We have a liquid of mass \(m\) and density \(\rho\) in a bent tube of cross-section \(S\). A light piston is attached to a spring of constant \(k\). Let the liquid be displaced by a small distance \(y\) downwards in the vertical arm.

When the liquid moves down by \(y\) in the vertical arm:

• The piston moves down by \(y\).The change in force from the spring is \(F_s = ky\).
• The liquid level in the inclined arm moves up by distance \(y\) along the tube.
• The vertical height increase in the inclined arm is \(h_2 = y \sin \theta\).
• The vertical height decrease in the vertical arm is \(h_1 = y\).

The total difference in vertical levels created by the displacement is \(h_{eff} = h_1 + h_2 = y + y \sin \theta\).
The pressure difference due to gravity is \(\Delta P = \rho g (y + y \sin \theta)\).
The restoring force due to gravity acting on the cross-section \(S\) is: \[ F_g = – (\Delta P) S = – \rho g S (1 + \sin \theta) y \]

Total restoring force \(F_{net} = F_s + F_g\): \[ F_{net} = – [k + \rho g S (1 + \sin \theta)] y \] Using Newton’s Second Law \(F = ma\): \[ m \frac{d^2y}{dt^2} = – [k + \rho g S (1 + \sin \theta)] y \] This is the equation of Simple Harmonic Motion (SHM) with stiffness \(K_{eff} = k + \rho g S (1 + \sin \theta)\).

The angular frequency is \(\omega = \sqrt{\frac{K_{eff}}{m}}\). The time period \(T\) is: \[ T = 2\pi \sqrt{\frac{m}{k + \rho g S (1 + \sin \theta)}} \]