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(C) The time period of a spring-mass system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.For the two springs individually,we have $t_1 = 2 \pi \sqrt{\fr

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$\sqrt{t_1^2 + t_2^2}$

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(C) The time period of a spring-mass system is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.For the two springs individually,we have $t_1 = 2 \pi \sqrt{\frac{m}{k_1}}$ and $t_2 = 2 \pi \sqrt{\frac{m}{k_2}}$.Squaring both sides,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 $\frac{1}{k_2} = \frac{t_2^2}{4 \pi^2 m}$.For a series combination of springs,the equivalent spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$.The time period $t$ for the series combination is $t = 2 \pi \sqrt{\frac{m}{k_{eq}}}$,so $t^2 = 4 \pi^2 \frac{m}{k_{eq}}$.Substituting the expression for $\frac{1}{k_{eq}}$,we get $t^2 = 4 \pi^2 m \left( \frac{1}{k_1} + \frac{1}{k_2} \right) = 4 \pi^2 m \left( \frac{t_1^2}{4 \pi^2 m} + \frac{t_2^2}{4 \pi^2 m} \right) = t_1^2 + t_2^2$.Therefore,$t = \sqrt{t_1^2 + t_2^2}$.

When a block of mass $m$ is suspended separately by two different springs having time periods $t_1$ and $t_2$,respectively. If the same mass is connected to the series combination of both springs,then its time period is given by:

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