A particle is executing S.H.M. of amplitude 'A'. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is

Step-by-step solution

The potential energy at displacement x from equilibrium is PE = 1 2 kx^2 , where k is the force constant. Maximum potential energy occurs at amplitude A , giving PE_ max = 1 2 kA^2 . Given that PE = 1 2 PE_ max , substitution yields: 1 2 kx^2 = 1 2 ( 1 2 kA^2 ) Simplifying both sides gives 1 2 kx^2 = 1 4 kA^2 . Canceling the common factor 1 2 k produces x^2 = 1 2 A^2 , so x = A 2 . The displacement when potential energy is half its maximum is D .