JEE Mains · Physics · STD 11 - 13. oscillations
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
- A \(\frac{2 \pi R }{ g }\)
- B \(\frac{ g }{2 \pi R }\)
- C \(\frac{1}{2 \pi} \sqrt{\frac{g}{R}}\)
- D \(2 \pi \sqrt{\frac{ R }{ g }}\)
Answer & Solution
Correct Answer
(D) \(2 \pi \sqrt{\frac{ R }{ g }}\)
Step-by-step Solution
Detailed explanation
Force along the tunnel \(F =-\left(\frac{ GMmr }{ R ^{3}}\right) \cos \theta\) \(F =-\frac{ gm }{ R } x \left(\frac{ GM }{ R ^{2}}= g , r \cos \theta= x \right)\) \(a=-\frac{g}{R} x\) \(\omega^{2}=\frac{g}{R} \quad T=2 \pi \sqrt{\frac{R}{g}}\)
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