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(A) First,analyze the spring arrangement. There is one spring of constant $k$ in parallel with a combination of two springs in parallel (each $k$) whi

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$12$

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(A) First,analyze the spring arrangement. There is one spring of constant $k$ in parallel with a combination of two springs in parallel (each $k$) which are in series with another spring of constant $k$.$1$. The two springs in parallel at the top have an equivalent spring constant $k_p = k + k = 2k$.$2$. This combination is in series with the spring below it (constant $k$). The equivalent constant $k_s$ for this branch is given by $\frac{1}{k_s} = \frac{1}{2k} + \frac{1}{k} = \frac{1+2}{2k} = \frac{3}{2k}$,so $k_s = \frac{2k}{3}$.$3$. This branch is in parallel with the single spring of constant $k$ on the left. Thus,the total equivalent spring constant $k_{eq} = k + k_s = k + \frac{2k}{3} = \frac{5k}{3}$.The time period $T$ of a spring-mass system is given by $T = 2\pi \sqrt{\frac{M}{k_{eq}}}$.Substituting $k_{eq} = \frac{5k}{3}$:$T = 2\pi \sqrt{\frac{M}{5k/3}} = 2\pi \sqrt{\frac{3M}{5k}} = \pi \sqrt{4 \cdot \frac{3M}{5k}} = \pi \sqrt{\frac{12M}{5k}}$.Comparing this with the given expression $\pi \sqrt{\frac{\alpha M}{5K}}$,we get $\alpha = 12$.

The time period of simple harmonic motion of mass $M$ in the given figure is $\pi \sqrt{\frac{\alpha M}{5 K}}$,where the value of $\alpha$ is . . . . . . .

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