In a spring-block system as shown in the figu | vedclass

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(C) The two springs on the left are in parallel,so their equivalent spring constant is $K_p = K + K = 2K$.This combination is in series with the third

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$1.414$

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(C) The two springs on the left are in parallel,so their equivalent spring constant is $K_p = K + K = 2K$.This combination is in series with the third spring of constant $K$. The equivalent spring constant $K_{eq}$ is given by:$\frac{1}{K_{eq}} = \frac{1}{2K} + \frac{1}{K} = \frac{1+2}{2K} = \frac{3}{2K}$$K_{eq} = \frac{2K}{3} = \frac{2 	imes 9 \pi^2}{3} = 6 \pi^2 \ Nm^{-1}$.The mass of the block is $m = 3 \ kg$.The time period of oscillation is $T = 2 \pi \sqrt{\frac{m}{K_{eq}}}$.Substituting the values: $T = 2 \pi \sqrt{\frac{3}{6 \pi^2}} = 2 \pi \sqrt{\frac{1}{2 \pi^2}} = 2 \pi 	imes \frac{1}{\pi \sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \ s$.$T = 1.414 \ s$.

In a spring-block system as shown in the figure,if the spring constant $K = 9 \pi^2 \ Nm^{-1}$,then the time period of oscillation is (in $s$)

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