The correct option is
C π2√3MkIn the situation given in the problem, the cylinder is in its equilibrium position when springs are unstrained. When it is slightly rolled and released, it starts executing
SHM and due to friction, the cylinder is in pure rolling motion. Now during oscillation, let us consider the centre of the cylinder is at a distance
x from the mean position and moving with a speed
v as shown in figure.
As the cylinder is in pure rolling, its angular speed of rotation can be given as
ω=vR.
The total energy of the system at this instant,
ET=12Mv2+12Icylω2+12k(2x)2×2
[springs are compressed and elongated by
x each]
=12Mv2+12(12MR2)(vR)2+12k(2x)2×2
[
Icyl=12MR2]
Differentiating with respect to time, we get
dETdt=12M(2vdvdt)+12(12MR2)1R2(2vdvdt)+4k(2xdxdt)=0
⇒Mva+12Mva+8kxv=0
⇒a=−163kxM
Comparing the equation with basic differential equation of
SHM,
We get, the angular frequency of
SHM as,
ω=√16k3M
Thus, the time period of this oscillation is
T=2πω=2π√3M16k=π2√3Mk