Find the time period of `m` if pulley `P` is light and small.

Explanation

Let the displacement of block at any time be x and stretch of spring be x,, then
`(x' + 0)/(2) = x' = 2x`

Method I: force method

At equailibrum
`Block, T = m g` (i)
pulley, `2 T = T` (ii)
Spring, `kx_(0) = mg`
Let the block is further displaced by small amount released.
Equation of motion at any time

`T_(1) - mg = ma`
`2 T_(1) - mg = ma` (ii)
`2K (x_(0) + x') - mg = ma` (iii) From Eqs, (iii)and (iv)
`2 Kx' = ma implies a = (2 Kx')/(m) = ((4 k)/(m)) x implies ((4 k)/(m)) x`
`omega^(2) = (4 k)/(m) implies omega = sqrt ((4 k)/(m))`
Method II : Energy method
`(1)/(2) m v^(2) + (1)/(2) k (x_(0) + x')^(2) - mg x =` constant
`(1)/(2) m 2 v a + (1)/(2) k 2 (x_(0) + x') (dx')/(dt) - mg (dx)/(dt) = 0`
`mva + k (x_(0) + x') (dx')/(dt) - mg v = 0`
`mv a + k (x_(0) + 2 x') 2 v - mg v= 0`
`ma + 2 k (x_(0) + 2 x') - mg v = 0`
`ma + 4 k x = 0` (as `mg = 2 Ks_(0))`
`a = - ((4 K)/(m)) x, implies omega = sqrt((4k)/(m))`
Method IIi : Net resoting force on `m` = Net increase in tension string connecting `m` and pulley `= Delta T`

. If increase in the tension connecting the block and the pulley is `Delta T`, then increase in the tension connecting spring and pulley will be `Delta T = Delta T//2`
. If x is the displacement of the block beyond equalibrium position, then aditional stretch of spring will be `x = 2 x`
`Delta T = ma impliesDelta T = ma`
`2kx' = 4 kx`
`a = - ((4k)/(m)) x implies omega = sqrt((4k)/(m))`