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General SciencePhysics: Oscillations and Waves

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The time period of a simple pendulum of length 1 m is approximately (g = 9.8 m/s²):

1.0 s

2.0 s

3.14 s

6.28 s

Correct Answer

🔬 SC • Science Concept & PrincipleGeneral SciencePhysics: Oscillations and Waves

Option A

0 s

Quick Summary:

The time period of a simple pendulum is determined by its length and the acceleration due to gravity. For a length of $1 \text{ m}$ and $g = 9.8 \text{ m/s}^2$ , the calculated time period is approximately $2.0 \text{ s}$ .

🔬SCScience SolutionConcept & Principle

💡 Explanation

💡 Everyday Analogy (Real-World Intuition)

A seconds pendulum, which is a pendulum with a time period of exactly 2 seconds, is used in mechanical clocks to tick once every second.

🧠 Memory Mnemonic / Shortcut Aid

Remember '2-PI-ROOT-L-BY-G' to recall the formula.

🔢 Key Formulas

$T = 2\pi \sqrt{\frac{L}{g}}$ — The standard formula for the time period of a simple pendulum, where $L$ is length and $g$ is gravity.

⚙️ Working Principle

A simple pendulum undergoes simple harmonic motion for small oscillations. The time period $T$ is given by $T = 2\pi \sqrt{\frac{L}{g}}$ . Given $L = 1 \text{ m}$ and $g \approx 9.8 \text{ m/s}^2$ , substituting these values yields $T = 2\pi \sqrt{\frac{1}{9.8}} \approx 2 \times 3.14159 \times 0.319 \approx 2.006 \text{ s}$ .

📌 Key Points

▸
The time period is independent of the mass of the bob.
▸
A pendulum with $T = 2 \text{ s}$ is called a 'seconds pendulum'.
▸
The period increases with an increase in length ( $T \propto \sqrt{L}$ ).

✅ Advantages

▸
Simple structure for time measurement.
▸
Useful for determining local $g$ values.

❌ Disadvantages / Limitations

▸
Valid only for small amplitudes.
▸
Air resistance and friction cause damping over time.

🛠️ Applications / Uses

▸
Mechanical clocks
▸
Metrology to measure gravity

📄 Additional Information

▸
Standard value: $\pi \approx 3.14159$ and $\sqrt{g} \approx \pi$ is a common approximation used in quick calculations where $g \approx 9.86$ or $\pi^2 \approx 9.87$ .
▸
Option A is incorrect as it is half the actual period; Option C is just $\pi$ ; Option D is $2\pi$ .

📊 Diagram / Illustration

✅

B is correct — Using the formula $T = 2\pi \sqrt{\frac{L}{g}}$ with $L = 1$ and $g = 9.8$ , the calculated period is approximately $2.0 \text{ s}$ .

Core Concepts Used

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Simple Harmonic Motion Pendulum Dynamics Time Period

💡 EXAM TIP

In competitive exams, remember that $\pi^2 \approx 9.87 \approx g$ . If $g$ is taken as $\pi^2$ , the formula simplifies significantly: $T = 2\pi \sqrt{\frac{L}{\pi^2}} = 2\sqrt{L}$ .

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