Question

A cylindrical body of mass m and cross section A is floating in a liquid of density $\rho_{L}$ such that its axis is vertical. If body is displaced by a small displacement 'x' vertically, find the time period of oscillation of the body :

• A. $2\pi \sqrt{\frac{m}{\rho_{L}Ag}}$
• B. $3\pi \sqrt{\frac{m}{\rho_{L}Ag}}$
• C. $4\pi \sqrt{\frac{m}{\rho_{L}Ag}}$
• D. $5\pi \sqrt{\frac{m}{\rho_{L}Ag}}$

Hint:

For any floating body undergoing vertical oscillations, the restoring force constant 'k' is always equal to the product of the liquid density, the cross-sectional area at the waterline, and the acceleration due to gravity ($k = \rho A g$). You can use this standard result directly to save time in the exam.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A floating cylinder in equilibrium experiences a buoyant force equal to its weight. When it is pushed down by a small distance 'x', the submerged volume increases, leading to an increased buoyant force. This excess buoyant force acts as a restoring force, causing the cylinder to execute Simple Harmonic Motion (SHM).
Step 2: Key Formula or Approach:
1. The restoring force for a displaced floating body is the excess buoyant force: $F = - \Delta B = - (\text{extra submerged volume}) \times \rho_{L} \times g$.
2. The extra submerged volume is $A \times x$.
3. For SHM, $F = -k x$, where $k$ is the restoring force constant.
4. The time period of SHM is $T = 2\pi \sqrt{\frac{m}{k}}$.
Step 3: Detailed Explanation:
Let the cylinder be in equilibrium with a submerged height 'h'.
At equilibrium, Buoyant force = Weight of the cylinder
$\rho_{L} A h g = m g$
When the cylinder is displaced downwards by a distance 'x', the new submerged height is $(h + x)$.
The new buoyant force is $B' = \rho_{L} A (h + x) g$.
The net restoring force $F$ acting on the cylinder is:
$F = \text{Weight} - B'$
$F = m g - \rho_{L} A (h + x) g$
Substituting $m g = \rho_{L} A h g$:
$F = \rho_{L} A h g - \rho_{L} A h g - \rho_{L} A x g$
$F = - \rho_{L} A x g$
Comparing this with the SHM equation $F = -k x$, we get the effective spring constant:
$k = \rho_{L} A g$
Now, the time period of oscillation $T$ is:
$T = 2\pi \sqrt{\frac{m}{k}}$
$T = 2\pi \sqrt{\frac{m}{\rho_{L} A g}}$
Step 4: Final Answer:
The time period of oscillation of the body is $2\pi \sqrt{\frac{m}{\rho_{L}Ag}}$.