# The system shown find the angular frequency (w) oscillation for S.H.M .

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

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 , $2x_{1}+x_{2}=x$
$\large\frac{4T}{K_{1}}+\large\frac{T}{K_{2}}=x$
$T(\large\frac{4}{K_{1}}+\large\frac{1}{K_{2}})=x$
$T=\large\frac{x}{(\large\frac{4}{K_{1}}+\large\frac{1}{K_{2}})}$
$T=ma$
$ma=\large\frac{x}{(\large\frac{4}{K_{1}}+\large\frac{1}{K_{2}})}$
$a=\large\frac{K_{2}K_{1}}{(4K_{2}+K_{1})}\times\large\frac{x}{m}$
$w=\sqrt{\large\frac{K_{2} K_{1}}{(4K_{2}+K_{1})\;m}}$
edited Mar 9, 2014 by yamini.v