A particle executes S.H.M. starting from the mean position. Its amplitude is 'a' and its periodic time is 'T'. At a certain instant, its speed ' u ' is half that of maximum speed V_ max . The displacement of the particle at that instant is

Step-by-step solution

In simple harmonic motion, the velocity at displacement x is u = a^2 - x^2 , with maximum velocity V_ max = a at x = 0 . Given u = V_ max 2 , substitution yields a^2 - x^2 = a 2 . Canceling and squaring both sides gives a^2 - x^2 = a^2 4 . Solving for x^2 : x^2 = a^2 - a^2 4 = 3a^2 4 . Thus x = 3 a 2 , corresponding to option D .