JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 8)
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 :
$$2\pi \sqrt {{R \over g}} $$
$${g \over {2\pi R}}$$
$${{2\pi R} \over g}$$
$${1 \over {2\pi }}\sqrt {{g \over R}} $$
Explanation
_26th_February_Morning_Shift_en_8_2.png)
Value of g on the particle of mass m,
$$g = {{GMd} \over {{R^3}}}$$
Force acting on the particle towards the center of the earth,
F = mg
Force along the tunnel = F cos$$\theta$$ = F1
= mg cos$$\theta$$
= m . $${{GMd} \over {{R^3}}}$$$$\left( {{x \over d}} \right)$$
= $${{GMm} \over {{R^3}}}$$ . x
= $${{{g_s}m} \over R}.\,x$$ [as gs = $${{GM} \over {{R^2}}}$$ at earth surface]
$$ \therefore $$ acceleration along the tunnel
$$\alpha$$ = $${{{F_1}} \over m}$$
= $${{{g_s}} \over R}.x$$
Time period = $${{2\pi {} } \over \omega } = 2\pi {\sqrt {{x \over \alpha }} } $$ [as $${\omega ^2} = {\alpha \over x}$$]
$$ = 2\pi {\sqrt {{R \over {{g_s}}}} } $$
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