A horizontal rod of mass m and length L is pivoted smoothly at one end. The rod's other end is supported by a spring of force constant k . The rod is rotated (in vertical plane) by a small angle θ from its horizontal equilibrium position and released. The angular frequency of the subsequent simple harmonic motion is:

Step-by-step solution

The restoring torque on the rod at the position shown is, τ = - k x   L = I   α ⇒ - k   L θ   L = m L 2 3 α ⇒ α = - 3   k m θ The Differential equation fo angular SHM is α = - ω 2 θ . Comparing the above equations, Angular frequency, ω = 3   k m .