Just hanging around , Medium part2

Let $y_1(t)$ and $y_2(t)$ be distances of pulley and mass respectively from their initial positions (wheh both springs are not deformed). We say that $y_1(t)$ (and $y_2(t)$ too) is greater then zero iff it is below its initial position. We can easily observe that the extension of spring $k$ is equal $y_1$ and for $2k$ it is $y_2-y_1$ . If we denote by $N_1$ tension in above string and $N_2$ is the second one, we get from second dynamics law: $mg+N_2-2N_1=m\ddot{y}_1,$ $mg-N_2=m\ddot{y}_2.$ It is easy to notice, that due to assumed signs there is $N_1=2ky_1$ and $N_2=2k(y_2-y_1)$ . So we have a system of differential equations. Observe that we don't need to solve it but only to find angular velocity $\omega$ . The angular velocity doesn't depend on constant components, therefore we can consider the following system $N_2-2N_1=m\ddot{y}_1,$ $-N_2=m\ddot{y}_2$ and therefore $2k(y_2-y_1)-4kx_1=m\ddot{y}_1,$ $-2k(y_2-y_1)=m\ddot{y}_2.$ In general the motion is a superposition of simple harmonic motions (normal modes) of both bodies. That's why we look for solutions of the form $y_1=Ae^{i\alpha t}$ and $y_2=Be^{i\alpha t}$ (to be honest we are not sure that such a motion exists but we can try it and indeed there will be a general solution). Moreover we substitute $\omega^2=\frac{k}{m}$ and we have $2\omega^2(B-A)-4\omega^2A=-\alpha^2 A,$ $-2\omega^2(B-A)=-\alpha^2B.$ After writing it in a matrix form and computing the determinant of $2\times2$ matrix which must be zero (for nonzero A,B) we have $\alpha^4-8\alpha^2\omega^2+8\omega^4=0$ so $\alpha^2=(4\pm2\sqrt{2})\omega^2.$ We observe that the motion is a linear superposition of $e^{i\alpha}$ for all $\alpha$ but since we have SHO, there is only one $\alpha^2$ allowed. Now it easy to conclude (due to additional conditions) that there must be $\alpha=\omega\sqrt{4-2\sqrt{2}}$ what (with $\alpha=2\pi\frac{1}{T}$ ) lead us to $T=2\pi\sqrt{\frac{1+\sqrt{2}}{2\sqrt{2}}\frac{m}{k}}$ and for $a=1+\sqrt{2}$ , $b=2\sqrt{2}$ we get $a^2+b^2=11+2\sqrt{2}$ which is a solution.