A block of mass m = 1 kg hangs by means of a string which goes over a pulley of mass M = 4m and moment of inertia $\mathrm{I}=\frac{1}{4}{\mathrm{MR}}^2$, as shown in the diagram. The string does not slip relative to the pulley. If the force constant of the spring K = 900 N/m, then find the angular frequency of small oscillations (in rad/s). 

At the time of equilibrium : ${\mathrm{Kx}}_0=2\mathrm{mg}+\mathrm{Mg}$        …(i)

If the block is depressed by x, the pulley owing to the constraint is depressed by $\frac{\mathrm{x}}{2}$. If the acceleration of the block is a, then the acceleration of the pulley should be $\frac{\mathrm{a}}{2}$.

Writing equation of motion

For block : $\mathrm{T}-\mathrm{mg}=\mathrm{ma}$           …(ii)

For pulley : $\mathrm{K}\left({\mathrm{x}}_0+\frac{\mathrm{x}}{2}\right)-\mathrm{T}-{\mathrm{T}}^{\mathrm{\prime }}-\mathrm{Mg}=\frac{\mathrm{Ma}}{2}$               …(iii)

The acceleration of point 'P' should be zero as it is directly connected to a fixed point.

Hence, $\mathrm{\alpha R}=\frac{\mathrm{a}}{2}\Rightarrow \mathrm{\alpha }=\frac{\mathrm{a}}{2\mathrm{R}}$            …(iv)

Now writing torque equation for pulley : ${\mathrm{T}}^{\mathrm{\prime }}\mathrm{R}-\mathrm{TR}=\mathrm{I\alpha }=\mathrm{I}\cdot \frac{\mathrm{a}}{2\mathrm{R}}$ or ${\mathrm{T}}^{\mathrm{\prime }}-\mathrm{T}=\frac{\mathrm{Ia}}{2{\mathrm{R}}^2}$                …(v)

Using above equations, we get $\mathrm{a}=\frac{\mathrm{Kx}}{\left[4\mathrm{m}+\mathrm{M}+\frac{\mathrm{I}}{{\mathrm{R}}^2}\right]}$

As the acceleration vector and displacement vectors both are opposite to each other,

hence above equation can be written in vector form as,

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{K}}{\left[4\mathrm{m}+\mathrm{M}+\frac{\mathrm{I}}{{\mathrm{R}}^2}\right]}\cdot \overrightarrow{\mathrm{x}}$

Comparing with equation : $\overrightarrow{\mathrm{a}}=-{\mathrm{\omega }}^2\overrightarrow{\mathrm{x}}$;

we get ${\mathrm{\omega }}^2=\frac{\mathrm{K}}{\left(4\mathrm{m}+\mathrm{M}+\frac{\mathrm{I}}{{\mathrm{R}}^2}\right)}\text{ or }\mathrm{\omega }=\sqrt{\frac{\mathrm{K}}{\left(4\mathrm{m}+\mathrm{M}+\frac{\mathrm{I}}{{\mathrm{R}}^2}\right)}}$

$\Rightarrow \mathrm{\omega }=\sqrt{\frac{900}{\left(4\times 1+4\times 1+\frac{4{\mathrm{mR}}^2}{4{\mathrm{R}}^2}\right)}}=10 \mathrm{rad}/\mathrm{s}$