When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.

$$\therefore -2kx+f=-M_{ac}$$ ..... (i)

Where $$a_c$$ = acceleration of centre of mass.

The torque acting on the disc about its centre of mass C is

$$\tau = f \times R = I \times {a_c}$$

$$\therefore$$ $$f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} \times {{{a_c}} \over R}$$

[$$\therefore$$ $$I = {1 \over 2}M{R^2},{a_c} = R{a_c}$$ for rolling without slipping]

$$\therefore$$ $$f = {1 \over 2}M{a_c}$$ ...... (ii)

From eq. (i) and (ii),

$$ \Rightarrow - 2kx + f = - 2f$$

$$ \Rightarrow f = {{2k} \over 3} \times x$$

Frictional force depends on x. As x increase friction increase. Frictional force (f) is maximum at $$x = A$$.

Where A = amplitude of S.H.M

Maximum frictional force,

$${f_{\max }} = {{2k} \over 3} \times A$$

The force should be almost equal to the limiting friction ($$\mu mg$$) for rolling without slipping.

$$\therefore$$ $$\mu mg = {{2k} \over 3} \times A$$ ..... (iv)

For S.H.M

Velocity amplitude $$ = A\omega $$

$$\therefore$$ $${V_0} = A\omega $$

$$\therefore$$ $${V_0} = {{3m\,mg} \over {2k}}w$$ (from (iv))

$$\therefore$$ $${V_0} = {{3\mu mg} \over {2k}} \times \sqrt {{{4k} \over {3M}}} $$ (As $$ \omega = \sqrt {{{4k} \over {3M}}} $$)

$$\therefore$$ $${V_0} = mg\sqrt {{{3M} \over k}} $$