The cylinder is floating. When it is pushed down by a small displacement $y$, it displaces an additional volume of liquid equal to $s \cdot y$.
Because the liquid is incompressible and contained in the vessel, this displaced volume must cause the liquid level in the annular region (area $S-s$) to rise by a height $h$.
$$ \text{Volume Down} = \text{Volume Up} $$ $$ s \cdot y = (S – s) \cdot h $$ Therefore, the rise in liquid level is: $$ h = \frac{s}{S – s} y $$

The buoyant force is determined by the total volume of the cylinder submerged. Compared to the equilibrium position, the cylinder is deeper by $y$, AND the water level around it is higher by $h$.
Total increase in submerged depth = $y + h$.
Substituting $h$: $$ \text{Total extra depth} = y + \frac{s}{S – s} y = y \left( 1 + \frac{s}{S – s} \right) $$ $$ = y \left( \frac{S – s + s}{S – s} \right) = y \left( \frac{S}{S – s} \right) $$

The additional buoyant force acts upwards, serving as the restoring force. $$ F_{restoring} = – (\text{density}) \cdot g \cdot (\text{extra submerged volume}) $$ $$ F = – \rho g \cdot s \cdot (\text{Total extra depth}) $$ $$ F = – \rho g s \left[ y \left( \frac{S}{S – s} \right) \right] $$ Applying Newton’s Second Law ($F_{net} = ma$): $$ m a = – \left( \frac{\rho g s S}{S – s} \right) y $$ $$ a = – \left[ \frac{\rho g s S}{m(S – s)} \right] y $$

This equation is in the standard form for Simple Harmonic Motion, $a = -\omega^2 y$.
Identifying the term for $\omega^2$: $$ \omega^2 = \frac{\rho g s S}{m(S – s)} $$ Taking the square root gives the angular frequency: $$ \omega = \sqrt{\frac{\rho g s S}{m(S – s)}} $$