The equivalent spring constant of the system | vedclass

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(B) In the given figure,the two springs with spring constant $K_1$ are connected in parallel.The equivalent spring constant of these two springs in pa

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$\frac{2K_1K_2}{2K_1 + K_2}$

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(B) In the given figure,the two springs with spring constant $K_1$ are connected in parallel.The equivalent spring constant of these two springs in parallel is $K_p = K_1 + K_1 = 2K_1$.Now,this equivalent spring $K_p$ is connected in series with the spring having spring constant $K_2$.The equivalent spring constant $K_{eq}$ for two springs in series is given by $\frac{1}{K_{eq}} = \frac{1}{K_p} + \frac{1}{K_2}$.Substituting $K_p = 2K_1$,we get $\frac{1}{K_{eq}} = \frac{1}{2K_1} + \frac{1}{K_2} = \frac{K_2 + 2K_1}{2K_1K_2}$.Therefore,$K_{eq} = \frac{2K_1K_2}{2K_1 + K_2}$.

The equivalent spring constant of the system as shown in the figure will be:

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The effective spring constant of the two-spring system as shown in the figure will be:

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