At equilibrium, the excess pressure inside the bubble $P_0$ is balanced by the surface tension. A soap bubble has two surfaces, so the Young-Laplace equation gives:

$$P_0 = \frac{4\alpha}{r_0}$$

Consider a small radial displacement $x$, so the radius becomes $r = r_0 + x$. The total restoring force can be found by considering the net force on the oscillating mass. The net force is the difference between the outward pressure force and the inward surface tension force.

$$F_{net}(r) = P(4\pi r^2) - \alpha \frac{d A}{dr} = P(4\pi r^2) - \alpha \frac{d(8\pi r^2)}{dr} = P(4\pi r^2) - 16\pi\alpha r$$

For an isothermal process, $PV = P_0V_0$, where $V = \frac{4}{3}\pi r^3$.

$$P = P_0 \frac{V_0}{V} = P_0 \left(\frac{r_0}{r}\right)^3$$ $$F_{net}(r) = P_0 \left(\frac{r_0}{r}\right)^3 (4\pi r^2) - 16\pi\alpha r = 4\pi P_0 \frac{r_0^3}{r} - 16\pi\alpha r$$

Let $r = r_0 + x$ and expand for small $x$:

$$F_{net}(x) \approx 4\pi P_0 r_0^3 \frac{1}{r_0(1+x/r_0)} - 16\pi\alpha(r_0+x)$$ $$F_{net}(x) \approx 4\pi P_0 r_0^2(1-x/r_0) - 16\pi\alpha r_0 - 16\pi\alpha x$$

The equilibrium terms cancel: $4\pi P_0 r_0^2 - 16\pi\alpha r_0 = 4\pi(\frac{4\alpha}{r_0})r_0^2 - 16\pi\alpha r_0 = 0$. The remaining restoring force is:

$$F_{restore} = - (4\pi P_0 r_0^2)\frac{x}{r_0} - 16\pi\alpha x = -4\pi P_0 r_0 x - 16\pi\alpha x$$

Substitute $P_0=4\alpha/r_0$:

$$F_{restore} = -4\pi (\frac{4\alpha}{r_0}) r_0 x - 16\pi\alpha x = -16\pi\alpha x - 16\pi\alpha x = -32\pi\alpha x$$

This corresponds to a simple harmonic motion with an effective spring constant $k_{eff} = 32\pi\alpha$. The period of oscillation is:

$$T = 2\pi\sqrt{\frac{m}{k_{eff}}} = 2\pi\sqrt{\frac{m}{32\pi\alpha}}$$