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(D) For a spring-mass system,the time period is given by $t = 2\pi \sqrt{\frac{m}{K}}$.For the individual springs,we have $t_1 = 2\pi \sqrt{\frac{m}{K

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$t^{-2} = t_1^{-2} + t_2^{-2}$

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(D) For a spring-mass system,the time period is given by $t = 2\pi \sqrt{\frac{m}{K}}$.For the individual springs,we have $t_1 = 2\pi \sqrt{\frac{m}{K_1}}$ and $t_2 = 2\pi \sqrt{\frac{m}{K_2}}$.Squaring these,we get $t_1^2 = 4\pi^2 \frac{m}{K_1} \implies K_1 = \frac{4\pi^2 m}{t_1^2}$ and $t_2^2 = 4\pi^2 \frac{m}{K_2} \implies K_2 = \frac{4\pi^2 m}{t_2^2}$.In the given figure,the springs are in parallel. The equivalent spring constant is $K_{eq} = K_1 + K_2$.The time period for the combined system is $t = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$.Squaring this,$t^2 = 4\pi^2 \frac{m}{K_1 + K_2} \implies \frac{1}{t^2} = \frac{K_1 + K_2}{4\pi^2 m} = \frac{K_1}{4\pi^2 m} + \frac{K_2}{4\pi^2 m}$.Substituting the expressions for $K_1$ and $K_2$,we get $\frac{1}{t^2} = \frac{1}{t_1^2} + \frac{1}{t_2^2}$,which can be written as $t^{-2} = t_1^{-2} + t_2^{-2}$.

$A$ mass $m$ is suspended separately by two different springs of spring constant $K_1$ and $K_2$ giving the time periods $t_1$ and $t_2$ respectively. If the same mass $m$ is connected by both springs as shown in the figure,then the time period $t$ is given by the relation:

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