Assuming \(x_{2}>x_{1},\dot{x}_{2}>\dot{x}_{1},\ddot{x}_{2}>\ddot{x}_{1}\) and all are positive, the free body diagram for the cylinder and the cart is
Equation of motion for cylinder. \(\sum F_{x}\) \begin{equation} -2k\left ( x_{2}-x_{1}\right ) -F=m\ddot{x}_{2}\tag{1} \end{equation} And taking moment around C.G. of cylinder, using anti-clock wise as positive\begin{align*} -Fr & =-I_{cg}\alpha \\ Fr & =I_{cg}\alpha \end{align*}
Since we assumed no slip, then \(\left ( \ddot{x}_{2}-\ddot{x}_{1}\right ) =\alpha r\) and the above becomes\begin{align} Fr & =I_{cg}\frac{\left ( \ddot{x}_{2}-\ddot{x}_{1}\right ) }{r}\nonumber \\ F & =I_{cg}\frac{\left ( \ddot{x}_{2}-\ddot{x}_{1}\right ) }{r^{2}}\nonumber \\ & =\frac{1}{2}mr^{2}\frac{\left ( \ddot{x}_{2}-\ddot{x}_{1}\right ) }{r^{2}}\nonumber \\ & =\frac{1}{2}m\left ( \ddot{x}_{2}-\ddot{x}_{1}\right ) \tag{2} \end{align}
Using (2) in (1) gives EQM for \(x_{2}\)\begin{align} m\ddot{x}_{2}+2k\left ( x_{2}-x_{1}\right ) +\frac{1}{2}m\left ( \ddot{x}_{2}-\ddot{x}_{1}\right ) & =0\nonumber \\ \frac{3}{2}m\ddot{x}_{2}-\frac{1}{2}m\ddot{x}_{1}+2kx_{2}-2kx_{1} & =0\tag{3} \end{align}
For EQM for \(x_{1}\), resolving forces in \(x\) direction gives\[ -kx_{1}+2k\left ( x_{2}-x_{1}\right ) +F=2m\ddot{x}_{1}\] Using \(F\) found in (2) into the above gives\[ -kx_{1}+2k\left ( x_{2}-x_{1}\right ) +\frac{1}{2}m\left ( \ddot{x}_{2}-\ddot{x}_{1}\right ) =2m\ddot{x}_{1}\] Simplifying\begin{align} 2m\ddot{x}_{1}-\frac{1}{2}m\left ( \ddot{x}_{2}-\ddot{x}_{1}\right ) +kx_{1}-2k\left ( x_{2}-x_{1}\right ) & =0\nonumber \\ -\frac{1}{2}m\ddot{x}_{2}+\frac{5}{2}m\ddot{x}_{1}+3kx_{1}-2kx_{2} & =0\tag{4} \end{align}
Writing (3,4) in matrix form gives (note. Using (4) for top row and then use (3) for second row)\begin{equation} \begin{bmatrix} \frac{5}{2}m & -\frac{1}{2}m\\ -\frac{1}{2}m & \frac{3}{2}m \end{bmatrix}\begin{Bmatrix} \ddot{x}_{1}\\ \ddot{x}_{2}\end{Bmatrix} +\begin{bmatrix} 3k & -2k\\ -2k & 2k \end{bmatrix}\begin{Bmatrix} x_{1}\\ x_{2}\end{Bmatrix} =\begin{Bmatrix} 0\\ 0 \end{Bmatrix} \tag{5} \end{equation} If we had picked (3) for top row and then (4) for second row, the result will be\begin{equation} \begin{bmatrix} -\frac{1}{2}m & \frac{3}{2}m\\ \frac{5}{2}m & -\frac{1}{2}m \end{bmatrix}\begin{Bmatrix} \ddot{x}_{1}\\ \ddot{x}_{2}\end{Bmatrix} +\begin{bmatrix} -2k & 2k\\ 3k & -2k \end{bmatrix}\begin{Bmatrix} x_{1}\\ x_{2}\end{Bmatrix} =\begin{Bmatrix} 0\\ 0 \end{Bmatrix} \tag{6} \end{equation} Since So (5) and (6) are equivalent. To verify both (5) and (6) give the same eigenvalues, here is a check
