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

Step 1: Understanding the Question:
The question asks for the time period of a simple pendulum, given its length and the value of acceleration due to gravity.
Step 2: Key Formula or Approach:
The time period (\(T\)) of a simple pendulum is calculated using the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \] where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity.
Step 3: Detailed Explanation:
We are given the following values:
- Length of the pendulum, \(L = 1 \text{ m}\).
- Acceleration due to gravity, \(g = 9.8 \text{ m/s}^2\).
Substitute these values into the time period formula:
\[ T = 2\pi \sqrt{\frac{1}{9.8}} \] Now, we calculate the value inside the square root:
\[ \frac{1}{9.8} \approx 0.10204 \] Take the square root of this value:
\[ \sqrt{0.10204} \approx 0.31944 \] Finally, multiply by \(2\pi\) (using \(\pi \approx 3.14159\)):
\[ T \approx 2 \times 3.14159 \times 0.31944 \] \[ T \approx 6.28318 \times 0.31944 \approx 2.007 \text{ s} \] Step 4: Final Answer:
The calculated time period is approximately \(2.007 \text{ s}\), which rounds to \(2.01 \text{ s}\).