 Solution

At $\omega =\sqrt{3}{\omega }_0$, $s=3$, so the periodic amplitudes becomeThe 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 $T{y}_1/(2L)=-Th/L$, 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 $\sqrt{3}{\omega }_0$.

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.