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For the long filament, take the time-periodic solution of the dimensionless bending equation, after transients have decayed. Write $y=\mathrm{Re}[Y(x)e^{it}]$. Then $Y^{(4)}=-iY$, so the spatial exponents satisfy ${\lambda }^4=-i$. Define $C=\cos (\pi /8)$ and $S=\sin (\pi /8)$. The two roots with negative real part are ${\lambda }_1=-C+iS$ and ${\lambda }_2=-S-iC$; the other roots violate decay at infinity.

Hence $Y=A{e}^{{\lambda }_{1}x}+B{e}^{{\lambda }_{2}x}$. The driven displacement gives $A+B=y_0$, while the zero bending moment gives ${\lambda }_1^{2}A+{\lambda }_2^{2}B=0$. Since ${\lambda }_2^2=-{\lambda }_1^2$, $A=B=y_0/2$. The oscillatory bending of a moment-free semi-infinite filament isHolding each wave phase constant gives its dimensionless phase velocity:The first travels toward the actuator and attenuates over length $1/C$; the second travels away and attenuates over the longer length $1/S$. These are spatially damped phase patterns in an overdamped bending equation, not two undamped inertial beam waves. Their combination satisfies both displacement and zero-moment conditions at the driven end.