Simple Harmonic Motion

Concept:The small oscillations of a rod submerged in two immiscible liquids result from the balance between gravitational torque and buoyant torques from both liquid layers.Explanation:The rod has density $\rho$, length $L$, and mass $M = \rho A L$.The lower liquid (density $6\rho$) occupies $y = 0$ to $y = L/2$; the upper liquid (density $2\rho$) occupies $y = L/2$ to $y = L$.For a small angular displacement $\theta$:Gravitational torque (restoring):$\tau_g = -Mg(L/2)\sin\theta \approx -\frac12 \rho A g L^2 \theta$.Buoyant force in lower liquid:$F_{B1} = (6\rho)A(L/2)g = 3\rho A L g$, acting at $y_1 = L/4$.Torque: $\tau_{B1} = +F_{B1} y_1 \theta = +\frac34 \rho A g L^2 \theta$.Buoyant force in upper liquid:$F_{B2} = (2\rho)A(L/2)g = \rho A L g$, acting at $y_2 = 3L/4$.Torque: $\tau_{B2} = +F_{B2} y_2 \theta = +\frac34 \rho A g L^2 \theta$.Net torque: $\tau_{\text{net}} = \tau_g + \tau_{B1} + \tau_{B2} = \left(-\frac12 + \frac34 + \frac34\right)\rho A g L^2 \theta = \rho A g L^2 \theta$ (opposing displacement).Thus, restoring torque: $\tau_{\text{restoring}} = -\rho A g L^2 \theta$.Moment of inertia about hinge: $I = \frac13 M L^2 = \frac13 \rho A L^3$.Using $I \frac{d^2\theta}{dt^2} = \tau_{\text{restoring}}$:$\frac13 \rho A L^3 \frac{d^2\theta}{dt^2} = -\rho A g L^2 \theta$$\Rightarrow \frac{d^2\theta}{dt^2} + \frac{3g}{L}\theta = 0$.Hence, $\omega = \sqrt{3g/L}$ and $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{3g}} = \frac{2\pi}{\sqrt{3}}\sqrt{\frac{L}{g}}$.Answer:$n = \sqrt{3} \approx 1.73$