Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance $(R / 2)$ from the earth's centre, where ' $R$ ' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period :
Correct Option: 1
(1)
$\cos \theta=\frac{x}{d}$
If displaced from equilibrium position,
$F_{\text {restoring }}=\left(\frac{\mathrm{GMmd}}{R^{3}}\right) \cos \theta$
$\mathrm{F}_{\text {Res. }}=\frac{\mathrm{GMmd}}{\mathrm{R}^{3}} \cdot \frac{\mathrm{x}}{\mathrm{d}}=\frac{\mathrm{GMmx}}{\mathrm{R}^{3}}$
$a_{R}=\frac{G M x}{R^{3}} \quad G M_{e}=g R^{2}$
$\mathrm{T}=2 \pi \sqrt{\mid \frac{\mathrm{x}}{\mathrm{a}}}$
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{X}}{\mathrm{GMx}}} \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{gR}^{2}}}$
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$