JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Elasticity & SHM

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) $2 π \frac{R}{g}$ ✓ Correct
(B) $\frac{g}{2 π R}$
(C) $\frac{1}{2 π} \frac{g}{R}$
(D) $\frac{2 π R}{g}$

Answer: (A)

Solution

Inside the earth, at distance

r

from the centre, the gravitational field is

\frac{g r}{R}

pointing towards the centre. Measuring the displacement

x

along the tunnel from its mid-point, the component of this field along the tunnel is

\frac{g r}{R} \cdot \frac{x}{r} = \frac{g x}{R}

, while the perpendicular component is taken up by the frictionless wall. The motion along the tunnel is therefore governed by

a = - \frac{g}{R} x

which is simple harmonic with

ω = \frac{g}{R}

. The perpendicular distance of the chord from the centre cancels out, so the period is

T = 2 π \frac{R}{g}

for any chord.

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Elasticity & SHM

Solution

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