MEDIUM · THE SIMPLE PENDULUM
LENGTH FROM PERIOD
A clockmaker needs a pendulum that completes exactly one full oscillation every 3.0 seconds (T = 3.0 s). Using g = 9.80665 m/s² and the small-angle formula, calculate the required pendulum length L. Express your answer in metres.
Linked equations:T = 2\pi\sqrt{\frac{L}{g}}
§ 01
Step-by-step solution
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Step 1
Rearrange T = 2π√(L/g). Isolate √(L/g) first by dividing both sides by 2π. What is τ = T/(2π)?
Hint
This normalised value τ equals √(L/g). You will square it in the next step to find L/g.
Step 2
Square τ and multiply by g to find L = g × τ². What is L in metres?
Solution walkthrough
Starting from the period formula T = 2π√(L/g), we solve for L algebraically before plugging in numbers.
Step 1: divide both sides by 2π to isolate the radical: τ = T/(2π) = 3.0/(2π) ≈ 0.4775 s. This quantity τ equals √(L/g).
Step 2: square both sides to get τ² = L/g, then multiply by g: L = g × τ² = 9.80665 × (0.4775)² = 9.80665 × 0.22801 ≈ 2.2356 m.
Step 3: verify by substituting back: T = 2π√(2.2356/9.80665) = 2π × 0.4775 = 3.0000 s. The round-trip confirms the algebra.
This inverse problem — given a desired period, find the length — is what clockmakers actually solve. Every grandfather clock is built around this calculation. Note how sensitive L is to precision in g: local gravity variations across Earth (about ±0.5%) would require recalibrating the pendulum length by a similar fraction to keep perfect time.
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