Question

What is the time period of a simple pendulum of length \(1\ \text{m}\) if \(g = 9.8\ \text{m/s}^2\)?

• A. \(1\ \text{s}\)
• B. \(2\ \text{s}\)
• C. \(2.01\ \text{s}\)
• D. \(3\ \text{s}\)

Hint:

For a simple pendulum, the time period depends only on the length \(L\) and gravity \(g\), not on the mass of the bob.

Answer 1

The Correct Option is C

Concept: The time period of a simple pendulum is given by the formula \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \(L\) = length of the pendulum, \(g\) = acceleration due to gravity.

Step 1: Substitute the given values. \[ L = 1\ \text{m}, \qquad g = 9.8\ \text{m/s}^2 \] \[ T = 2\pi \sqrt{\frac{1}{9.8}} \]

Step 2: Simplify the expression. \[ \sqrt{\frac{1}{9.8}} \approx 0.319 \] \[ T = 2\pi \times 0.319 \]

Step 3: Calculate the time period. \[ T \approx 2 \times 3.1416 \times 0.319 \] \[ T \approx 2.01\ \text{s} \] Thus, the time period of the pendulum is \[ \boxed{2.01\ \text{s}} \]