End-driven two-bead string (${\omega }_0^2=T/(2ML)$)

Place two equal masses $M$ on a massless mechanical string with spans $L,2L,L$ and uniform tension $T$. Fix one end and drive the other with displacement $h(t)$. Linear transverse force balance gives ${\ddot{y}}_1+{\omega }_0^2(3y_1-y_2)=0$ and ${\ddot{y}}_2+{\omega }_0^2(3y_2-y_1)=2{\omega }_0^{2}h$. The mass-normalized stiffness matrix has symmetric and antisymmetric normal modes with squared angular frequencies $2{\omega }_0^2$ and $4{\omega }_0^2$. The mode coordinates $(y_1+y_2)/2$ and $(y_1-y_2)/2$ separate the forced equations.

For a cosine drive of amplitude $h_0$, the periodic response of an end-driven two-bead string has signed amplitudes $A_1=2h_0/[(2-s)(4-s)]$ and $A_2=(3-s)A_1$, where $s={\omega }^2/{\omega }_0^2$. At $s=3$, $A_2=0$ while $A_1=-2h_0$. Equal and opposite transverse forces from the driven end and moving first bead cancel at the second bead. This antiresonance differs from a natural resonance: it is a zero of one measured response between the two natural frequencies. Free normal modes remain unless initial conditions exclude them.