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(A) The time period of a particle of mass $m$ attached to a spring of constant $k$ is given by $T = 2\pi \sqrt{\frac{m}{k}}$.For the first spring,$T_1

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$T=\sqrt{T_1^2+T_2^2}$

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(A) The time period of a particle of mass $m$ attached to a spring of constant $k$ is given by $T = 2\pi \sqrt{\frac{m}{k}}$.For the first spring,$T_1 = 2\pi \sqrt{\frac{m}{k_1}}$,so $T_1^2 = 4\pi^2 \frac{m}{k_1}$.For the second spring,$T_2 = 2\pi \sqrt{\frac{m}{k_2}}$,so $T_2^2 = 4\pi^2 \frac{m}{k_2}$.When two springs are connected in series,the effective spring constant $k$ is given by $\frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2}$.The time period $T$ for the series combination is $T = 2\pi \sqrt{\frac{m}{k}}$.Squaring both sides,$T^2 = 4\pi^2 \frac{m}{k} = 4\pi^2 m \left( \frac{1}{k_1} + \frac{1}{k_2} \right)$.Substituting the expressions for $T_1^2$ and $T_2^2$,we get $T^2 = 4\pi^2 \frac{m}{k_1} + 4\pi^2 \frac{m}{k_2} = T_1^2 + T_2^2$.Therefore,$T = \sqrt{T_1^2 + T_2^2}$.

$A$ particle connected to the end of a spring executes $S$.$H$.$M$. with period $T_1$. While the corresponding period for another spring is $T_2$. If the period of oscillation with two springs in series is $T$,then

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