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(B) For Fig. $1$,the time period of oscillation for a single spring-mass system is given by:$T_a = 2\pi \sqrt{\frac{M}{k}}$For Fig. $2$,the two spring

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$2$

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(B) For Fig. $1$,the time period of oscillation for a single spring-mass system is given by:$T_a = 2\pi \sqrt{\frac{M}{k}}$For Fig. $2$,the two springs are connected in series. The equivalent spring constant $k_{eq}$ for two springs in series is given by:$\frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} \Rightarrow k_{eq} = \frac{k}{2}$The time period of oscillation for the series combination is:$T_b = 2\pi \sqrt{\frac{M}{k_{eq}}} = 2\pi \sqrt{\frac{M}{k/2}} = 2\pi \sqrt{\frac{2M}{k}}$Now,find the ratio $\frac{T_b}{T_a}$:$\frac{T_b}{T_a} = \frac{2\pi \sqrt{\frac{2M}{k}}}{2\pi \sqrt{\frac{M}{k}}} = \sqrt{\frac{2M/k}{M/k}} = \sqrt{2}$Given $\frac{T_b}{T_a} = \sqrt{x}$,we have $\sqrt{x} = \sqrt{2}$,which implies $x = 2$.

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