= Solution
The original diagram has three horizontal spans $L$, $2L$ and $L$. In the small-slope approximation the massless string segments are straight and their transverse <force> components are <tension> times slope. The horizontal <tension> components cancel to first order.
For the first bead, the left segment contributes $-Ty_1/L$ and the middle segment contributes $T(y_2-y_1)/(2L)$. For the second bead, the middle contribution is $T(y_1-y_2)/(2L)$ and the driven right segment contributes $T(h-y_2)/L$. <Newton's second law> therefore gives
$$
M\ddot y_1=\frac{T}{2L}(y_2-3y_1),\qquad M\ddot y_2=\frac{T}{2L}(y_1-3y_2+2h).
$$
With $\omega_0^2=T/(2ML)$ and $h=h_0\cos\omega t$, the coupled <equations of motion> are
$$
\boxed{\begin{aligned}
\ddot y_1+3\omega_0^2y_1-\omega_0^2y_2&=0,\\
\ddot y_2+3\omega_0^2y_2-\omega_0^2y_1&=2\omega_0^2h_0\cos\omega t.
\end{aligned}}
$$
The factor two in the driving term follows from the right span having length $L$, while the middle span has length $2L$. Treating all three spans as equal would change both the coupling and the <normal mode> <frequencies>.
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