A particle of mass m is attached to three springs A, B and C of equal force constant k [Fig. (8)]. If the particle

is pushed towards any one of the springs, then the time period of its oscillation will be

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$2 π \sqrt{(m / k)}$

$2 π \sqrt{(\frac{2 m}{k})}$

$2 π \sqrt{(m / 2 k)}$

$2 π \sqrt{m / 3 k}$

answer is C.

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Detailed Solution

Let the mass m is pushed down by a distance x. The total force will be
$F = kx + kxcos 60^{\circ} + kxcos 60^{\circ} = 2 kx m \frac{d^{2} x}{{dt}^{2}} = 2 kx or \frac{d^{2} x}{{dt}^{2}} = (\frac{2 k}{m}) x T = 2 π \sqrt{(\frac{m}{2 k})}$

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