Let T₁ and T₂ be time periods of two springs A and B when a mass m is suspended from t

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13.Oscillations

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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

$T = T_1+ T_2$

$T = \frac{T_1T_2}{T_1+ T_2}$

$T^2 = T_1^2 + T_2^2$

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

Solution

$\mathrm{T}_{1}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{1}}} \quad$ 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:

$\frac{1}{\mathrm{T}^{2}}=\frac{1}{\mathrm{T}_{1}^{2}}+\frac{1}{\mathrm{T}_{2}^{2}}$

Standard 11

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Solution

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