Frequency of Small Oscillations of Pivoted Rod with Springs

To find the frequency of small oscillations of the system, we analyze the restoring torque acting on the rod when it is displaced by a small angle $\theta$ from its equilibrium position.

1. Restoring Torque:
   The uniform rod of mass $M$ and length $L$ is pivoted at its center. Therefore, the distance from the pivot to each end of the rod is:
   $r = \frac{L}{2}$

   When the rod is displaced by a small angle $\theta$, the displacement of each end from its equilibrium position is:
   $x = r\theta = \frac{L}{2}\theta$

   Since the springs are attached to the opposite ends, one spring is extended by $x$ while the other is compressed by $x$. Each spring exerts a restoring force perpendicular to the rod given by:
   $F = kx = k\left(\frac{L}{2}\theta\right)$

   Both forces create restoring torques about the central pivot in the same rotational sense:
   $\tau = - 2 \left(F \times \frac{L}{2}\right) = - 2 \left(k\frac{L}{2}\theta \cdot \frac{L}{2}\right) = -\frac{kL^2}{2}\theta$

2. Moment of Inertia:
   The moment of inertia $I$ of a uniform rod of mass $M$ and length $L$ about an axis passing through its center of mass is:
   $I = \frac{ML^2}{12}$

3. Equation of Motion:
   Applying Newton's second law for rotation, $\tau = I \alpha = I \frac{d^2\theta}{dt^2}$:
   $\frac{ML^2}{12}\frac{d^2\theta}{dt^2} = -\frac{kL^2}{2}\theta$

   Dividing both sides by $L^2$:
   $\frac{M}{12}\frac{d^2\theta}{dt^2} + \frac{k}{2}\theta = 0$

   $\frac{d^2\theta}{dt^2} + \left(\frac{6k}{M}\right)\theta = 0$

4. Frequency of Oscillation:
   The angular frequency $\omega$ is:
   $\omega = \sqrt{\frac{6k}{M}}$

   Therefore, the frequency of oscillation $f$ is:
   $f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{6k}{M}}$

This corresponds to Option C.