A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constant k . The rod is gently pushed through a small angle θ in one direction and released. The frequency of oscillation is (the springs are fixed to the walls)

Step-by-step solution

Consider the situation below From the above diagram θ = x l ⁡ / 2 ⇒ x = L 2 θ In the above diagram the rod is displaced through an angle θ as in above diagram Restoring torque = - 2 kx × L 2 = - kxL angular acceleration, ⇒ a = − k × L 2 θ × L Ι = − 6 k M θ = − ω 2 θ ω = 6 K M f = 1 2 π 6 K M