JEE Main · PhysicsHard

$Two identical stars of mass M orbit their common center of mass in a circular path of radius R . A small planet of mass m is placed at the center of mass and given a small displacement x along the axis perpendicular to the orbital plane. The frequency of oscillation of the planet is:$

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Correct answer: B — $2 GM / R^{3}$

Solution

$The distance of each star from the planet at displacement x is d = R^{2} + x^{2} .$
$The net restoring force is F = - 2 \cdot (GM m / d^{2}) \cdot cos (θ), where cos (θ) = x / d .$
$For small x, F \approx - 2 GM m x / R^{3} .$
$The angular frequency is ω = 2 GM / R^{3} .$
$Hence the answer is (B).$