Consider the forces acting on the entire piston-rod assembly.
Downward Forces: 1. Gravity: $mg$ 2. Atmospheric pressure on top piston: $p_0 A_2$ 3. Gas pressure on bottom piston: $P A_1$ (acts downwards because gas is above it)
Upward Forces: 1. Atmospheric pressure on bottom piston: $p_0 A_1$ 2. Gas pressure on top piston: $P A_2$ (acts upwards because gas is below it)

Net Equilibrium Equation: Total Upward Force = Total Downward Force $$ P A_2 + p_0 A_1 = P A_1 + p_0 A_2 + mg $$ Grouping terms by pressure: $$ P(A_2 – A_1) = p_0(A_2 – A_1) + mg $$

Let the assembly be displaced downwards by a small distance $x$.
The upper piston moves down into the gas (Volume decreases by $A_2 x$). The lower piston moves down out of the gas (Volume increases by $A_1 x$).
Net Volume Change: $$ dV = -A_2 x + A_1 x = -(A_2 – A_1)x $$
Assuming an Isothermal Process ($PV = \text{const}$): $$ dP = – \frac{P dV}{V} = – \frac{P [-(A_2 – A_1)x]}{V} = \frac{P(A_2 – A_1)x}{V} $$
The restoring force is the change in the net upward gas force. The net gas force is $F_{gas} = P(A_2 – A_1)$. $$ F_{restoring} = – dP(A_2 – A_1) $$ $$ F_{restoring} = – \left[ \frac{P(A_2 – A_1)x}{V} \right] (A_2 – A_1) $$ $$ F_{restoring} = – \frac{P(A_2 – A_1)^2}{V} x $$

The effective spring constant is $k = \frac{P(A_2 – A_1)^2}{V}$. $$ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{P(A_2 – A_1)^2}{mV}} $$ From the Ideal Gas Law, $V = \frac{nRT}{P}$. Substitute this into the equation: $$ \omega = \sqrt{\frac{P(A_2 – A_1)^2}{m (nRT/P)}} = \sqrt{\frac{P^2 (A_2 – A_1)^2}{m nRT}} $$ $$ \omega = \frac{P(A_2 – A_1)}{\sqrt{m n R T}} $$ Finally, substitute the equilibrium condition $P(A_2 – A_1) = p_0(A_2 – A_1) + mg$ into the numerator: $$ \omega = \frac{p_0(A_2 – A_1) + mg}{\sqrt{n m R T}} $$