The maximum potential energy of a block executing simple harmonic motion is 25 J . A is amplitude of oscillation. At A 2 , the kinetic energy of the block is

Step-by-step solution

The maximum potential energy of a block executing simple harmonic motion is given by U max = 1 2 m ω 2 A 2 = 25   J As, kinetic energy of particle executing SHM is K E = 1 2 m ω 2 A 2 - x 2 . So, kinetic energy of block K E   at   A 2 = 1 2 m v 1 2 = 1 2 m ω 2 A 2 - A 2 4 ⇒ K E = 1 2 m ω 2 3 A 2 4 = 3 4 1 2 m ω 2 A 2 ⇒ K E = 3 4 × 25 = 18 . 75   J