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(B) The period of a mass $m$ attached to a spring of constant $k$ is given by $t = 2 \pi \sqrt{\frac{m}{k}}$.Thus,$t_1 = 2 \pi \sqrt{\frac{m}{k_1}}$ a

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$T^2 = t_1^2 + t_2^2$

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(B) The period of a mass $m$ attached to a spring of constant $k$ is given by $t = 2 \pi \sqrt{\frac{m}{k}}$.Thus,$t_1 = 2 \pi \sqrt{\frac{m}{k_1}}$ and $t_2 = 2 \pi \sqrt{\frac{m}{k_2}}$.Squaring these,we get $t_1^2 = 4 \pi^2 \frac{m}{k_1}$ and $t_2^2 = 4 \pi^2 \frac{m}{k_2}$,which implies $\frac{1}{k_1} = \frac{t_1^2}{4 \pi^2 m}$ and $\frac{1}{k_2} = \frac{t_2^2}{4 \pi^2 m}$.When two springs are connected in series,the effective spring constant $k_{eff}$ is given by $\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$.The period of the system in series is $T = 2 \pi \sqrt{\frac{m}{k_{eff}}}$,so $T^2 = 4 \pi^2 \frac{m}{k_{eff}}$.Substituting the expressions for $\frac{1}{k_1}$ and $\frac{1}{k_2}$ into the series formula:$\frac{T^2}{4 \pi^2 m} = \frac{t_1^2}{4 \pi^2 m} + \frac{t_2^2}{4 \pi^2 m}$.Multiplying by $4 \pi^2 m$,we obtain $T^2 = t_1^2 + t_2^2$.

$A$ particle at the end of a spring executes simple harmonic motion with a period $t_1$,while the corresponding period for another spring is $t_2$. If the period of oscillation with the two springs in series is $T$,then

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