The given setup is in equilibrium springs were in natural length before force was applied. If I have to replace the above setup of 1 block and 2 springs by 1 block and 1 spring, What will be the spring constant of that new spring, such that the block is displaced by the same amount as previously.

$K_{1} = K_{2} + K_{3}$

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$\frac{K_{3} (K_{1} + K_{2})}{K_{1} + K_{2} + K_{3}}$

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$\frac{K_{1} K_{2} + K_{2} K_{3}}{K_{1} + K_{2} + K_{3}}$

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$\frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{K_{3}}$

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Solution

The correct option is B
$\frac{K_{3} (K_{1} + K_{2})}{K_{1} + K_{2} + K_{3}}$

Spring1 and spring 2 will have same expansion so they are in parallel.

2 parallel spring with spring constant $K_{1}$ & $K_{2}$ can be replaced with 1 spring of constant ( $K_{1} + K_{2}$ ).

Now this new spring will be in series with the spring 3.

So for series equivalent is

$\frac{1}{K_{1} + K_{2}} + \frac{1}{K_{3}} = \frac{1}{K_{n e t}}$

$K_{n e t} = \frac{K_{3} (K_{1} + K_{2})}{K_{1} + K_{2} + K_{3}}$

Alternate solution

$X_{3} = \frac{F}{K_{2}}$ .....(i) $K_{3} X_{3} = (K_{1} + K_{2}) X_{1} = F$ $\Rightarrow X_{1} = \frac{F}{K_{1} + K_{2}}$ ..........(ii)

$F = K_{0} X_{n e t}$

$X_{0} = \frac{F}{K_{o n e t}}$ .......(iii)

$X_{0} = X_{1} + X_{3}$

$\Rightarrow \frac{F}{K_{0 n e t}} = \frac{F}{K_{1} + K_{2}} + \frac{F}{K_{2}}$

$\frac{1}{K_{n e t}} = \frac{1}{K_{1} + K_{2}} + \frac{1}{K_{3}}$

$K_{n e t} = \frac{K_{3} (K_{1} + K_{2})}{K_{1} + K_{2} + K_{3}}$

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