Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance $(\frac{R}{2})$ 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:

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$2 π \sqrt{\frac{R}{g}}$

$\frac{2 π R}{g}$

$\frac{g}{2 π R}$

$\frac{1}{2 π} \sqrt{\frac{g}{R}}$

answer is A.

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Detailed Solution

$At a displacement ’y’ from’O’$

$\begin{array}{l} \Rightarrow m a = - m (\frac{G M r}{R^{3}} \sin θ) \\ \Rightarrow m a = - \frac{G M m}{R^{3}} y \\ \Rightarrow a = - (\frac{G M}{R^{3}}) y = - (\frac{g}{R}) y \\ Comparing with a = - ω^{2} y \\ \Rightarrow ω = \sqrt{\frac{g}{R}} \\ \Rightarrow T = \frac{2 π}{ω} = 2 π \sqrt{\frac{R}{g}} \end{array}$

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