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 :
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 :
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Inside the earth the force is \(\frac{mg}{R}r\) towards the centre; its component along the chord is proportional to the displacement from the chord's midpoint.
Updated On: Nov 4, 2026
- \(2\pi\sqrt{\frac{R}{g}}\)
- \(\frac{g}{2\pi R}\)
- \(\frac{2\pi R}{g}\)
- \(\frac{1}{2\pi}\sqrt{\frac{g}{R}}\)
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The Correct Option is A
Solution and Explanation
Inside the earth, the gravitational force on mass \(m\) at distance \(r\) from the centre is \(F=\frac{mg}{R}r\), directed towards the centre.
Let the tunnel be at perpendicular distance \(d=\frac R2\) from the centre and let \(x\) be the displacement from the chord midpoint. The component of \(F\) along the tunnel is \(\frac{mg}{R}r\cdot\frac{x}{r}=\frac{mg}{R}x\), directed towards the midpoint (restoring).
So \(a=-\frac gR x\), \(\omega^2=\frac gR\) and \(T=2\pi\sqrt{\frac Rg}\) (independent of \(d\)).
Answer: (1) \(2\pi\sqrt{\frac{R}{g}}\)
Let the tunnel be at perpendicular distance \(d=\frac R2\) from the centre and let \(x\) be the displacement from the chord midpoint. The component of \(F\) along the tunnel is \(\frac{mg}{R}r\cdot\frac{x}{r}=\frac{mg}{R}x\), directed towards the midpoint (restoring).
So \(a=-\frac gR x\), \(\omega^2=\frac gR\) and \(T=2\pi\sqrt{\frac Rg}\) (independent of \(d\)).
Answer: (1) \(2\pi\sqrt{\frac{R}{g}}\)
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