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Q.

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

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a

12π2kM

b

12π6kM

c

12π24kM

d

12πkM

answer is C.

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Detailed Solution

The rod will execute angular oscillations
Torque method:
Let rod be turned by small angle θ
Extension in each of the springs =  Lθ2
Force applied by each spring =  kLθ2
Torque on the system =  2kLθ2L2=kL2θ2
 Iα=kL2θ2
ML212α=kL2θ2 
 α=6kMθ
Angular 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=dω0dt
Energy of the oscillating system
E=12kx2+12kx2+12Iω02 , where x is the extension in each spring.
But for small extension, x=Lθ2 
So,  E=212kLθ22+12ML212ω02
Energy 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 
 

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