Let $T_1$ and $T_2$ be the time periods of two springs $A$ and $B$ when a mass $m$ is suspended from them separately. Now both the springs are connected in parallel and same mass $m$ is suspended with them. Now let $T$ be the time period in this position. Then
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$\mathrm{T}_{1}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{1}}} \quad$ or $\mathrm{k}_{1}=\frac{4 \pi^{2} \mathrm{m}}{\mathrm{T}_{1}^{2}}$
${\mathrm{T}_{2}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{2}}}} {\text { or } \quad \mathrm{k}_{2}=\frac{4 \pi^{2} \mathrm{m}}{\mathrm{T}_{2}^{2}}}$
${\mathrm{Now} \quad \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{2}}}} {\text { or } \quad \mathrm{k}=\frac{4 \pi^{2} \mathrm{m}}{\mathrm{T}_{2}^{2}}}$
In parallel $\mathrm{k}=\mathrm{k}_{1}+\mathrm{k}_{2}$
Substituting the values of $\mathrm{k}, \mathrm{k}_{1}$ and $\mathrm{k}_{2}$ we get:
$\frac{1}{\mathrm{T}^{2}}=\frac{1}{\mathrm{T}_{1}^{2}}+\frac{1}{\mathrm{T}_{2}^{2}}$
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