Normal modes of a spring-mounted pendulum cart ($mM{L}^2{\omega }^4-[kM{L}^2+(m+M)MgL]{\omega }^2+kMgL=0$)

A cart of mass $m$ moves horizontally on a spring of stiffness $k$, carrying a simple pendulum of bob mass $M$ and length $L$. Let $x$ be cart displacement and $\varphi$ the small angle from downward vertical. The quadratic Lagrangian isThe kinetic cross term comes from the bob's horizontal velocity $\dot{x}+L\dot{\varphi }$ to first order. The generalized eigenvalue problem has mass matrix $\left(\begin{array}{cc}m+M& ML\\ ML& M{L}^2\end{array}\right)$ and restoring matrix $\mathrm{diag}(k,MgL)$. Their characteristic determinant gives the displayed polynomial in ${\omega }^2$. For $m=M=L=g=1,k=2$, the squared angular frequencies are $2\mp \sqrt{2}$, with mode shapes $(x,\varphi )=(1,\pm \sqrt{2})$. The bob and cart move in phase in the lower normal mode and in opposite phases in the higher normal mode.

For a bob of mass $m$ on a massless rod of length $\ell$, let a massless support slide horizontally against a spring of stiffness $k$. With support displacement $s=\ell \eta$ and angle $\theta$ from downward vertical, the quadratic Lagrangian, divided by $m{\ell }^2$, is $(\dot{\eta }+\dot{\theta }{)}^2/2-(k/m){\eta }^2/2-(g/\ell ){\theta }^2/2$. Subtracting its two Euler-Lagrange equations gives $\eta =mg\theta /(k\ell )$. There is one finite normal mode, with the displayed frequency. A stiff spring fixes the support and recovers the simple pendulum frequency. The missing second finite frequency reflects a singular kinetic matrix in small oscillations, not an omitted oscillation branch.