# Assume that a tunnel is dug along a chord of the earth,

Question:

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 :

1. (1) $2 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$

2. (2) $\frac{1}{2 \pi} \sqrt{\frac{g}{R}}$

3. (3) $\frac{2 \pi \mathrm{R}}{\mathrm{g}}$

4. (4) $\frac{g}{2 \pi R}$

Correct Option: 1

Solution:

(1)

$\cos \theta=\frac{x}{d}$

If displaced from equilibrium position,

$F_{\text {restoring }}=\left(\frac{\mathrm{GMmd}}{R^{3}}\right) \cos \theta$

$\mathrm{F}_{\text {Res. }}=\frac{\mathrm{GMmd}}{\mathrm{R}^{3}} \cdot \frac{\mathrm{x}}{\mathrm{d}}=\frac{\mathrm{GMmx}}{\mathrm{R}^{3}}$

$a_{R}=\frac{G M x}{R^{3}} \quad G M_{e}=g R^{2}$

$\mathrm{T}=2 \pi \sqrt{\mid \frac{\mathrm{x}}{\mathrm{a}}}$

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{X}}{\mathrm{GMx}}} \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{gR}^{2}}}$

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$

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