= Solution
With the stated parameters, the <mass> and restoring <matrices> are
$$
A=\begin{pmatrix}2&1\\1&1\end{pmatrix},\qquad
B=\begin{pmatrix}2&0\\0&1\end{pmatrix}.
$$
The <generalized eigenvalue problem> gives
$$
\det(B-\omega^2A)=(2-2\omega^2)(1-\omega^2)-\omega^4
=\omega^4-4\omega^2+2=0.
$$
Therefore the two normal <angular frequencies> are
$$
\boxed{\omega_{\mathrm{low}}=\sqrt{2-\sqrt2},\qquad
\omega_{\mathrm{high}}=\sqrt{2+\sqrt2}}.
$$
The second row of the modal equation is $-\omega^2x+(1-\omega^2)\phi=0$. It gives the <mode shapes>
$$
\boxed{\begin{aligned}
\omega_{\mathrm{low}}:&\quad (x,\phi)=a(1,\sqrt2)\cos(\omega_{\mathrm{low}}t+\delta),\\
\omega_{\mathrm{high}}:&\quad (x,\phi)=a(1,-\sqrt2)\cos(\omega_{\mathrm{high}}t+\delta).
\end{aligned}}
$$
Each <normal mode> has arbitrary small amplitude $a$ and phase $\delta$. The bob's horizontal displacement is $X=x+\phi$ to first order. Thus in the lower mode, bob and cart move in phase, with bob amplitude $1+\sqrt2$ times the cart amplitude. In the higher mode they move oppositely, because $X=(1-\sqrt2)x$; the bob amplitude is $\sqrt2-1$ times the cart amplitude. Its vertical displacement is quadratic in the angle and is absent at this order.
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