1. Gravitational field inside the Earth

Assuming Earth has uniform density, the gravitational field at a distance $r$ from the centre is proportional to $r$:

$g_r = g\frac{r}{R}$

and it is directed toward the centre.

So, for a particle at position vector $\vec r$ inside Earth, acceleration is

$\vec a = -\frac{g}{R}\vec r$

This is the key SHM relation inside a uniform Earth.

2. Motion along the chord

A tunnel is dug along a chord at perpendicular distance $R/2$ from the centre.

Let $x$ be the displacement of the particle along the chord measured from the midpoint of the chord.

At any point in the tunnel, the particle is at distance

$r = \sqrt{x^2 + \left(\frac{R}{2}\right)^2}$

from the centre.

But we do not need the full magnitude. We only need the component of gravitational acceleration along the tunnel.

Since

$\vec a = -\frac{g}{R}\vec r$

the component along the chord is simply proportional to the coordinate $x$ along that chord:

$a_x = -\frac{g}{R}x$

because the perpendicular part is balanced by the tunnel wall's normal reaction, and only the along-tunnel component causes motion.

Thus,

$\frac{d^2x}{dt^2} = -\frac{g}{R}x$

This is the standard equation of SHM.

3. Identify angular frequency

Comparing with

$\frac{d^2x}{dt^2} = -\omega^2 x$

we get

$\omega^2 = \frac{g}{R}$

so

$\omega = \sqrt{\frac{g}{R}}$

4. Time period

The time period of SHM is

$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{R}{g}}$

5. Match with options

$T = 2\pi \sqrt{\frac{R}{g}}$

This matches Option A.

6. Comparison with stored answer

Stored correct answer: A

Derived answer: A

They agree.