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

The gravitational force on a particle of mass $m$ at a distance $x$ from the centre of the earth is given by $F = -\dfrac{GMm}{R^3} x$.

The acceleration of the particle is $a = -\dfrac{GM}{R^3} x = -\dfrac{g}{R} x$.

This shows that the particle executes simple harmonic motion (SHM) with angular frequency $\omega = \sqrt{\dfrac{g}{R}}$.

The velocity $v$ of a particle in SHM is related to its position $x$ and amplitude $A$ by the equation $v^2 = \omega^2 (A^2 - x^2)$.

Given that at the surface $x = R$, the velocity is $v = \sqrt{gR}$, we can find the amplitude $A$:

$gR = \dfrac{g}{R} (A^2 - R^2)$

$R^2 = A^2 - R^2 \Rightarrow A = \sqrt{2}R$

The position of the particle at any time $t$ can be written as $x = A \cos(\omega t + \phi)$.

At $t = 0$, the particle is at $x = R$ and moving towards the centre, so:

$R = \sqrt{2}R \cos\phi \Rightarrow \cos\phi = \dfrac{1}{\sqrt{2}} \Rightarrow \phi = \dfrac{\pi}{4}$

The particle reaches the other end of the tunnel when $x = -R$:

-R = \sqrt{2}R \cos \left(\omega t +\dfrac{\pi}{4}\right)

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

The smallest positive phase corresponding to this position is \dfrac{3\pi}{4}.

$\omega t +$\dfrac{\pi}{4} = \dfrac{3\pi}{4}

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

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