Step-by-step Solution:

1. Identify the type of motion: The rod is pivoted at its center and subjected to restoring forces from the springs when displaced. This will result in rotational oscillations. For a small angular displacement θ, the motion is Angular Simple Harmonic Motion (SHM).

2. Equation of Angular SHM: The equation of motion for angular SHM is given by τ = Iα, where τ is the net restoring torque, I is the moment of inertia, and α is the angular acceleration (α = d²θ/dt²). For SHM, the torque is proportional to the angular displacement: τ = -Cθ, where C is the torsional constant. The angular frequency ω is given by ω² = C/I.

3. Calculate the Moment of Inertia (I): The object is a uniform rod of mass M and length L, pivoted at its center. The moment of inertia of a uniform rod about an axis passing through its center and perpendicular to its length is: $I = {1 \over {12}}ML^2$

4. Calculate the Restoring Torque (τ): Let the rod be displaced by a small angle θ from its equilibrium position in the horizontal plane.

   • The pivot is at the center, so each end is at a distance L/2 from the pivot.
   • For a small angle θ, the horizontal displacement x of each end of the rod is approximately the arc length: x ≈ (L/2)θ.
   • One spring gets compressed by this distance x, and the other gets stretched by the same distance x.
   • The restoring force exerted by each spring is F = kx = k(L/2)θ.
   • These forces are applied at the ends of the rod, at a distance L/2 from the pivot. For small angles, the forces are nearly perpendicular to the rod.
   • The torque produced by each spring is $τ_spring = Force × lever arm$. The lever arm is L/2.
   • $τ_spring = F × (L/2) = (k(L/2)θ) × (L/2) = k(L²/4)θ$.
   • Both forces create a torque that tends to restore the rod to its equilibrium position. So, they act in the same rotational direction. The total restoring torque τ is the sum of the torques from both springs. $τ = - (τ_{spring1} + τ_{spring2}) = - (k{L^2 \over 4}θ + k{L^2 \over 4}θ) = - {kL^2 \over 2}θ$ The negative sign indicates that the torque is a restoring torque, acting opposite to the angular displacement θ.

5. Determine the Angular Frequency (ω): Now, we use the equation of motion Iα = τ: ${1 \over {12}}ML^2 α = - {kL^2 \over 2}θ$ The angular acceleration is α: $α = - \left( {{kL^2/2} \over {ML^2/12}} \right)θ = - \left( {{12kL^2} \over {2ML^2}} \right)θ = - \left( {{6k} \over M} \right)θ$ Comparing this with the standard SHM equation α = -ω²θ, we get: $ω^2 = {{6k} \over M} \implies ω = \sqrt{{{6k} \over M}}$

6. Calculate the Frequency of Oscillation (f): The frequency f is related to the angular frequency ω by f = ω / (2π). $f = {1 \over {2π}}ω = {1 \over {2π}} \sqrt{{{6k} \over M}}$

7. Conclusion: The frequency of oscillation matches option C.

Option Analysis:

• A: ${1 \over {2 \pi }}\sqrt {{{2k} \over M}}$ - This would be the frequency if the motion was linear SHM of mass M against two springs in parallel, which is incorrect as the motion is rotational.
• B: ${1 \over {2 \pi }}\sqrt {{k \over M}}$ - Incorrect. Does not correspond to any simple physical model of this system.
• C: ${1 \over {2 \pi }}\sqrt {{{6k} \over M}}$ - This matches our derived result.
• D: ${1 \over {2 \pi }}\sqrt {{{24k} \over M}}$ - This would result from a calculation error, for example, if the moment of inertia was incorrectly taken or the lever arm was mistaken.

The correct option is C.