A particle starts oscillating simple harmonically form its mean position with time period T . At time t= T 12 , the ratio of the potential energy to kinetic energy of the particle is ( 30^ = 60^ =0.5, 30^ = 60^ = 3 2 )

Step-by-step solution

Let, the equation of the particle performing SHM is given by x=A t , where = 2 T At t= T 12 , x=A ( 2 T T 12 )=A ( 6 )= A 2 The potential energy of the particle at T= T 12 The kinetic energy of the particle at t= T 12 aligned & K= 1 2 m ^2 x^2 (A^2-x^2 )= 1 2 m ^2 [A^2- ( A 2 )^2 ] & K= 1 2 m ^2 ( 3 A^2 4 )---(2) aligned From equation (1) and (2) K U = 3 1