Figure shows a system consisting of a massless pulley, a spring of force constant k and ablock of mass m. If the block is slightly displaced vertically down from its equillibrium position and then released, find the period of its vertical oscillation in cases (a) & (b).

Explanation

Let us assume that in equillibrium condition spring is `x_(0)` elongate from its natural length
In equilibrium `T_(0) = mg and kx_(0) = 2T_(0)`
`implies kx_(0) = 2 mg " " cdots (1)`
If the mass m moves down a distance x from its equilibrium position then pulley will move down by `(x)/(2)` . So the extra force in spring will be `(kx)/(2)` . From figure `F _("net") = mg -T = mg -(k)/(2) (x_(0) + (x)/(2))`
`F_("net") = mg - (kx_(0))/(2)- (kx)/(2)`
From eq. (1)
`F_("net") = (-kx)/(4) " " cdots (3)`
Now compare eq . (3) with F = - `K_(S.H.M) x ` then `K_(S.H.M)= (K)/(4)`
`implies T = 2pir T = 2pi sqrt((m)/(K_(S.H.M.)))=2pisqrt((4m)/(k))`

CASE (B) :
in this situiation of the mass m moves down distance x form its equilibrium position , then pulley will alspo move by x and so the spring will stetch by 2x .