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(B) For two identical springs with spring constant $k$,when joined in series,the equivalent spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \

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$2$

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(B) For two identical springs with spring constant $k$,when joined in series,the equivalent spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}$,so $k_{eq} = \frac{k}{2}$.The time period in series is $T_{1} = 2 \pi \sqrt{\frac{m}{k_{eq}}} = 2 \pi \sqrt{\frac{2m}{k}}$.When joined in parallel,the equivalent spring constant $k'$ is $k' = k + k = 2k$.The time period in parallel is $T_{2} = 2 \pi \sqrt{\frac{m}{k'}} = 2 \pi \sqrt{\frac{m}{2k}}$.The ratio of the time periods is $\frac{T_{1}}{T_{2}} = \frac{2 \pi \sqrt{\frac{2m}{k}}}{2 \pi \sqrt{\frac{m}{2k}}} = \sqrt{\frac{2m}{k} \cdot \frac{2k}{m}} = \sqrt{4} = 2$.

Find the ratio of time periods of two identical springs if they are first joined in series and then in parallel and a mass $m$ is suspended from them.

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