A particle is executing linear S.H.M. starting from mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is

Step-by-step solution

For a particle performing simple harmonic motion with amplitude A and angular frequency , whose displacement from mean position is x , the force constant is given by k = m ^2 . The potential energy at displacement x is PE = 1 2 k x^2 , while the kinetic energy is KE = 1 2 m ^2 (A^2 - x^2) = 1 2 k (A^2 - x^2) . At the point where the displacement is half the amplitude, x = A/2 , the potential energy becomes PE = 1 2 k (A/2)^2 = 1 8 k A^2 and the kinetic energy becomes KE = 1 2 k (A^2 - (A/2)^2) = 1 2 k (3A^2/4) = 3