Rods with Springs!

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Dotted lines represent equilibrium config. Solid lines represent instantenous config.

Let the rods be rotated by a small angle $\theta$ so that they make an angle $45° - \theta$ with $AC$ It is easy to see that in this process the springs at A and C would be compressed by $a= L(\cos(45°-\theta) - \cos 45°)$ and the springs at B and D would be extended by $b= L(\sin 45° - \sin(45°-\theta))$ . At this instant potential energy of the system could be written as

$PE = (2 * \frac{1}{2}Ka^2) + (2 * \frac{1}{2}Kb^2)$

Plugging values of $a$ and $b$ and simplifying we get

$PE = 2KL^2(1-\cos \theta)$

At this instant the rods are in pure rotation about their respective Instantenous centre of Rotation (IC)

Moment of inertia of each rod about respective IC

$I = \frac{ML^2}{12} + M(\frac{L}{2})^2 = \frac{ML^2}{3}$ (parallel axis theorem)

The Kinetic Energy of the system at this instant would be

$KE = 4* 0.5 * I\omega^2$

Total Energy $TE = PE + KE = const.$

$TE = 2KL^2(1-\cos \theta) + 2*I\omega^2 = const.$

Differentiating wrt. Time we get $2KL^2 (\sin\theta)\dot{\theta} + \frac{4ML^2}{3}\omega\dot{\omega} = 0$

As $\omega = \dot{\theta} and \dot{\omega} = \ddot{\theta}$ on simplifying we get

$KL^2 \sin\theta + (2/3)ML^2 \ddot{\theta} = 0$

On further simplification and using $\sin\theta \approx \theta$ we get

$\ddot{\theta} = -\frac{3K}{2M} \theta$

So Angular frequency = $\sqrt{\frac{3K}{2M}}$