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:

Consider a tunnel dug along a chord of the Earth at a perpendicular distance $$\frac{R}{2}$$ from the Earth's centre. Let a particle be at a distance $$x$$ from the midpoint of the tunnel along the tunnel.

The distance of the particle from the centre of the Earth is $$r = \sqrt{x^2 + \frac{R^2}{4}}$$. The gravitational force on the particle at distance $$r$$ from the centre (assuming uniform density) is directed towards the centre and has magnitude $$F = \frac{mg}{R}r$$.

The component of this force along the tunnel (towards the midpoint) is $$F_{\text{along}} = F \cdot \frac{x}{r} = \frac{mg}{R} \cdot r \cdot \frac{x}{r} = \frac{mg}{R} x$$.

This is a restoring force proportional to the displacement $$x$$, confirming simple harmonic motion. The equation of motion is $$m\ddot{x} = -\frac{mg}{R}x$$, giving $$\ddot{x} = -\frac{g}{R}x$$.

The angular frequency is $$\omega = \sqrt{\frac{g}{R}}$$, and the time period is $$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{R}{g}}$$.

Note that this time period is independent of the perpendicular distance of the chord from the centre, and is the same as for a tunnel through the centre of the Earth.

Hence the correct answer is Option 2: $$2\pi\sqrt{\frac{R}{g}}$$.