Seconds pendulum standard: What is the conventional length of a seconds pendulum (time period 2 s) under standard conditions?

Introduction / Context:

A seconds pendulum has a time period of 2 seconds (1 second for a swing in one direction). Its length under standard gravity is a classical benchmark in elementary mechanics and metrology.

Given Data / Assumptions:

• Simple pendulum approximation (point mass, light string) under standard gravity g ≈ 9.81 m/s^2.
• Small angular oscillations (small-angle approximation).

Concept / Approach:

The period of a simple pendulum is T = 2 * π * sqrt(L / g). For a seconds pendulum, T = 2 s. Solve for L to find the benchmark length commonly quoted in handbooks.

Step-by-Step Solution:

Verification / Alternative check:

Using g = 9.80665 m/s^2 (standard gravity) gives L ≈ 0.9936 m, still ≈ 99.4 cm. Slight local variations in g cause minor deviations but the conventional answer remains ~99.4 cm.

Why Other Options Are Wrong:

• 99.0 cm and 100 cm are rough guesses, not the standard derived value.
• 101–101.10 cm are too long for T = 2 s under standard g.

Common Pitfalls:

• Forgetting to square the ratio T/(2π) when solving for L.

Final Answer:

99.4 cm