Question 13 of 14advanced🔍 AnalyzeLong Answer5 marks

A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ_r. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period T = 2π hρ/ρ1g\sqrt{hρ/ρ₁g} where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

Correct Answer

Let the cylindrical cork have base area A, height h, and density ρ. It floats in a liquid of density ρᵣ.

Step 1: Equilibrium Position At equilibrium, let x₀ be the height of cork submerged in the liquid. By Archimedes' principle, the buoyant force equals the weight of the cork.

Weight of cork = mg = ρ × A × h × g Buoyant force = ρᵣ × A × x₀ × g

At equilibrium: ρ × A × h × g = ρᵣ × A × x₀ × g Therefore, x₀ = (ρ/ρᵣ) × h

Step 2: Displacement from Equilibrium When the cork is depressed by a small distance y from its equilibrium position, the submerged height becomes (x₀ + y).

The additional buoyant force acting upward = ρᵣ × A × y × g

Step 3: Equation of Motion This additional buoyant force acts as the restoring force. By Newton's second law:

F = ma -ρᵣ × A × y × g = m × (d²y/dt²)

The negative sign indicates the force is opposite to displacement y.

Step 4: Simple Harmonic Motion Form Rearranging: d²y/dt² = -(ρᵣ × A × g / m) × y

This is of the form d²y/dt² = -ω²y, which represents simple harmonic motion.

Thus, ω² = ρᵣ × A × g / m

Step 5: Time Period Since m = ρ × A × h, we get:

ω² = ρᵣ × A × g / (ρ × A × h) = ρᵣ × g / (ρ × h)

T = 2π/ω = 2πρ×h/ρr×g\sqrt{ρ × h / ρᵣ × g}

T = 2πhρ/ρ1g\sqrt{hρ/ρ₁g}

Hence proved.

Exercise: EXERCISES | Q: 13.17 | (Page 19)
For More Understanding

Explanation

This question tests the application of Archimedes' principle and SHM concepts. The key insight is that when a floating body is displaced from equilibrium, the change in buoyant force acts as a linear restoring force, which is the defining characteristic of SHM. The derivation follows the standard approach: find equilibrium, determine restoring force for small displacement, apply Newton's second law, and identify the angular frequency ω to get the time period.

Solution Steps

  1. Step 1: Apply equilibrium condition using Archimedes' principle to find submerged height x₀ = (ρ/ρᵣ)h

  2. Step 2: Determine additional buoyant force when displaced by y: F = -ρᵣ × A × y × g

  3. Step 3: Apply Newton's second law: -ρᵣ × A × y × g = m(d²y/dt²)

  4. Step 4: Identify SHM form: d²y/dt² = -ω²y where ω² = ρᵣAg/m

  5. Step 5: Substitute m = ρAh and derive T = 2πhρ/ρ1g\sqrt{hρ/ρ₁g}

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