Simple Harmonic (1).

Consider the origin to be placed at the bottom left of the system. The location of the fixed end of the left spring is $(0,R)$ and that of the right spring is $(L,2R)$ where L is the separation between the two fixed ends. Let the natural length of the first spring be $L_1$ and that of the second spring on the right be $L_2$ . At equilibrium:

$L = L_1 + L_2$

Consider a general time $t$ where the disc rotates counterclockwise by an angle $\theta$ where theta. Let the coordinates of the COM of the disk be $(x,R)$ . Therefore, the coordinates of the point of attachment of the second spring with the disk are:

$x_2= x + R\sin{\theta}$ $y_2= R + R\cos{\theta}$

Also, since the disk purely rolls:

$\dot{x} = R\dot{\theta}$

Integrating:

$x = R\theta +C$

When $\theta=0$ , then the system is in equilibrium and therefore $x = L_1$ . Then:

$x= R\theta +L_1$

Now, the kinetic energy of the purely rolling disk is:

$\mathcal{T} = \frac{m}{2}\dot{x}^2 + \frac{1}{2}\left(\frac{mR^2}{2}\right)\dot{\theta}^2 = \frac{3mR^2\dot{\theta}^2}{4}$

The potential energy of the system is:

$\mathcal{V} = \frac{k_1}{2} (x-L_1)^2 + \frac{k_2}{2}\left(\sqrt{(L-x_2)^2 + (2R-y_2)^2} - L_2\right)^2$

Plugging in all expressions and simplifying gives:

$\mathcal{V} = \frac{k_1R^2\theta^2}{2}+\frac{k_2}{2}\left(\sqrt{(L_2 -R\theta - R\sin{\theta})^2 + (R- R\cos{\theta})^2} - L_2\right)^2$

Now, if we were to consider small oscillations, one must consider the case when $\theta$ is small. By applying this approximation:

$\sin{\theta} \approx \theta$ $\cos{\theta} \approx 1$

Plugging these expressions in the expressions for potential energy and simplifying gives:

$\mathcal{V} = \frac{k_1R^2\theta^2}{2} + \frac{4k_2R^2\theta^2}{2}$

Now, the total energy of this system is conserved. This means:

$E = T + V$

Since energy is conserved:

$\frac{dE}{dt} = 0 = \left(\frac{3mR^2\ddot{\theta}}{2} + (k_1 + 4k_2)R^2\theta\right)\dot{\theta}$

Sine $\dot{\theta} \ne 0$ , then:

$\frac{3mR^2\ddot{\theta}}{2} + (k_1 + 4k_2)R^2\theta=0$

$\therefore \omega = \frac{2(k_1+4k_2)}{3m}$