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  constants k. The springs are fixed to rigid supports as  shown in the figure and the rod is free to oscillate in the  horizontal plane. The rod is gently pushed through a small  angle  θ in one direction and released. The frequency of oscillation is

detailed solution

Correct option is C

The rod will execute angular oscillationsTorque method:Let rod be turned by small angle θ. Extension in each of the springs =  Lθ2Force applied by each spring =  kLθ2Torque on the system =  −2kLθ2L2=kL2θ2 Iα=−kL2θ2⇒ML212α=−kL2θ2  α=−6kMθAngular acceleration is proportional, and directed opposite, to angular  displacement. Hence SHM.⇒f=12παθ=12π6kM Energy method:Let θ be angular displacement at a given time, and ω0 be the angular velocity of the  rod at that time. So, ω0=dθdt   and    α=dω0dtEnergy of the oscillating systemE=12kx2+12kx2+12Iω02 , where x is the extension in each spring.But for small extension, x=Lθ2 So,  E=212kLθ22+12ML212ω02Energy is constant. So, dEdt=0  ⇒kL242θdθdt+12ML2122ω0dω0dt=0    ⇒k2θ+M12α=0    ⇒α=−6kMθ Angular acceleration is proportional, and directed opposite, to angular   displacement. Hence SHM.⇒f=12παθ=12π6kM