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Question

A cylindrical block of mass $M$ and area of cross section $A$ is floating in a liquid of density $\rho$ and with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation is ______.

The correct answer is
$2\pi\sqrt{\frac{\rho A}{Mg}}$

Physics: Deriving the Period of Oscillation

A cylindrical block of mass M and cross-sectional area A floats in a liquid (density ρ). When displaced vertically and released, it oscillates. We need to find the oscillation period.

Equilibrium Analysis

At equilibrium, the block's weight (Mg) equals the buoyant force. Let the submerged depth be h. The buoyant force is $ F_B = A h \rho g $. Equating forces:

$ M g = A h \rho g $

Restoring Force for Oscillation

If the block is depressed by a small distance x, the submerged depth increases to (h + x). The new buoyant force is $ F_B' = A(h+x)\rho g $. The net upward force, acting as the restoring force, is the difference between the new buoyant force and the weight:

$ F_{restore} = F_B' - Mg $ $ F_{restore} = A(h+x)\rho g - A h \rho g $ $ F_{restore} = A x \rho g $ This force opposes the displacement x, so the restoring force is $ F_{restore} = - A x \rho g $.

Simple Harmonic Motion (SHM)

Using Newton's second law ($ F=Ma $, where $ a = \frac{d^2x}{dt^2} $):

$ M a = - A x \rho g $ $ M \frac{d^2x}{dt^2} = - A x \rho g $ The equation for SHM is $ \frac{d^2x}{dt^2} = -\omega^2 x $. Comparing, the square of the angular frequency ($ \omega^2 $) is:

$ \omega^2 = \frac{A \rho g}{M} $

Period Calculation

The angular frequency is $ \omega = \sqrt{\frac{A \rho g}{M}} $. The period (T) is related to angular frequency by $ T = \frac{2\pi}{\omega} $.

$ T = \frac{2\pi}{\sqrt{\frac{A \rho g}{M}}} $ $ T = 2\pi \sqrt{\frac{M}{A \rho g}} $

Result

The derived period of oscillation is $ T = 2\pi \sqrt{\frac{M}{\rho A g}} $. This expression matches Option 3. The provided correct answer is Option A.

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Similar Questions

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Important Questions from Oscillations and Waves

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  5. The equation of a transverse wave is $y = y_0 \sin 2\pi(ft - \frac{x}{\lambda})$. If the maximum particle velocity be four times that of wave velocity then
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