The motion of a particle inside a smooth straight tunnel through the Earth is Simple Harmonic Motion (SHM). The force is given by $F = -kr$, and the time period $T = 2\pi\sqrt{R/g}$ is independent of the chord length.

We can analyze the time taken by looking at the Phase Angle ($\phi$) covered on the reference circle:

• Equilibrium Point (E): For any straight line segment, the point closest to the Earth’s center is the mean position where speed is maximum. The time to travel from the Surface (Rest) to E is always $t = T/4$, corresponding to a phase of $\pi/2$.

• $T_A$ (Standard): The particle travels from surface to surface. The midpoint is the equilibrium point ($v_{max}$). The total phase covered is $\pi$ (corresponding to time $T/2$).
• $T_C$ (Equivalent to A): The tunnel is bent such that the vertex corresponds exactly to the equilibrium point of the path segments (velocity is maximum at the bend). Thus, the particle completes exactly the same phase as A.
  $\Rightarrow T_C = T_A$
• $T_B$ (Faster): The tunnel is bent inwards such that the particle reaches the vertex before reaching the virtual equilibrium point of the line. It turns back “early” in the phase cycle.
  $\Rightarrow \text{Phase} < \pi \implies T_B < T_A$
• $T_D$ (Slower): The tunnel is bent outwards/shallowly. The particle passes the virtual equilibrium point and continues (slowing down) before reaching the bend. It travels “extra” phase.
  $\Rightarrow \text{Phase} > \pi \implies T_D > T_A$