Find the natural frequency of the system shown in figure. The pulleys are smooth and massless.

Explanation

Let `x` be the downward displacement of the block from the equilibrium position.
Using constant relation, the deformation in the springs will be `x'=2x`
The energy of the system is
`E=2((1)/(2)kx'^(2))+(1)/(2)mv^(2)`or `E=k(2x)^(2)+(1)/(2)mv^(2)=4kx^(2)+(1)/(2)mv^(2)`
differentiating w.r.t. time, we get
`(dE)/(dt)=8kx(dx)/(dt)+mv(dv)/(dt)=0`
or `(d^(2)x)/(dt^(2))+(8k)/(m)x=0`
`:. omega=sqrt((8k)/(m))` or `f=(1)/(pi)sqrt((2k)/(m))`