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

Let the centre of the earth be the origin $(0,0)$. The tunnel is excavated along a chord at a perpendicular distance $\dfrac{R}{2}$ from the centre, so we can assume the tunnel lies along the line $y = \dfrac{R}{2}$.

The position vector of the particle at a distance $x$ from the centre of the tunnel is given by $\vec{r} = x \hat{i} + \dfrac{R}{2} \hat{j}$.

The gravitational force on a particle of mass $m$ inside the earth at a position $\vec{r}$ is directed towards the centre of the earth and is given by:

$\vec{F}_g = -\dfrac{GMm}{R^3} \vec{r}$

Substituting the position vector of the particle:

\vec{F}_g = -\dfrac{GMm}{R^3} \left( x \hat{i} +\dfrac{R}{2} \hat{j} \right)

The force has two components:
1. Along the tunnel (x-direction): $F_{\parallel} =$-\dfrac{GMmx}{R^3} \hat{i}, which provides the restoring force for oscillatory motion.
2. Perpendicular to the tunnel (y-direction): $F_{\perp} =$-\dfrac{GMm}{R^3} \left( \dfrac{R}{2} \right) \hat{j} = -\dfrac{GMm}{2R^2} \hat{j}.

Since the particle is constrained to move only along the tunnel, the normal force $N$ exerted by the frictionless wall must balance the perpendicular component of the gravitational force.

Therefore, the magnitude of the normal force is:

$N = |F_{\perp}| =$\dfrac{GMm}{2R^2}