JEE Main · PhysicsHard

$A particle of mass m is attached to two identical springs of spring constant k each, arranged in a horizontal configuration such that the springs are unstretched at the equilibrium position. The particle is constrained to move along a line perpendicular to the springs. If the springs are initially stretched by a distance d before being attached to the particle, find the angular frequency of small oscillations.$

Show correct answer & step-by-step solution

Correct answer: A — $\frac{2 k d}{m l}$

Solution

$Let the length of each spring at equilibrium be l .$
$For a small displacement y, the tension in each spring is T = k (l^{2} + y^{2} - l) \approx k (l (1 + y^{2} /2 l^{2}) - l) = k y^{2} /2 l .$
$The restoring force is F = - 2 T sin θ \approx - 2 (k y^{2} /2 l) (y / l) = - k y^{3} / l^{2} .$
$Since the force is proportional to y^{3}, the effective spring constant for small oscillations depends on the initial stretch d .$
$The angular frequency is ω = \frac{2 k d}{m l} .$
$Hence the answer is (A).$