Let T_1 and T_2 be the time periods of two sp | vedclass

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(D) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.For spring $A$,$T_1 = 2\pi \sqrt{\frac{m}{k_1}} \implies k_1 = \

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

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(D) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.For spring $A$,$T_1 = 2\pi \sqrt{\frac{m}{k_1}} \implies k_1 = \frac{4\pi^2 m}{T_1^2}$.For spring $B$,$T_2 = 2\pi \sqrt{\frac{m}{k_2}} \implies k_2 = \frac{4\pi^2 m}{T_2^2}$.When connected in parallel,the effective spring constant is $k = k_1 + k_2$.The new time period $T$ is given by $T = 2\pi \sqrt{\frac{m}{k}} \implies k = \frac{4\pi^2 m}{T^2}$.Substituting the expressions for $k, k_1,$ and $k_2$ into the parallel combination formula:$\frac{4\pi^2 m}{T^2} = \frac{4\pi^2 m}{T_1^2} + \frac{4\pi^2 m}{T_2^2}$.Dividing both sides by $4\pi^2 m$,we get $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$.

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 the same mass $m$ is suspended with them. If $T$ is the new time period in this position,then:

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