1. Set up the motion variable

Let $x$ be the horizontal displacement of the disk’s centre from the equilibrium position.

Since the springs are attached symmetrically to the axle and the disk remains oriented with axle horizontal, each spring changes length by $x$. So total restoring force on the disk is $F_s=-2kx.$

Thus the translational restoring force is like an SHO with effective spring constant $2k$.

2. Include rolling constraint

The disk rolls without slipping, so translational and rotational motions are coupled.

For translation: $M a = -2kx + f$ where $f$ is the friction force.

For rotation about the centre: $fR=I\,\alpha$

For rolling without slipping: $a=-\alpha R$ (signs imply friction provides the needed torque).

For a uniform disk, $I=\frac12 MR^2.$

Hence $f = -\frac{I}{R^2}a = -\frac12 Ma$ (up to sign convention; in magnitude, $|f|=\frac12 M|a|$).

Substitute into translation equation in consistent magnitude form using effective inertia: $\left(M+\frac{I}{R^2}\right)a=-2kx.$ So $\left(M+\frac12 M\right)a=-2kx$ $\frac32 M a=-2kx$ $a=-\frac{4k}{3M}x.$

Therefore the motion is SHM with $\omega^2=\frac{4k}{3M}.$

3. Use the initial condition to get amplitude

Initially the disk is at equilibrium ($x=0$) and given speed $V_0$. So the SHM amplitude is $A=\frac{V_0}{\omega}.$

Hence maximum acceleration is $a_{\max}=\omega^2A=\omega V_0.$

Using $|f|=\frac12 M|a|$, the maximum required static friction is $f_{\max,req}=\frac12 M a_{\max}=\frac12 M\omega V_0.$

For no slipping, $f_{\max,req}\le \mu Mg.$ So $\frac12 M\omega V_0\le \mu Mg$ $V_0\le \frac{2\mu g}{\omega}.$

Now $\omega=\sqrt{\frac{4k}{3M}}=2\sqrt{\frac{k}{3M}}.$ Therefore