For a particle of mass 0.5 kg executing simple harmonic motion, the graph of its kinetic energy ( KE ) versus displacement ( x ) is a downward-opening parabola. The maximum kinetic energy is 4 J and the kinetic energy becomes zero at displacements of +0.2 m and -0.2 m. The time period of the oscillation is

Step-by-step solution

From the given information, the maximum kinetic energy is K_ max = 4 J. The kinetic energy becomes zero at the extreme positions, so the amplitude of the motion is A = 0.2 m. The maximum kinetic energy of a particle in SHM is given by K_ max = 1 2 m ^2A^2 . Substituting the given values: 4 = 1 2 0.5 ^2 (0.2)^2 4 = 0.25 ^2 0.04 4 = 0.01 ^2 ^2 = 400 = 20 rad/s The time period T of the oscillation is given by T = 2 . T = 2 20 = 10 s. Answer: 10 s