In the arrangement shown in the diagram, pulleys are small and springs are ideal. `k_(1)=k_(2)=k_(3)=k_(4)=10Nm^(-1)` are force constants of the springs and mass `m=10kg.` If the time period of small vertical oscillations of the block of mass m is given by `2pix` seconds, then find the value of x.

Answer

`4`

Explanation

`T=2pisqt(m/(K_(eq)))`
where `K_(eq)=1/(4[1/(k_(1))+1/(k_(2))+1/(k_(3))+1/(k_(4))])=10/16`
`T=2pisqrt(m/(K_(eq)))=2pix`
So `x=sqrt(10/(10//16))=4,x=4`