1. Set up the displacement

Let the disk’s centre move horizontally by a small displacement $x$ from equilibrium.

Initially, each spring has its natural length $L$, so at equilibrium there is no spring force.

When the disk shifts by $x$ to the right, both attachment points on the axle also shift by $x$ to the right. Since the wall ends are fixed, each spring changes length by $x$.

So force by each spring on the disk is restoring and equal to $F_s = -kx.$

Since there are two identical springs, $F_{\text{net}} = -2kx.$

2. Translational equation of motion

Let $f$ be the friction force at the point of contact with the ground.

Taking rightward as positive, $M\ddot x = -2kx + f.$

3. Rotational equation of motion

The springs are attached to the axle, so they exert no torque about the centre of the disk. Only friction provides torque.

For a thin uniform disk, $I_{\text{CM}} = \frac{1}{2}MR^2.$

If the disk rolls without slipping, then $\ddot x = -R\alpha$ (in magnitude, translational and rotational accelerations are related; sign handled below).

Torque about centre due to friction: $fR = I\alpha.$

Using $I = \frac12 MR^2$ and $\alpha = -\ddot x/R$, $fR = \frac12 MR^2\left(-\frac{\ddot x}{R}\right),$ so $f = -\frac12 M\ddot x.$

4. Substitute into translational equation

From $M\ddot x = -2kx + f,$ putting $f = -\frac12 M\ddot x$, $M\ddot x = -2kx - \frac12 M\ddot x.$

Thus, $\frac32 M\ddot x = -2kx.$

So, $\ddot x + \frac{4k}{3M}x = 0.$

This is SHM with $\omega = \sqrt{\frac{4k}{3M}}.$

5. Check options

The correct option is $\boxed{\sqrt{\frac{4k}{3M}}}$ which is Option D.

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So the derived answer agrees with the stored answer.