What is the time period T of a simple pendulum of length L in a gravitational field g for small oscillations?
Answer and explanation
Correct answer: D. 2 pi sqrt(L / g)
- Step 1: For small oscillations (sin theta approx theta), the restoring torque produces simple harmonic motion with angular frequency omega = sqrt(g / L). Step 2. Period T = 2 pi / omega = 2 pi sqrt(L / g). Answer: 2 pi sqrt(L / g). Why not the others: sqrt(g / L) inverts length and gravity; pi sqrt(L / g) misses factor 2; division by 2 pi gives frequency.
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