Question

A simple pendulum of length 1 m is oscillating with a small amplitude. If the acceleration due to gravity is \( 9.8 \, \text{m/s}^2 \), what is the time period of the pendulum?

• A. \( 1.0 \, \text{s} \)
• B. \( 2.0 \, \text{s} \)
• C. \( 3.0 \, \text{s} \)
• D. \( 4.0 \, \text{s} \)

Hint:

The time period of a simple pendulum depends only on its length and the acceleration due to gravity, not on the mass of the bob or the amplitude (for small oscillations). Use \( T = 2\pi \sqrt{\frac{L}{g}} \) for quick calculations.

Answer 1

The Correct Option is B

The formula for the time period \( T \) of a simple pendulum with small amplitude is \( T = 2\pi \sqrt{\frac{L}{g}} \). Given \( L = 1 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \), substituting these values yields \( T = 2\pi \sqrt{\frac{1}{9.8}} \). This simplifies to \( T \approx 2\pi \sqrt{0.102} \). Since \( \sqrt{0.102} \approx 0.319 \), the time period is approximately \( T \approx 2 \times 3.14 \times 0.319 \approx 2.004 \, \text{s} \). Rounded to one decimal place, the time period is \( 2.0 \, \text{s} \).