$(a)\;\sqrt{\large\frac{K_{2} K_{1}}{(K_{1}+K_{2})\;m}}\qquad(b)\;\sqrt{\large\frac{K_{2} K_{1}}{(4K_{2}+K_{1})\;m}}\qquad(c)\;\sqrt{\large\frac{K_{2} K_{1}}{( 2K_{2}+K_{1})\;m}}\qquad(d)\;None\;of\;these$

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Answer : (b) $\;\sqrt{\large\frac{K_{2} K_{1}}{(4K_{2}+K_{1})\;m}}$

Explanation :

Let us strech the spring x

Let the spring $\;K_{2}\;$ move by $\;x_{2}\;$ and $\;K_{1}\;$ move by $\;x_{1}\;$ then the mass (m) moves $\;2x_{1}+x_{2}=x$

As the system is in equilibrium forces are same and let the force be T

So

$2T=K_{1} x_{1} \quad \; T=K_{2}x_{2}$

$x_{1}=\large\frac{2T}{K_{1}} \quad \; x_{2}=\large\frac{T}{K_{2}}$

Therefore , $2x_{1}+x_{2}=x$

$\large\frac{4T}{K_{1}}+\large\frac{T}{K_{2}}=x$

$T(\large\frac{4}{K_{1}}+\large\frac{1}{K_{2}})=x$

$T=\large\frac{x}{(\large\frac{4}{K_{1}}+\large\frac{1}{K_{2}})}$

$T=ma$

$ma=\large\frac{x}{(\large\frac{4}{K_{1}}+\large\frac{1}{K_{2}})}$

$a=\large\frac{K_{2}K_{1}}{(4K_{2}+K_{1})}\times\large\frac{x}{m}$

$w=\sqrt{\large\frac{K_{2} K_{1}}{(4K_{2}+K_{1})\;m}}$

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