= Solution
For the long filament, take the time-periodic solution of the dimensionless bending equation, after transients have decayed. Write $y=\operatorname{Re}[Y(x)e^{it}]$. 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 $Y=Ae^{\lambda_1x}+Be^{\lambda_2x}$. The driven displacement gives $A+B=y_0$, while the zero bending moment gives $\lambda_1^2A+\lambda_2^2B=0$. Since $\lambda_2^2=-\lambda_1^2$, $A=B=y_0/2$. The <oscillatory bending of a moment-free semi-infinite filament> is
$$
\boxed{y(x,t)=\frac{y_0}{2}\left[e^{-Cx}\cos(t+Sx)+e^{-Sx}\cos(t-Cx)\right].}
$$
Holding each wave phase constant gives its dimensionless <phase velocity>:
$$
\boxed{v_1=-\frac1S\simeq-2.6131,\qquad v_2=\frac1C\simeq1.0824.}
$$
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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-334-filament-waves.png]
{title=Two oppositely traveling damped bending waves}
{description=The two components have different attenuation lengths and opposite phase velocities. Their sum gives the long-filament response to a periodically moved, moment-free endpoint.}
Back to article page
