A particle starts oscillating simple harmonically from 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

The equation of displacement for a particle executing simple harmonic motion starting from the mean position is given by: x = A ( t) Given t = T 12 and = 2 T , the displacement is: x = A ( 2 T T 12 ) = A ( 6 ) = A (30^ ) = A 2 The potential energy ( U ) of the particle is: U = 1 2 kx^2 = 1 2 k ( A 2 )^2 = 1 8 kA^2 The total energy ( E ) of the particle is: E = 1 2 kA^2 The kinetic energy ( K ) is the difference between total energy and potential energy: K = E - U = 1 2 kA^2 - 1 8 kA^2 = 3 8 kA^2 The ratio of potent