A solid uniform cylinder of mass m performs small oscillations in horizontal plane. If slightly displaced from its mean position shown in the figure. Initially springs are in natural lengths and cylinder does not slip on ground during oscillations due to friction between ground and cylinder. Force constant of each spring is k.The time period of oscillations are

A snapshot during oscillations of the cylinder is shown in the figure. Its displacement is x and velocity v, its angular velocity is $\omega =\frac{v}{R}$ (due to pure rolling)

As the centre of the cylinder is at a distance x from the initial position, the springs which are connected at a point on its rim must be compresses and stretched by a distance 2x.Thus, at the intermediate position total energy of the oscillating system can be given as

$$\begin{gathered} E_t=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{m{R}^2}{2}\right){\omega }^2+\left[\frac{1}{2}k{\left(2x\right)}^2\right]2 \\ =\frac{1}{2}m{v}^2+\frac{1}{4}m{v}^2+4k{x}^2 \\ =\frac{3}{4}m{v}^2+4k{x}^2 \\ =\frac{d{E}_t}{dt}=\frac{3}{4}m.2v.a+4k.2x.v=0 \left[E_{t}is\text{\hspace{0.17em}constant}\right] \\ \frac{3}{2}ma=-8k.x \\ \Rightarrow a=-\left(\frac{16x}{3m}\right)x \\ \omega =\sqrt{\frac{16k}{3m}} \\ \Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{3m}{16k}}=\frac{\pi }{2}\sqrt{\frac{3m}{k}} \end{gathered}$$