Find the time period of mass M when displaced from its equilibrium position and then released for the system shown in figure.
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Explanation

Here, for calculation of time period we have neglected acceleration of gravity. Because, it is constant at all places and doesn.t affect on resultant restoring force.

Let the equilibrium position, the spring has extended by an amount `x_(0)`. Let the mass be pulled through a distance x and then released. But, string is inextensible, hence the spring alone will contribute the total extension `x+x= 2x`. So that extension in the spring will be `2x+ x_(0)`.
When the extension of spring is `x_(0)`, (M is not suspended) the restoring force in spring.
`F= 2kx_(0),"........."(1)" "[therefore F= T+T" and "T= kx_(0)]`
When mass is suspended, the net extension in spring `2x+x_(0)` and restoring force in spring,
`F. = 2k(2x+ x_(0))= 4kx + 2kx_(0)" ""........"(2)`
Restoring force on system,
`f. = -(F. -F)`
`=F -F.`
`=2kx_(0)-4kx-2kx_(0)" "`[From equ. (1) and (2)]
`= -4kx`
but `F= Ma`
`therefore Ma = -4kx`
`therefore a= -(4k)/(M)*k" ""........."(1)`
`therefore a propto -x` where, `(4k)/(M)` is constant.
hence, motion of block is SHM
The equation compare to `a= -omega^(2)y`,
`omega^(2) = (4k)/(M)`
`therefore omega = sqrt((4k)/(M))`
`therefore (2pi)/(T)= sqrt((4k)/(M))`
`therefore T = (2pi)/(2) sqrt((M)/(k))`
`therefore T= pi sqrt((M)/(k))" or "T= 2pi sqrt((M)/(4k))`.