= Solution
In the negligible bead-mass limit the bead contribution to $\mathcal Q$ vanishes. Since the particle starts at rest, $\mathcal Q=0$, and hence $r^2\dot\theta+b\dot z=0$. Assume $b\ne0$ and a regular massless limit. Integrating along the trajectory gives
$$
\Delta z=-\frac1b\int_0^T r^2\dot\theta\,dt=-\frac1b\oint_C(x\,dy-y\,dx),
$$
where $C$ is the closed projection onto the plane perpendicular to the coil axis. The given final $r$ and $\theta$ ensure that this projection closes. <Green's theorem> identifies
$$
A=\frac12\oint_C(x\,dy-y\,dx)=\frac12\oint_C r^2\,d\theta,\qquad \boxed{\Delta z=-\frac{2A}{b}}.
$$
Thus \b[$A$ is the <signed area> enclosed by the transverse projection], positive for counterclockwise traversal viewed from positive $z$. Self-intersecting projections contribute their winding-weighted areas; it is neither an unsigned area nor an area of a surface in three dimensions. For $b>0$, a positive <signed area> gives negative axial displacement. This is <signed-area displacement under a helical constraint>: returning to the same projected coordinates can leave a nonzero axial displacement.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2016/paper-1-signed-area.png]
{title=Signed transverse area and the associated axial displacement for an illustrative closed projection}
The plotted loop illustrates the geometric integral only; it is not asserted to be a particular <spring> trajectory.
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