A system shown in the figure consists of a massless pulley, a spring of force constant $k$ and a block of mass $m$ . If the block is slightly displaced vertically downwards from its equilibrium position and released, then the period of vertical oscillations is

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In this situation, if mass m moves downwards a distance $x$ from the equilibrium position, the pulley will also move by x and so, the spring will stretch by $2 x$ .

Therefore, the spring force will be $2 k x$ . The restoring force on the block will be $4 k x$ .

Hence, $F = - 4 k x$

$\Rightarrow m g - 4 k x = m a$

$\Rightarrow m a = - 4 k x + m g$

$\Rightarrow$ $a = - \frac{4 k x}{m}$ $+ g \Rightarrow ω^{2} = | \frac{a}{x} |$

$T = 2 π \sqrt{| \frac{x}{a} |}$

$T = 2 π \sqrt{(\frac{m}{4 k})}$

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