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Reset $t=0$ when gravity is switched off. The subsequent overdamped relaxation of a clamped--free filament obeys ${\zeta }_{\perp }{h}_t=-B{h}_{xxxx}$ with the same homogeneous endpoint conditions. Separate a normal mode as $h=\varphi (x/L)e^{-\Gamma t}$. Writing $s=x/L$ givesThe two clamp conditions reduce the general hyperbolic/trigonometric solution toThe two free-end conditions have matrixIts determinant is $2(1+\cosh \alpha \cos \alpha )$. Hence the clamped--free bending mode values satisfyThere is no zero mode: at zero eigenvalue $\varphi$ is cubic, and the four clamp/free conditions force all coefficients to vanish. With the following normalization a mode isThe fourth-order operator is self-adjoint and positive under these endpoint conditions, since $\int f{g}^{\prime \prime \prime \prime }=\int f^{\prime \prime }{g}^{\prime \prime }$. Its orthogonal eigenfunction expansion givesFor 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 giveThus the dominant long-time answer, including its shape and amplitude, isThe next mode decays much faster. For other initial shapes with an accidentally zero first-mode projection, the lowest mode with $c_n\ne 0$ 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.