What would be the period of the free oscillations of the system shown here if mass `M_(1)` is pulled down a little force constant of the spring is `k`, mass of fixed pulley is negligible and movable pulley is smooth

Answer

`T = 2pisqrt((M_(2) + 4M_(1))/(k))`

Explanation

(i) Equilibrium position determination
`M_(1)g = T rarr` (from `FBD` of `M_(1)`)
`2T = M_(2)g + kx_(0)` (from `FBD` of `M_(2)`)
`:. 2M_(1)g = M_(2)g + kx_(0)`
`:. kx_(0) = 2M_(1)g - M_(2)g`
`(ii)` Displace block `M_(1)` by small disp, `x` by
At new displacement position
`-Mgx + (1)/(2)M_(1)V^(2) + (1)/(2)M_(2)((v)/(2))^(2) + M_(2)g((x)/(2)) + (1)/(2)K (x_(0) + (x)/
(2))^(2) x = C`
`-M_(1)g'(dx)/(dt)+(1)/(2)M_(1)2v(dv)/(dt)+(M_(2))/(8)2v(dv)/(dt)+(M_(2)g)/(2)(dx)/(dt)+(K)/(2)2(x_(0)+(x)/(2))((1)/(2)(dx)/(dt)) =0`
`rArr - M_(1)g+M_(1)a+(M_(2)a)/(4)+(M_(2)g)/(2)+(K)/(2)(x_(0)+(x)/(2))=0`, (where `a = (dv)/(dt)`)
`rArr - M_(1)g + (M_(2)g)/(2)+M_(1)a+(M_(2)a)/(4)+(Kx_(0))/(2)+(Kx)/(4) = 0` (from equilibrium `- M_(1)g+(M_(2)g)/(2)+(Kx_(0))/(2)=0`)
Hence, `(4M_(1) + M_(2))/(4) a= (-Kx)/(4)`
`:. a = -(K)/(4M_(1) + M_(2))x , omega^(2) = (K)/((4M_(1) + M_(2)))`
`omega = sqrt((K)/((4M_(1) + M_(2)))) , T = (2pi)/(omega)`
`:. T_(2) = 2pisqrt((4M_(1) + M_(2))/(K))`