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(D) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{K_{eq}}}$.For figure $(a)$: The effective spring constant is $K_{eq,a

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$T_{b}=2 \,T_{c}$

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(D) The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{K_{eq}}}$.For figure $(a)$: The effective spring constant is $K_{eq,a} = K$. Thus,$T_a = 2\pi \sqrt{\frac{m}{K}}$.For figure $(b)$: The two springs are in series. The effective spring constant is $\frac{1}{K_{eq,b}} = \frac{1}{K} + \frac{1}{K} = \frac{2}{K}$,so $K_{eq,b} = \frac{K}{2}$. Thus,$T_b = 2\pi \sqrt{\frac{m}{K/2}} = 2\pi \sqrt{\frac{2m}{K}} = \sqrt{2} T_a$.For figure $(c)$: The two springs are in parallel. The effective spring constant is $K_{eq,c} = K + K = 2K$. Thus,$T_c = 2\pi \sqrt{\frac{m}{2K}} = \frac{1}{\sqrt{2}} (2\pi \sqrt{\frac{m}{K}}) = \frac{T_a}{\sqrt{2}}$.Comparing the results: $T_a = \sqrt{2} T_c$,which can be written as $T_a = \frac{T_c}{1/\sqrt{2}}$ or $T_c = \frac{T_a}{\sqrt{2}}$.Looking at the options,$T_a = \frac{T_c}{1/\sqrt{2}}$ is not directly listed,but $T_a = \frac{T_c}{\sqrt{2}}$ is option $(b)$. Let's re-verify: $T_c = T_a / \sqrt{2} \implies T_a = \sqrt{2} T_c$. Wait,$T_a = T_c / (1/\sqrt{2}) = \sqrt{2} T_c$. Option $(b)$ states $T_a = \frac{T_c}{\sqrt{2}}$,which is $T_c = \sqrt{2} T_a$. This is incorrect. Let's check $T_b = \sqrt{2} T_a$. Option $(a)$ is $T_a = \sqrt{2} T_b$,which is incorrect. Let's re-evaluate: $T_b = \sqrt{2} T_a$ and $T_c = T_a / \sqrt{2}$. Therefore,$T_b = 2 T_c$. This matches option $(d)$.

All the springs in figures $(a)$,$(b)$,and $(c)$ are identical,each having a force constant $K$. $A$ mass $m$ is attached to each system. If $T_a, T_b$,and $T_c$ are the time periods of oscillations of the three systems in figures $(a)$,$(b)$,and $(c)$ respectively,then:

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