Analysis
We analyze the equilibrium shift when a mass $m$ is suspended from the lower piston. This pulls the system downwards. The spring extends, and the internal fluid pressure redistributes to balance the new forces.
Step-by-Step Derivation
1. Kinematics (Displacement):
Let the upper piston A ($S_1$) move down by $x$. Due to the incompressibility of water, the volume displaced must be equal. The lower piston B ($S_2$) moves down by $y$.
$$S_1 x = S_2 y \implies y = \frac{S_1}{S_2}x$$
The net change in the length of the water column is $\Delta h = y – x = x(\frac{S_1}{S_2} – 1)$.
2. Force Balance Changes:
On Piston A (Top): The spring extends by $x$, so the upward force decreases by $\Delta F_s = kx$. For equilibrium to be maintained, the fluid pressure force acting on A must change.
$$\Delta P_A S_1 = -kx \implies \Delta P_A = \frac{-kx}{S_1}$$
On Piston B (Bottom): The extra mass $m$ exerts a downward force $mg$. This is balanced by the decrease in fluid pressure at B.
$$\Delta P_B S_2 = -mg \implies \Delta P_B = \frac{-mg}{S_2}$$
3. Pressure Relation:
The pressure at the bottom is related to the pressure at the top by the hydrostatic law: $P_B = P_A + \rho g h$.
Taking changes ($\Delta$):
$$\Delta P_B = \Delta P_A + \rho g (\Delta h)$$
Substituting the values:
$$\frac{-mg}{S_2} = \frac{-kx}{S_1} + \rho g (y – x)$$
$$\frac{-mg}{S_2} = \frac{-kx}{S_1} + \rho g x \left( \frac{S_1}{S_2} – 1 \right)$$
Multiply the entire equation by $S_1 S_2$ to clear denominators:
$$mg S_1 = kx S_2 – \rho g x (S_1 – S_2) S_1$$
$$mg S_1 = x [ k S_2 – \rho g S_1 (S_1 – S_2) ]$$