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

To determine the time period of oscillation for a solid sphere floating in water with half of its volume submerged, we need to analyze the dynamics of its simple harmonic motion (SHM). The properties of oscillation will be governed by Archimedes' Principle and Hooke's Law analog for buoyancy force.

Step-by-Step Solution:

1. Firstly, consider the object in equilibrium. When the sphere is floating with half of it submerged, the buoyant force exactly equals the weight of the sphere:
2. Where \(F_b\) is the buoyant force and \(mg\) is the weight of the sphere.
3. If the sphere is displaced vertically downward by a small distance \(x\), the volume of water displaced slightly changes due to the increased submerged depth. Thus, the additional restoring buoyant force is: \(\Delta F_b = \rho_w g A x\)
   • \(\rho_w\) is the density of water
   • \(A\) is the cross-sectional area of the sphere where it intersects the water surface.
4. For SHM, the restoring force is proportional to the displacement: \(\Delta F_b = k x\) Where \(k = \rho_w g A\) is the effective spring constant.
5. The mass of the sphere is \(m = \frac{4}{3}\pi R^3 \rho_s\), where \(\rho_s\) is the density of the sphere.
6. The acceleration for SHM is given by \(a = \frac{k}{m}x = \frac{\rho_w g A}{\rho_s \frac{4}{3}\pi R^3}x\). Simplifying for time period \(T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{R}{g} \cdot \frac{2}{3}}\).
7. Hence, the correct expression for the time period is \(T = 2\pi \sqrt{\frac{2R}{3g}}\).

Therefore, the accurate option is \(2\pi \sqrt{\frac{2R}{3g}}\), which matches the given correct answer.