I.E Irodov Exercise 4.54

I will alter the nomenclature a bit. Let $k$ be the spring constant, and let $x$ be the spring displacement from equilibrium. The system Lagrangian has two kinetic energy terms (for the mass and the pulley), and two potential energy terms (for the mass and the spring).

$L = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \Big( \frac{\dot{x}}{R} \Big)^2 + m g x - \frac{1}{2} k x^2 \\ = \frac{1}{2} \Big( m + \frac{I}{R^2} \Big) \dot{x}^2 + m g x - \frac{1}{2} k x^2$

Equation of motion:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x}}} = \frac{\partial{L}}{\partial{x}}$

Evaluating results in:

$\Big( m + \frac{I}{R^2} \Big) \ddot{x} = - kx + m g$

The oscillation comes from the homogeneous equation, which is:

$\ddot{x} = - \frac{k}{m + \frac{I}{R^2}} x$

This corresponds to simple harmonic motion with the following angular frequency. Note that this applies to all oscillations; not just small ones.

$\omega = \sqrt{\frac{k}{m + \frac{I}{R^2}}}$