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Three identical massless springs each of force constant $k$ are attached to a block of mass $M$ with two massless pulleys as shown. The time period of vertical oscillations of the block is

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$2 π \sqrt{\frac{M}{6 k}}$

$2 π \sqrt{\frac{3 M}{k}}$

$2 π \sqrt{\frac{6 M}{k}}$

$2 π \sqrt{\frac{M}{k}}$

answer is A.

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Detailed Solution

The equivalent system is shown in figure.

The equivalent force constant

$\frac{1}{k_{e}} = \frac{1}{k} + \frac{1}{k / 4} + \frac{1}{k}$

or $k_{e} = k / 6$

Thus $T = 2 π \sqrt{\frac{M}{k / 6}} = 2 π \sqrt{\frac{6 M}{k}}$

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