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A frictionless tunnel is bored along a chord of the Earth at a perpendicular distance R/2 from the Earth's centre (R = Earth's radius). A particle released in the tunnel undergoes simple harmonic motion. What is its time period? (g = surface gravity)

1. 2*pi*sqrt(R/g)
2. 2*pi*R/g
3. (1/2*pi)*sqrt(g/R)
4. 2*pi*sqrt(2R/g)

Correct answer: 2*pi*sqrt(R/g)

Solution

Inside a uniform Earth, gravity at distance r from the centre is g*r/R directed toward the centre. Projecting this force along the tunnel gives a restoring force proportional to displacement with omega² = g/R, independent of the chord's offset. Thus the period is T = 2*pi*sqrt(R/g), the same as a diametric tunnel.

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