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

Explanation

For the calculation purpse, in this situation we will neglect gravity because it is constant throughout and will not effect the net restoring force.
Let in the equilibrium position, the spring has extended by an amount `x_(0)`

Now, if the mass is given a further displacement downwards by an amount x. The string and spring both should increase in length by x.
But, string is inextensible, hence the spring alone will contribute the total extension `x+x=2x`, to lower the mass down by x from initial equilibrium mean position `x_(0)`. So, net extension in the spring `(=2x+x_(0))`.
Now force on the mass before bulling (in the `x_(o)`extension caes)
`F=2T`
But `T=kx_(0)` [where k is spring constant]
`:.F=2kx_(0)" " ....(i)`
When the mass is lowered down further by x,
`F'=2T['`
But new spring length `=(2x+x_(0))`
`:.F'=2k(2x+x_(0))" ".....(ii)`
Restoring force on the system.
`F_("restoring")=-[F'-F]`

Using Eqs. (i) and (ii), we get
`F_("restoring")=-[2k(2x+x_(0))-2kx_(0)]`
`=-[2xx2kx+2kx_(0)-2kx_(0)]`
`=-4kx`
or `Ma=-4kx`
where, a= acceleration " " (As, F =ma)
`rArra=-((4k)/(M))x`
`k_(1)` M being constant.
`:.aprop-x`
Hence, motion is SHM.
Comparing the above accleration expression with standard SHM equation `a=-omega^(2)x,` we get
`omega^(2)=(4k)/(M)rArromega=sqrt((4k)/(M)`
:. Time period `T=(2pi)/(omega)=(2pi)/(sqrt((4k)/(M)))=2pisqrt((M)/(4k))`