Answer : (d) None of these

Explanation :

Let the springs $\;K_{1} , K_{2} , K_{3} , K_{4} ,K_{5}\;$ move by $\;x_{1} ,x_{2} ,x_{3} ,x_{4} ,x_{5}\;.$

Then the mass 'm' moves by

$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x\;(assume)$

As the system is in equilibrium , the tension in the springs be T

By considering the F.B.D's

$2T=K_{4}x_{4} \qquad \; therefore\; , x_{4}=\large\frac{2T}{K_{4}} \;$ similarly for $\;K_{2}\;$ also

and $\;K_{5}x_{5}=T\;,K_{1}x_{1}=T\;,K_{3}x_{3}=T$

$x=\large\frac{T}{K_{1}}+\large\frac{4T}{K_{2}}+\large\frac{T}{K_{3}}+\large\frac{4T}{K_{4}}+\large\frac{T}{K_{5}}$

$x=T\;(\large\frac{1}{K_{1}}+\large\frac{4}{K_{2}}+\large\frac{1}{K_{3}}+\large\frac{4}{K_{4}}+\large\frac{1}{K_{5}})$

Let $\;(\large\frac{1}{K_{1}}+\large\frac{4}{K_{2}}+\large\frac{1}{K_{3}}+\large\frac{4}{K_{4}}+\large\frac{1}{K_{5}})=f$

$x=Tf$

$T=\large\frac{x}{f} \qquad (T=ma)$

$ma=\large\frac{x}{f}$

$a=\large\frac{x}{mf}$

$w=\sqrt{\large\frac{1}{mf}}$

$T=\large\frac{2\pi}{w}=2 \pi \sqrt{mf}$