 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 areTheir determinant is $(3-s{)}^2-1=(2-s)(4-s)$, giving the signed displacement amplitudesA negative sign denotes a phase reversal relative to the prescribed cosine; physical amplitude magnitudes are $|A_j|$. In the quasistatic limit,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 $\sqrt{2}{\omega }_0$ 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.