Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance $\frac{R}{2}$ from the earth's centre, where $R$ is the radius of the earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period:
Answer: (B) $2\pi\sqrt{\frac{R}{g}}$
Inside the earth, gravity at distance $r$ from the centre is $\dfrac{GMm}{R^3}r = \dfrac{mg}{R}r$, directed towards the centre.
Let $x$ be the distance of the particle from the midpoint of the chord. The component of gravity along the tunnel is
$$F = -\frac{mg}{R}r\cdot\frac{x}{r} = -\frac{mg}{R}x$$
This is SHM with $\omega^2 = g/R$, so $T = 2\pi\sqrt{\dfrac{R}{g}}$, independent of the distance of the chord from the centre.
Solution by Sreeraj P, M.Sc Physics