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Find the ratio of time periods of two identical springs if they are first joined in series $\&$ then in parallel $\&$ a mass $m$ is suspended from them :
$4$
$2$
$1$
$3$
Solution
$\mathrm{T}_{1}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{\mathrm{eq}}}}(\text { in series })$
$\frac{1}{\mathrm{k}_{\mathrm{eq}}}=\frac{1}{\mathrm{k}}+\frac{1}{\mathrm{k}}=\frac{2}{\mathrm{k}}$
$\therefore \quad k_{e q}=\frac{k}{2}$
$\therefore \quad \mathrm{T}_{1}=2 \pi \sqrt{\frac{2 \mathrm{m}}{\mathrm{k}}}$
$\mathrm{T}_{2}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}^{\prime}}}(\text { in parallel })$
But $k^{\prime}=k+k=2 k$
$\mathrm{T}_{2}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$
$\therefore \frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}=\frac{2 \pi \sqrt{\frac{2 \mathrm{m}}{\mathrm{k}}}}{2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}}=2$
