For a system executing simple harmonic motion,

Step-by-step solution

For a system executing simple harmonic motion, the total mechanical energy is conserved. Let the displacement of the particle from the mean position be x and the amplitude be A . The kinetic energy of the system is given by: K = 1 2 m ^2(A^2 - x^2) Assuming the potential energy to be zero at the mean position, the potential energy is given by: U = 1 2 m ^2 x^2 At the mean position ( x = 0 ): K_ max = 1 2 m ^2 A^2 U_ min = 0 At the extreme positions ( x = A ): K_ min = 0 U_ max = 1 2 m ^2 A^2 Comparing the maximum a