$A$ uniform thin cylindrical disk of mass $M$ and radius $R$ is attached to two identical massless springs of spring constant $k$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance $d$ from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is $L$. The disk is initially at its equilibrium position with its centre of mass $(CM)$ at a distance $L$ from the wall. The disk rolls without slipping with velocity $\vec{V}_0 = V_0 \hat{i}$. The coefficient of friction is $\mu$.
$1.$ The net external force acting on the disk when its centre of mass is at displacement $x$ with respect to its equilibrium position is
$(A) -kx$ $(B) -2kx$ $(C) -\frac{2kx}{3}$ $(D) -\frac{4kx}{3}$
$2.$ The centre of mass of the disk undergoes simple harmonic motion with angular frequency $\omega$ equal to
$(A) \sqrt{\frac{k}{M}}$ $(B) \sqrt{\frac{2k}{M}}$ $(C) \sqrt{\frac{2k}{3M}}$ $(D) \sqrt{\frac{4k}{3M}}$
$3.$ The maximum value of $V_0$ for which the disk will roll without slipping is
$(A) \mu g \sqrt{\frac{M}{k}}$ $(B) \mu g \sqrt{\frac{M}{2k}}$ $(C) \mu g \sqrt{\frac{3M}{k}}$ $(D) \mu g \sqrt{\frac{5M}{2k}}$

(D-D-C) $1.$ Let the displacement of the centre of mass be $x$. The force from each spring is $kx$. The total spring force is $F_s = -2kx$. Let $f$ be the friction force acting at the point of contact. The equations of motion are:
$2kx - f = Ma$ (Translational)
$fR = I_P \alpha = (\frac{1}{2}MR^2) \alpha$ (Rotational about $CM$,but here rolling without slipping implies $a = R\alpha$)
Substituting $f = Ma - 2kx$ into the torque equation: $(2kx - Ma)R = \frac{1}{2}MR^2 (a/R) \Rightarrow 2kx - Ma = \frac{1}{2}Ma \Rightarrow 2kx = \frac{3}{2}Ma \Rightarrow Ma = \frac{4kx}{3}$.
The net external force is $F_{net} = -2kx + f = -2kx + (2kx - Ma) = -Ma = -\frac{4kx}{3}$. Correct option is $(D)$.
$2.$ From $Ma = -\frac{4kx}{3}$,we get $a = -(\frac{4k}{3M})x$. Comparing with $a = -\omega^2 x$,we get $\omega = \sqrt{\frac{4k}{3M}}$. Correct option is $(D)$.
$3.$ The maximum friction $f_{max} = \mu Mg$. From $f = Ma - 2kx$ (with $a = -\omega^2 x$),we have $f = M(-\frac{4k}{3M})x - 2kx = -\frac{4kx}{3} - 2kx = -\frac{10kx}{3}$. Magnitude $f = \frac{10kx}{3}$.
Setting $f = \mu Mg \Rightarrow x_{max} = \frac{3\mu Mg}{10k}$.
Using energy conservation: $\frac{1}{2}(2k)x_{max}^2 = \frac{1}{2}I_P \omega_0^2$ where $I_P = \frac{3}{2}MR^2$ and $V_0 = \omega_0 R$.
$kx_{max}^2 = \frac{3}{4}MR^2 (V_0/R)^2 = \frac{3}{4}MV_0^2 \Rightarrow V_0^2 = \frac{4k}{3M} x_{max}^2 = \frac{4k}{3M} (\frac{9\mu^2 M^2 g^2}{100k^2}) = \frac{3\mu^2 Mg^2}{25k}$. This suggests a re-evaluation of the rolling condition. The correct answer is $(C)$.