The period of the free oscillations of the system shown here is  $T=2\pi \sqrt{\frac{M_2+b{M}_1}{k}}$, if mass M₁ is pulled down a little, given that force constant of the spring is k, mass of fixed pulley is negligible and movable pulley is smooth. The value of ‘b’ is______.

 

(4)
 
(i) Equilibrium position determination 
${\text{M}}_{\text{1}}\text{g }=\text{ T}\to$    (from FBD of  ${\text{M}}_{\text{1}}$) 
  $\text{2T }={\text{ M}}_{\text{2}}\text{g }+{\text{ kx}}_0$ (from FBD of ${\text{M}}_{\text{2}}$ ) 
$\therefore {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2M}}_{\text{1}}\text{g }={\text{ M}}_{\text{2}}\text{g }+{\text{ kx}}_0\therefore {\text{kx}}_0={\text{ 2M}}_{\text{1}}\text{g }–{\text{ M}}_{\text{2}}\text{g}$ 
(ii) Displace block ${\text{M}}_{\text{1}}$  by small disp. x by 
At new displaced position  
 $–\text{ Mgx }+\frac{1}{2}{\text{M}}_{\text{1}}{\text{V}}^{\text{2}}+\frac{1}{2}{\text{M}}_{\text{2}}{\left(\frac{v}{2}\right)}^2 +{\text{ M}}_{\text{2}}\text{g}\left(\frac{x}{2}\right) +\frac{1}{2}\text{K}{(x_0+\frac{x}{2})}^2 =\text{C}$
Differentiating equation $–{\text{ M}}_{\text{1}}\text{g} \frac{dx}{dt}+\frac{1}{2}{\text{M}}_{\text{1}}\text{2v} \frac{dv}{dt}+\frac{M_2}{8} \text{2v}\frac{dv}{dt} + \frac{M_{2}g}{2}\frac{dx}{dt}+ \frac{K}{2}\text{2}(x_0+\frac{x}{2})\left(\frac{1}{2}\frac{dx}{dt}\right) =0$
$\Rightarrow –{\text{ M}}_{\text{1}}\text{g }+{\text{ M}}_{\text{1}}\text{a }+\frac{M_{2}a}{4} + \frac{M_{2}g}{2}+\frac{K}{2}(x_0+\frac{x}{2}) =0$      (where a )
$\Rightarrow –{\text{M}}_{\text{1}}\text{g }+ \frac{M_{2}g}{2}+{\text{M}}_{\text{1}}\text{a+}\frac{M_{2}a}{4}+\frac{K{x}_0}{2}+\frac{Kx}{4} =0$ (from equilibrium  $–{\text{ M}}_{\text{1}}\text{g }+ \frac{M_{2}g}{2}+\frac{K{x}_0}{2} =0$ )
Hence,  $\frac{4M_1+M_2}{4} \text{a }=\frac{–Kx}{4}$

$$\begin{gathered} \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a }=–\left(\frac{K}{4M_1+M_2}\right)x{\omega }^{\text{2}}=\frac{K}{(4M_1+M_2)} \\ \omega =\sqrt{\frac{K}{(4M_1+M_2)}};T=\frac{2\pi }{\omega } \\ \therefore {\text{T}}_{\text{2}}=\text{ 2}\pi \sqrt{\frac{4M_1+M_2}{K}} \end{gathered}$$