The potential energy of a particle of mass 4   kg in motion along the x -axis is given by U = 4 1 - cos 4 x   J . The time period of the particle for small oscillation sin θ ≃ θ π K   s . The value of K is _____ .

Step-by-step solution

Given here, potential energy, U = 4 1 - cos 4 x   J Using the relation between conservative force and potential energy, F = - d U d x . We have, F = - 4 + sin 4 x 4 = - 16 sin 4 x For small θ , sin θ ≈ θ . Acceleration of particle is a = - 64 x m = - 64 x 4 = - 16 x As the oscillations are simple harmonic in nature, so a = - ω 2 x ⇒ ω 2 = 16 ⇒ ω = 4   rad   s - 1 Now, time period of oscillation is T = 2 π ω = π 2 . Thus, the value of K =