A particle executes simple harmonic motion with amplitude A . At a displacement of x = A 2 from the mean position, what is the ratio of its kinetic energy to its potential energy?

Step-by-step solution

Let the force constant of the simple harmonic motion be k . The potential energy U at a displacement x is given by: U = 1 2 k x^2 At x = A 2 : U = 1 2 k ( A 2 )^2 = 1 8 k A^2 The total mechanical energy E of the particle is: E = 1 2 k A^2 The kinetic energy K at this displacement is: K = E - U = 1 2 k A^2 - 1 8 k A^2 = 3 8 k A^2 The ratio of kinetic energy to potential energy is: K U = 3 8 k A^2 1 8 k A^2 = 3 1 Thus, the ratio is 3 : 1 . Answer: 3 : 1