Question

Find the frequency of small oscillations of the arrangement illustrated in the figure. The radius of the pulley is $R$, its moment of inertia relative to the rotation axis is $I$, the mass of the body is $m$, and the spring stiffness is $x$. The mass of the thread and the spring is negligible, the thread does not slide over the pulley, there is no friction in the axis of the pulley.

Answer 1

The physical system consists with a pulley and the block. Choosing an internal frame, let us direct the $x-$axis as shown in the figure.
Initially the system is an equilibrium position. Now from the condition of translation equilibrium for the block
$T_0=mg\left(1\right)$
Similarly for the rotational equilibrium of the pulley
$k\Delta /R=T_{0}R$
or $T_0=k\delta l\left(2\right)$
from Eqsn $\left(1\right)$ and $\left(2\right)$ $\Delta l=\frac{mg}{k}\left(3\right)$
Noe let us disturb the equilibrium of the system no matter in which way to analyse its motion. At an arbitrary position shown in the figure, from Newton's second law of motion for the block
$F_x=m{w}_z$
$mg-T=mw=m\ddot{x}\left(4\right)$
Similarly for the pulley
$N_x=I{\beta }_x$
$TR-k(\Delta l+x)R=I\ddot{t}heta\left(5\right)$
Bur $w=\beta R$ or, $\ddot{x}=R\ddot{\theta }\left(6\right)$
from $\left(5\right)$ and $\left(6\right)$ $TR-k(\Delta l+x)R=\frac{I}{R}\ddot{x}\left(7\right)$
Solving $\left(4\right)$ and $\left(7\right)$ using the initial condition of the problem
$-kRx=(mR+\frac{I}{R})\ddot{x}$
or, $\ddot{x}=-\left(\frac{k}{m+\frac{1}{R^2}}\right)x$
Hence the sought time period, $T=\frac{2\pi }{{\omega }_0}=2\pi \sqrt{\frac{m+l/R^2}{k}}$
Note: we may solve this problem by using the conservation of mechanical energy also