Question

A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is 1.6 m . The period of oscillation of the sphere is (acceleration due to gravity \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) )

• A. \(0.2\pi \text{ }\mathrm{s}\)
• B. \(0.4\pi \text{ }\mathrm{s}\)
• C. \(0.6\pi \text{ }\mathrm{s}\)
• D. \(0.8\pi \text{ }\mathrm{s}\)

Hint:

A small sphere in a concave bowl behaves like a simple pendulum whose length is the radius of curvature.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For small oscillations in a spherical bowl of radius \(R\), the restoring force gives \(T=2\pi\sqrt{\dfrac{R}{g}}\).

Step 2: Substitute
\(R=1.6\ \text{m},\ g=10\ \text{m/s}^2\).
\(T=2\pi\sqrt{\dfrac{1.6}{10}}=2\pi\sqrt{0.16}\).

Step 3: Calculation
\(T=2\pi\times0.4=0.8\pi\ \text{s}\).

Final Answer:
The period is 0.8 pi seconds. \[ \boxed{\text{(D) }0.8\pi\ \text{s}} \]