In figure the spring has a force constant k . The pulley is light and smooth, the spring and the string are light. The suspended block has a mass m kg . If the block is slightly displaced vertically down from its equilibrium position and released then the period of its vertical oscillations.

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TestHub · Accepted Answer

The constant force mg only shifts the mean position, so ignore it for the time period.

If the block goes down by x, the spring stretches by x/2, so the string tension is k(x/2)/2 each side: restoring force $\frac{k x}{4}$ .

$a = \frac{k}{4 m} x \Rightarrow ω^{2} = \frac{k}{4 m} \Rightarrow T = 2 π \frac{4 m}{k} = 4 π \frac{m}{k}$ so (B) is correct.

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