= Solution
Reset $t=0$ when gravity is switched off. The subsequent <overdamped relaxation of a clamped--free filament> obeys $\zeta_\perp h_t=-Bh_{xxxx}$ with the same homogeneous endpoint conditions. Separate a <normal mode> as $h=\phi(x/L)e^{-\Gamma t}$. Writing $s=x/L$ gives
$$
\phi''''=\alpha^4\phi,\qquad \Gamma=\frac{B\alpha^4}{\zeta_\perp L^4}.
$$
The two clamp conditions reduce the general hyperbolic/trigonometric solution to
$$
\phi(s)=C[\cosh(\alpha s)-\cos(\alpha s)]
+D[\sinh(\alpha s)-\sin(\alpha s)].
$$
The two free-end conditions have matrix
$$
\begin{pmatrix}\cosh\alpha+\cos\alpha&\sinh\alpha+\sin\alpha\\
\sinh\alpha-\sin\alpha&\cosh\alpha+\cos\alpha\end{pmatrix}
\begin{pmatrix}C\\D\end{pmatrix}=0.
$$
Its <determinant> is $2(1+\cosh\alpha\cos\alpha)$. Hence the <clamped--free bending mode> values satisfy
$$
\cos\alpha_n\cosh\alpha_n=-1,\qquad
\alpha_1=1.8751040687\ldots,\quad\alpha_2=4.6940911330\ldots.
$$
There is no zero mode: at zero eigenvalue $\phi$ is cubic, and the four clamp/free conditions force all coefficients to vanish. With the following normalization a mode is
$$
\phi_n(s)=\cosh(\alpha_ns)-\cos(\alpha_ns)
-\eta_n[\sinh(\alpha_ns)-\sin(\alpha_ns)],\qquad
\eta_n=\frac{\cosh\alpha_n+\cos\alpha_n}{\sinh\alpha_n+\sin\alpha_n}.
$$
The fourth-order operator is <self-adjoint> and positive under these endpoint conditions, since $\int f g''''=\int f''g''$. Its orthogonal <eigenfunction expansion> gives
$$
h(x,t)=\sum_{n\ge1}c_n\phi_n(x/L)
\exp\left[-\frac{B\alpha_n^4t}{\zeta_\perp L^4}\right],\qquad
c_n=\frac{\int_0^Lh(x,0)\phi_n(x/L)dx}{\int_0^L\phi_n(x/L)^2dx}.
$$
For the previously attained steady profile, the first coefficient is nonzero: $h_s$ is positive and the first mode can be chosen positive in the interior. Two applications of <integration by parts>, using both functions' endpoint conditions, also give
$$
c_1=\frac{wL^4}{B\alpha_1^4}
\frac{\int_0^1\phi_1(s)ds}{\int_0^1\phi_1(s)^2ds}.
$$
Thus the dominant long-time answer, including its shape and amplitude, is
$$
\boxed{h(x,t)\sim c_1\phi_1(x/L)e^{-t/\tau_1},\qquad
\tau_1=\frac{\zeta_\perp L^4}{B\alpha_1^4}
=\frac{\zeta_\perp L^4}{12.36236337\ldots\,B}.}
$$
The next mode decays much faster. For other initial shapes with an accidentally zero first-mode projection, the lowest mode with $c_n\ne0$ determines the asymptotic behaviour. The decay time has the correct units and exhibits the strong $L^4$ dependence characteristic of bending opposed by local viscous drag.
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