A block of mass `M_(1)` is constrained to move along with a moveable pulley of mass `M_(2)` which is connected to a sprin of force constant k, as shown in the figure. If the mass of the fixed pulley is negligible and friction is absent everywhere, then the period of small oscillations of the system is

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))`