A simple pendulum has a period of 2.0 seconds at Earth's surface. What would be its period if it were placed on the Moon, where the gravitational acceleration is approximately g / 6?

Answer and explanation

Correct answer: A. 4.90 seconds (2 * sqrt(6) s)

  1. Step 1: Recall pendulum period formula: T = 2 pi sqrt(L / g).
  2. Step 2: On the Moon, g_moon = g / 6.
  3. Step 3: New period: T_moon = 2 pi sqrt(L / (g / 6)) = sqrt(6) (2 pi sqrt(L / g)) = 2.0 sqrt(6).
  4. Step 4: Calculate: 2.0 * 2.449 approx 4.90 s.
Answer
4.90 seconds (2 * sqrt(6) s).
Why not the others: '12.0 seconds' multiplies period by 6 directly; '0.82 seconds' divides period by sqrt(6); '2.0 seconds' assumes period is independent of gravity.

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