A particle executes simple harmonic motion. Which of the following describes the graph showing the variation of its potential energy ( PE ) with the magnitude of its velocity ( v ) as it moves from the extreme position to the mean position?

Step-by-step solution

The potential energy ( PE ) of a particle in SHM is given by PE = 1 2 m ^2x^2 . The velocity v of the particle is related to its displacement x by the equation v^2 = ^2(A^2 - x^2) , where A is the amplitude. Rearranging this for x^2 , we get x^2 = A^2 - v^2 ^2 . Substituting this into the potential energy equation: PE = 1 2 m ^2 (A^2 - v^2 ^2 ) = 1 2 m ^2A^2 - 1 2 mv^2 This equation is of the form y = -cx^2 + k (where c and k are positive constants). This represents a downward-opening parabola with its maximum valu