The ratio of kinetic energy at the mean position to potential energy at A 2 of a particle performing SHM is

Step-by-step solution

Kinetic energy K = 1 2 m ω 2 (A 2 - y 2 ) At mean position y = 0 K = 1 2 m ω 2 ( A 2 ) Potential energy U = 1 2 m ω 2 y 2 ...(i) U a t y = A 2 U = 1 2 m A 2 4 ω 2 ...(ii) Dividing Eq. (i) by Eq. (ii), we get K E U = 4 1