Time Period of Small Oscillations of Thin Rod in Immiscible Liquids

To find the time period of small oscillations of the thin rod hinged at the bottom, we analyze the torques acting on the rod when it is displaced by a small angle $\theta$ from its vertical equilibrium position.

1. Forces and Torques Acting on the Rod

Let $A$ be the uniform cross-sectional area of the rod.

• Weight of the rod ($W$): The total mass of the rod is: $M = \rho A L$ The force of gravity acts vertically downwards at the center of mass of the rod, located at a distance $\frac{L}{2}$ from the hinge along the rod.

  The torque due to gravity about the hinge attempts to increase $\theta$ (destabilizing torque): $\tau_g = + M g \left(\frac{L}{2}\right) \sin\theta \approx \rho A L g \left(\frac{L}{2}\right) \theta = \frac{1}{2} \rho A L^2 g \theta$

• Buoyant force from Liquid 1 (bottom half, $0 \le y \le L/2$): The density of Liquid 1 is $6\rho$. The volume of this segment is $A \left(\frac{L}{2}\right)$. The buoyant force is: $F_{B1} = (6\rho) A \left(\frac{L}{2}\right) g = 3 \rho A L g$ This force acts vertically upwards at the center of this segment, which is at a distance $\frac{L}{4}$ from the hinge.

  The torque due to $F_{B1}$ about the hinge acts to restore the rod to the vertical position: $\tau_{B1} = - F_{B1} \left(\frac{L}{4}\right) \sin\theta \approx - (3 \rho A L g) \left(\frac{L}{4}\right) \theta = -\frac{3}{4} \rho A L^2 g \theta$

• Buoyant force from Liquid 2 (top half, $L/2 \le y \le L$): The density of Liquid 2 is $2\rho$. The volume of this segment is $A \left(\frac{L}{2}\right)$. The buoyant force is: $F_{B2} = (2\rho) A \left(\frac{L}{2}\right) g = \rho A L g$ This force acts vertically upwards at the center of this segment, which is at a distance $\frac{L}{2} + \frac{L}{4} = \frac{3L}{4}$ from the hinge.

  The torque due to $F_{B2}$ about the hinge acts to restore the rod: $\tau_{B2} = - F_{B2} \left(\frac{3L}{4}\right) \sin\theta \approx - (\rho A L g) \left(\frac{3L}{4}\right) \theta = -\frac{3}{4} \rho A L^2 g \theta$

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2. Net Restoring Torque and Equation of Motion

The net torque about the hinge is given by: $\tau_{\text{net}} = \tau_g + \tau_{B1} + \tau_{B2}$ $\tau_{\text{net}} = \left( \frac{1}{2} \rho A L^2 g - \frac{3}{4} \rho A L^2 g - \frac{3}{4} \rho A L^2 g \right) \theta$ $\tau_{\text{net}} = \left( \frac{1}{2} - \frac{6}{4} \right) \rho A L^2 g \theta = - \rho A L^2 g \theta$

The moment of inertia of the thin rod about the hinge at one of its ends is: $I = \frac{1}{3} M L^2 = \frac{1}{3} (\rho A L) L^2 = \frac{1}{3} \rho A L^3$

Using Newton's second law for rotation, $I \ddot{\theta} = \tau_{\text{net}}$: $\frac{1}{3} \rho A L^3 \ddot{\theta} = -\rho A L^2 g \theta$ $\ddot{\theta} + \frac{3g}{L} \theta = 0$

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3. Time Period of Oscillations

This equation is of the standard simple harmonic motion (SHM) form $\ddot{\theta} + \omega^2 \theta = 0$, where: $\omega = \sqrt{\frac{3g}{L}}$

The time period of small oscillations $T$ is: $T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{3g}{L}}} = \frac{2\pi}{\sqrt{3}} \sqrt{\frac{L}{g}}$

Comparing this with the given expression $T = \frac{2\pi}{n} \sqrt{\frac{L}{g}}$, we get: $n = \sqrt{3} \approx 1.732$

Thus, the value of $n$ is $\mathbf{1.73}$ (or $\sqrt{3}$).