Suppose a tunnel is excavated along a chord of the earth at a...

Let the centre of the earth be $O$ and the centre of the tunnel be $C$. The distance $OC = R/2$.

When the particle is at a distance $x$ from $C$, its distance from the centre of the earth is $r = \sqrt{x^2 + (R/2)^2}$.

The gravitational force acting on the particle of mass $m$ at a distance $r$ from the centre of the earth is directed towards $O$ and is given by:

$F = \dfrac{GMmr}{R^3}$

The component of this force along the tunnel acts as the restoring force. From the geometry of the right-angled triangle formed by $O, C$, and the particle, the component of the force along the tunnel is:

$F_x = -F \cos \theta = -F \left(\dfrac{x}{r}\right)<br><br>Substituting the value of$F$, we get:<br><br>$F_x = -\left(\dfrac{GMmr}{R^3}\right) \left(\dfrac{x}{r}\right) = -\dfrac{GMm}{R^3}$x$

Since the restoring force is proportional to the displacement $x$, the particle executes simple harmonic motion with an effective force constant $k =$\dfrac{GMm}{R^3}.

The time period of the simple harmonic motion is:

$T = 2\pi$\sqrt{\dfrac{m}{k}} ==2\pi \sqrt{\dfrac{R^3}{GM}}