A student measures L = 1.000 m and times 30 oscillations in 60.2 s. Find g. Also estimate the error if the timing is off by 0.2 s. Step 1: T = 60.2/30 = 2.0067 s. Step 2: g = 4π²L/T² = 4 × 9.8696 × 1.000 / (2.0067)² = 39.478 / 4.0268 ≈ 9.804 m/s². Step 3: Error estimate: g ∝ 1/T², so Δg/g ≈ 2ΔT/T = 2 × (0.2/30)/2.0067... more simply ΔT = 0.2/30 = 0.0067 s per oscillation; Δg/g ≈ 2 × 0.0067/2.0067 ≈ 0.0067 → Δg ≈ 0.066 m/s². Answer: g ≈ 9.80 m/s², ±0.07 from timing alone. Timing many swings is what makes this humble setup so accurate.