= Solution
At $\omega=\sqrt3\omega_0$, $s=3$, so the periodic amplitudes become
$$
\boxed{A_1=-2h_0,\qquad A_2=0}.
$$
The second bead is stationary in this particular motion, although the endpoint oscillates. The first bead moves at twice the endpoint amplitude and in opposite <phase>. This is an <antiresonance> of the second-bead response, not a natural <resonance> of the complete two-bead system.
The <force> balance makes the cancellation explicit. At $y_2=0$, the middle segment supplies transverse <force> $Ty_1/(2L)=-Th/L$, while the end segment supplies $Th/L$; these cancel on the second bead. The first bead then oscillates between two effectively fixed neighbours with restoring coefficient $3T/(2L)$, whose single-coordinate natural <angular frequency> is $\sqrt3\omega_0$.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015-resit-driven-bead-antiresonance.png]
{title=Periodic bead responses and the stationary second bead at antiresonance}
As elsewhere in the undamped problem, this exact nodal motion presumes initial conditions selecting the periodic particular response. Additional free <normal modes> would move the second bead as well; they do not decay automatically.
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