 T T 12 )= asin ( 6 )= a 2 Then the ratio of the kinetic energy to the potential energy at x= a 2 will be given by K.E. P.E. = 1 2 k (a^2-x^2 ) 1 2 k x^2 = a^2- ( a 2 )^2 ( a 2 )^2 = 3 1