Question

Consider a solid sphere of radius \(R\) floating in a pond with half of the sphere submerged. The sphere is pushed vertically downwards at the topmost point and released, such that it executes a simple harmonic motion. Acceleration due to gravity is \(g\). What is the time period of oscillation?

• A. \(2\pi \sqrt{\frac{2R}{3g}}\)
• B. \(2\pi \sqrt{\frac{R}{g}}\)
• C. \(2\pi \sqrt{\frac{3R}{2g}}\)
• D. \(2\pi \sqrt{\frac{2R}{g}}\)

Hint:

For any floating body of constant cross-section \(A\) at the waterline, the effective SHM spring constant is \(k = \rho_{\text{fluid}} A g\).
Using \(m = \rho_{\text{body}} V\), the time period can be directly solved as \(T = 2\pi \sqrt{\frac{V_{\text{sub}}}{A g}}\).
For a half-submerged sphere, \(V_{\text{sub}}/A = \frac{2/3 \pi R^3}{\pi R^2} = \frac{2R}{3}\), giving the time period immediately.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question involves a floating solid sphere in equilibrium that is displaced vertically.
The displacement changes the buoyant force, creating a linear restoring force that drives simple harmonic motion.

Step 2: Key Formula or Approach:

• Floating equilibrium condition:
\[ W = F_B \implies \rho_s V_s g = \rho_w V_{\text{sub}} g \]

• Time period of SHM:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

• Restoring force for a small displacement \(y\).:
\[ F = -\rho_w A g y \implies k = \rho_w A g \]
where \(A\) is the cross-sectional area at the waterline.

Step 3: Detailed Explanation:

• Since half of the sphere is submerged at equilibrium, we have:
\[ V_{\text{sub}} = \frac{1}{2} V_s = \frac{1}{2} \left(\frac{4}{3}\pi R^3\right) = \frac{2}{3}\pi R^3 \]

• The mass of the sphere is:
\[ m = \rho_s V_s = \rho_w V_{\text{sub}} = \rho_w \left(\frac{2}{3}\pi R^3\right) \]

• At the waterline of a half-submerged sphere, the cross-sectional area of the interface is a circle of radius \(R\).:
\[ A = \pi R^2 \]

• When the sphere is pushed down by a small vertical displacement \(y\), the extra volume submerged is \(\Delta V \approx A y = \pi R^2 y\).

• The restoring force due to the increased buoyancy is:
\[ F_{\text{restoring}} = -\rho_w (\Delta V) g = -\rho_w (\pi R^2 g) y \]

• Thus, the effective spring constant is:
\[ k = \rho_w \pi R^2 g \]

• Now, we calculate the period of oscillation \(T\).:
\[ T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{\frac{2}{3}\pi \rho_w R^3}{\rho_w \pi R^2 g}} = 2\pi \sqrt{\frac{2R}{3g}} \]

Step 4: Final Answer:
The time period of oscillation is \(2\pi \sqrt{\frac{2R}{3g}}\).