Correct Answer - D
(d) When the disc is at a
disctance x from the mean position (equilibrium
position), the forces
acting on the disc are
shown in the figure
`:. -2kx + f =- Ma_c. …(1) Mean where `a_c = acceleration of center positon`
of mass Also the torque acting
on the disc about its center of
mass C is `pi = fxxR = I xx alpha_c`
`:. f= (Ialpha)/(R ) = ((1)/(2)MR^2)/(R ) xx(a_c)/(R )`
`[:. I = (1)/(2)MR^2 and a_c = Ralpha_c for rolling without slipping ]`
`:.f=(1)/(2)Ma_c ...(ii)`
From (i) &(ii)`
`-2kx+(1)/(2)Ma_c = -Ma_c`
`rArr (3)/(2)Ma_c = 2kx rArr Ma_c = (4kx)/(3)`
`rArr` Net external force acting on the disce when its centre of
mass is at displacement x with respect to the equilibrium
position `=(4kx)/(3)` directed towards the euqilibrium.