= Solution
Away from the natural <resonances>, seek a periodic particular solution $y_j=A_j\cos\omega t$. Let $s=\omega^2/\omega_0^2$. The amplitude equations are
$$
(3-s)A_1-A_2=0,\qquad-A_1+(3-s)A_2=2h_0.
$$
Their determinant is $(3-s)^2-1=(2-s)(4-s)$, giving the signed displacement amplitudes
$$
\boxed{A_1=\frac{2h_0}{(2-s)(4-s)},\qquad A_2=\frac{2(3-s)h_0}{(2-s)(4-s)}}.
$$
A negative sign denotes a <phase> reversal relative to the prescribed cosine; physical amplitude magnitudes are $|A_j|$. In the quasistatic limit,
$$
\boxed{A_1\longrightarrow h_0/4,\qquad A_2\longrightarrow3h_0/4}.
$$
These are the values of a straight string interpolating between endpoint displacements $0$ and $h_0$ at bead positions $L$ and $3L$ along a total length $4L$.
Because damping is explicitly absent, this periodic response is not a unique attracting late-time state. Arbitrary homogeneous oscillations at $\sqrt2\omega_0$ and $2\omega_0$ persist unless the initial conditions eliminate them. At either natural <resonance>, the drive produces secular growth rather than a finite bounded periodic response. Thus “steady state” here refers to the selected particular solution away from the poles.
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