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(C) For a spring-mass system,the time period is given by $T = 2\pi \sqrt{\frac{m}{K}}$,where $K$ is the effective spring constant.In figure $(A)$,the

Vedclass · Accepted Answer

$A-Q, B-P, C-S, D-R$

Vedclass · Answer

(C) For a spring-mass system,the time period is given by $T = 2\pi \sqrt{\frac{m}{K}}$,where $K$ is the effective spring constant.In figure $(A)$,the springs are in parallel,so the effective spring constant is $K = k_1 + k_2$. Thus,$T = 2\pi \sqrt{\frac{m}{k_1 + k_2}}$. This matches $(Q)$.In figure $(B)$,the springs are in series,so $\frac{1}{K} = \frac{1}{k_1} + \frac{1}{k_2}$,which gives $K = \frac{k_1 k_2}{k_1 + k_2}$. Thus,$T = 2\pi \sqrt{\frac{m(k_1 + k_2)}{k_1 k_2}}$. This matches $(P)$.In figure $(C)$,the mass is between two springs. When the mass moves by $x$,one spring is compressed by $x$ and the other is stretched by $x$. The restoring force is $F = -(k+k)x = -2kx$. Thus,$K = 2k$,and $T = 2\pi \sqrt{\frac{m}{2k}}$. This matches $(S)$.In figure $(D)$,the springs are at $45^{\circ}$ to the vertical. If the mass is pulled down by $y$,each spring stretches by $y' = y \cos 45^{\circ}$. The restoring force component along the vertical is $F = -2(ky') \cos 45^{\circ} = -2k(y \cos 45^{\circ}) \cos 45^{\circ} = -2ky(1/2) = -ky$. Thus,$K = k$,and $T = 2\pi \sqrt{\frac{m}{k}}$. This matches $(R)$.Therefore,the correct matching is $A-Q, B-P, C-S, D-R$.

In the following questions,match Column-$I$ with Column-$II$ and choose the correct options.

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