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Chapter 10: Problem 1368

A rectangular block having mass \(\mathrm{m}\) and cross sectional area A is floating in a liquid having density \(\rho\). If this block in its equilibrium position is given a small vertical displacement, its starts oscillating with periodic time \(\mathrm{T}\). Then in this case \(\ldots \ldots\) (A) \(\mathrm{T} \propto(1 / \sqrt{\mathrm{m}})\) (B) \(T \propto \sqrt{\rho}\) (C) \(\mathrm{T} \propto(1 / \sqrt{\mathrm{A}})\) (D) \(\mathrm{T} \propto(1 / \sqrt{\rho})\)

Short Answer

Expert verified
The short answer is: (D) \(\mathrm{T} \propto\frac{1}{\sqrt{\rho}}\) The period of oscillation T of the rectangular block is inversely proportional to the square root of the liquid's density ρ.

Step by step solution

01

Understand buoyant force and equilibrium position

When the block is in equilibrium, its weight is balanced by the buoyant force exerted by the liquid. The buoyant force is equal to the weight of the liquid displaced by the submerged part of the block. It can be calculated using the following formula: \[F_\text{b} = V_\text{submerged} \cdot \rho \cdot g\] where \(F_\text{b}\) = buoyant force, \(V_\text{submerged}\) = volume of the submerged part of the block, \(\rho\) = density of the liquid, and \(g\) = acceleration due to gravity.
02

Calculate the volume of the submerged part of the block in equilibrium

In equilibrium, the weight of the block is equal to the buoyant force, thus we can write: \[mg = V_\text{submerged} \cdot \rho \cdot g\] Solving for \(V_\text{submerged}\): \[V_\text{submerged} = \frac{m}{\rho}\] Since the cross-sectional area of the block is A, the depth h of the submerged part in equilibrium position can be calculated as: \[h = \frac{V_\text{submerged}}{A} = \frac{m}{\rho A}\]
03

Calculate buoyant force for the displaced block

When the block is slightly displaced from its equilibrium position by a distance x (small vertical displacement), the new depth of the submerged part is \(h + x\). The volume of the submerged part becomes \((h + x)A\) and the buoyant force is: \[F_\text{b} = (h + x)A \rho g\]
04

Set up the equation of motion for the oscillating block

As the block is slightly displaced, there will be a net force acting on it which causes an acceleration: \[F_\text{net} = (h + x)A \rho g - mg\] \[ma = (ha + xA) \rho g - mg\] where ma is the mass times acceleration. Dividing both sides by m and rearranging: \[a = g\left(\frac{h}{m}A \rho + xA \rho - 1\right)\] Considering only the displacement x part: \[a = gxA\rho\] Since the acceleration is proportional to the displacement x, this is a simple harmonic motion with the angular frequency \(\omega\), given by: \[\omega^2 = gA\rho\]
05

Calculate the period T of oscillation

The period of oscillation T is related to the angular frequency \(\omega\) by: \[T = \frac{2\pi}{\omega}\] Replacing \(\omega^2\) with \(gA\rho\): \[T = \frac{2\pi}{\sqrt{gA\rho}}\] From the given options, we find that (D) best represents this relationship: \[\mathrm{T} \propto\frac{1}{\sqrt{\rho}}\]

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