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 \(g = 10\text{ m/s}^2\))

• A. \(0.2π\)
• B. \(0.4π\)
• C. \(0.8π\)
• D. \(π\)

Hint:

The sphere acts like a pendulum of length equal to the radius.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A small sphere rolling in a watch glass (a shallow bowl of radius \(R\)) behaves like a simple pendulum of length \(R\) for small oscillations.

Step 2: Formula:
\[ T=2\pi\sqrt{\frac Rg} \]

Step 3: Substitute:
\[ T=2\pi\sqrt{\frac{1.6}{10}}=2\pi\sqrt{0.16}=2\pi\times0.4=0.8\pi\ \text{s} \]

Step 4: Choose:
Option (C).

Final Answer:
The period is 0.8 pi s. \[ \boxed{0.8\pi} \]