Let T_1 and T_2 be the time periods of two&nb | vedclass

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Question

$\mathrm{T}_{1}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{1}}} \quad$ or  $\mathrm{k}_{1}=\frac{4 \pi^{2} \mathrm{m}}{\mathrm{T}_{1}^{2}}$

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Vedclass · Accepted Answer

$\frac{1}{T^2} =\frac{1}{T_1^2}+\frac{1}{T_2^2}$

Vedclass · Answer

$\mathrm{T}_{1}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{1}}} \quad$ or  $\mathrm{k}_{1}=\frac{4 \pi^{2} \mathrm{m}}{\mathrm{T}_{1}^{2}}$

${\mathrm{T}_{2}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{2}}}}  {	ext { or } \quad \mathrm{k}_{2}=\frac{4 \pi^{2} \mathrm{m}}{\mathrm{T}_{2}^{2}}}$

${\mathrm{Now} \quad \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{2}}}}  {	ext { or } \quad \mathrm{k}=\frac{4 \pi^{2} \mathrm{m}}{\mathrm{T}_{2}^{2}}}$

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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