All the springs in figures (a), (b), and (c) | vedclass

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

Vedclass · Accepted Answer

$T_b = 2T_c$

Vedclass · Answer

(D) The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{K_{eq}}}$, where $K_{eq}$ is the equivalent spring constant.For figure $(a)$: The system has a single spring with constant $K$. Thus, $K_{eq,a} = K$. The time period is $T_a = 2\pi \sqrt{\frac{m}{K}}$.For figure $(b)$: Two springs are connected in series. The equivalent spring constant is given by $\frac{1}{K_{eq,b}} = \frac{1}{K} + \frac{1}{K} = \frac{2}{K}$, so $K_{eq,b} = \frac{K}{2}$. The time period is $T_b = 2\pi \sqrt{\frac{m}{K/2}} = 2\pi \sqrt{\frac{2m}{K}} = \sqrt{2} T_a$.For figure $(c)$: Two springs are connected in parallel. The equivalent spring constant is $K_{eq,c} = K + K = 2K$. The time period is $T_c = 2\pi \sqrt{\frac{m}{2K}} = \frac{1}{\sqrt{2}} (2\pi \sqrt{\frac{m}{K}}) = \frac{T_a}{\sqrt{2}}$.Comparing the results: $T_a = \sqrt{2} T_c$, which implies $T_c = \frac{T_a}{\sqrt{2}}$. Also, $T_b = \sqrt{2} T_a$. From $T_c = \frac{T_a}{\sqrt{2}}$ and $T_b = \sqrt{2} T_a$, we get $T_b = 2 T_c$. Thus, option $(d)$ is correct.

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 periodic times of oscillation of the three systems in figures $(a)$, $(b)$, and $(c)$ respectively, then:

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