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Potential energy of a particle of mass 2kg is given as $\frac{x^{3}}{6} - 2 x$ . Initially particle is at equilibrium position, then time period for small oscillation about stable equilibrium position is

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$\frac{4 π}{3} \sec$

$\frac{π}{2} \sec$

$2 π \sec$

$\frac{π}{3} \sec$

answer is A.

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Detailed Solution

At stable equilibrium, the force acting on the body is given by
$d F = - 2 d x$
$K = 2$
$T = 2 π \sqrt{\frac{m}{K}} = 2 π \sqrt{\frac{2}{2}} = 2 π \sec$

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