A uniform thin cylindrical disc 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 Fig. 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 horizontal plane. The unstretched length of each spring is `L`. The disc is initially at its equilibrium position with its centre of mass (`CM`) at a distance `L` from the wall. The disc rolls without slipping with velocity `V_(0)=V_(0)hati`. The coefficient of friction is `mu`.

The net external force acting on the disc when its centre of mass is at displacement `x` with respect to its equilibrium position is

Options

A. `-kx`
B. `-2kx`
C. `-(2kx)/(3)`
D. `-(4kx)/(3)`

Answer

D

Explanation

`f-2kx=Ma`,`fR=-Iprop` implies `fR=-(1//2)MR^(2)(a//R)`
Solve to get net force =`Ma=-(4kx)/(3)`
Negative sign is because this force will act in the backward direction.