 Solution

In the uniform-Earth chord oscillator, neglect the mass removed by the narrow tunnel and any rotation of the Earth. Inside a uniform sphere, the shell theorem gives $\mathbf{F}=-(4\pi G\rho m/3)\mathbf{r}$. On the constrained chord $\mathbf{r}=(d,y)$, its along-tunnel component isThe transverse component is balanced by the tunnel constraint. With release from $y_s=\sqrt{R^2-d^2}$ at rest,The far end is $-y_s$, so $t_{\mathrm{transit}}=\pi /{\omega }_0\approx 2530\mathrm{s}=42.2\min$, independent of chord offset in this ideal model. Here $\rho =5515\mathrm{kg}{\mathrm{m}}^{-3}$ and $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ give ${\omega }_0=1.24171\times 10^{-3}{\mathrm{s}}^{-1}$.

With linear drag, $\ddot{y}+(b/m)\dot{y}+{\omega }_0^{2}y=0$. Inserting a complex exponential $e^{-(\gamma +i\omega )t}$ givesso in the underdamped regimeReal physical solutions combine conjugate modes. Release from rest specifically gives $y=y_s{e}^{-\gamma t}[\cos \omega t+(\gamma /\omega )\sin \omega t]$; a lone exponential with prefactor $y_s$ does not impose that initial velocity.

The critical gravity-tunnel settling time follows from critical damping, the fastest non-oscillatory release:Interpret the 1% tolerance relative to the initial displacement from equilibrium. With $s={\omega }_{0}t$, $(1+s)e^{-s}=0.01$ has $s=6.63835$. HenceThe equilibrium is reached only asymptotically; using a pure $e^{-{\omega }_{0}t}$ would miss the critical-damping polynomial factor and underestimate this time.