What is the period of a simple pendulum of length 0.98 m on Earth where g = 9.8 m/s^2?
Answer and explanation
Correct answer: B. 2.0 seconds
- Step 1: The period of a simple pendulum for small oscillations is T = 2 pi sqrt(L / g).
- Step 2: T = 2 pi sqrt(0.98 / 9.8) = 2 pi sqrt(0.10) approx 2 3.1416 0.3162 approx 2.0 s.
Answer
2.0 seconds.
Why not the others: '1.0 second' divides 2.0 by 2; '6.28 seconds' computes 2*pi directly without multiplying by sqrt(0.1); '0.50 seconds' inverts the period.
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