Suppose a tunnel is excavated across the earth (radius = R )...

The acceleration of a particle at a distance $x$ from the centre of the earth inside a tunnel is given by $a = -\dfrac{g}{R}x$.

This represents simple harmonic motion (SHM) with an angular frequency $\omega = \sqrt{\dfrac{g}{R}}$.

The general equation for the position of a particle in SHM is:

$x(t) = x_0 \cos(\omega t) + \dfrac{v_0}{\omega} \sin(\omega t)$

Let the particle be thrown from one end of the tunnel, so its initial position is $x_0 = -R$. It is thrown along the length of the tunnel with an initial speed $v_0 = \sqrt{gR}$.

Substituting these initial conditions into the equation of motion:

$x(t) = -R \cos(\omega t) + \dfrac{\sqrt{gR}}{\sqrt{g/R}} \sin(\omega t)$

$x(t) = -R \cos(\omega t) + R \sin(\omega t)$

To traverse the entire length of the tunnel, the particle must reach the other end, which corresponds to $x(t) = R$. Substituting this into the equation:

$R = -R \cos(\omega t) + R \sin(\omega t)$

$\sin(\omega t) - \cos(\omega t) = 1$

Multiplying and dividing by $\sqrt{2}$:

$\sqrt{2} \left( \sin(\omega t) \dfrac{1}{\sqrt{2}} - \cos(\omega t) \dfrac{1}{\sqrt{2}} \right) = 1$

$\sqrt{2} \sin\left(\omega t - \dfrac{\pi}{4}\right) = 1$

$\sin\left(\omega t - \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$

The smallest positive angle satisfying this is:

$\omega t - \dfrac{\pi}{4} = \dfrac{\pi}{4}$

$\omega t = \dfrac{\pi}{2}$

Substituting $\omega = \sqrt{\dfrac{g}{R}}$:

$t = \dfrac{\pi}{2} \sqrt{\dfrac{R}{g}}$

Answer: $\dfrac{\pi}{2} \sqrt{\dfrac{R}{g}}$