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

Let the mass of the Earth be $M$ and its radius be $R$. The particle is released from a height $R$ above the surface, so its initial distance from the center of the Earth is $2R$.

Using the principle of conservation of mechanical energy from the point of release to the surface of the Earth:

$E_i = E_f$

$-\dfrac{GMm}{2R} = -\dfrac{GMm}{R} + \dfrac{1}{2}mv^2$

$\dfrac{1}{2}mv^2 = \dfrac{GMm}{2R}$

$v = \sqrt{\dfrac{GM}{R}} = \sqrt{gR}$

This is the velocity of the particle as it enters the tunnel. Inside the Earth, the gravitational force at a distance $x$ from the center is $F = -\dfrac{GMm}{R^3}x$.

The acceleration is $a = -\dfrac{g}{R}x$, which shows that the particle executes Simple Harmonic Motion (SHM) with angular frequency $\omega = \sqrt{\dfrac{g}{R}}$.

Using the velocity-position relation for SHM, $v^2 = \omega^2(A^2 - x^2)$, we can find the amplitude $A$ of the motion. At the surface ($x = R$), the velocity is $v = \sqrt{gR}$:

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

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

The time taken to traverse the tunnel is the time to travel from $x = R$ to $x = -R$. Using the equation of SHM $x = A\cos(\omega t + \phi)$:

At the entrance ($t = 0$), $x = R$ and the particle is moving towards the center:

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

At the exit, $x = -R$:

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

The next phase where the cosine is $-\dfrac{1}{\sqrt{2}}$ is $\dfrac{3\pi}{4}$:

$\omega t + \dfrac{\pi}{4} = \dfrac{3\pi}{4} \Rightarrow \omega t = \dfrac{\pi}{2}$

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

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