Standard length – second’s pendulum What is the approximate length of a second’s pendulum at standard gravity (period ≈ 2 s), expressed in centimeters?

Difficulty: Easy

Correct Answer: 99.4 cm

Explanation:


Introduction / Context:
A second’s pendulum is a simple pendulum whose period is about 2 seconds. Its length provides a classic calibration for early clocks and an instructive example of the small-angle pendulum formula.



Given Data / Assumptions:

  • Target period T ≈ 2 s.
  • Small-angle approximation holds: T = 2π√(L/g).
  • Standard gravity g ≈ 9.81 m/s^2 at sea level at mid-latitudes.


Concept / Approach:
Rearrange the period formula to find the length L that yields T = 2 s. Convert meters to centimeters to compare with options.



Step-by-Step Solution:

Start with T = 2π√(L/g).Solve for L: L = g * (T / (2π))^2.Compute T/(2π) ≈ 2 / 6.283 ≈ 0.3183.Square: ≈ 0.1013. Multiply by g: 9.81 * 0.1013 ≈ 0.993 m.Convert to cm: ≈ 99.3–99.4 cm, matching 99.4 cm.


Verification / Alternative check:
Local variations in g change L slightly (about ±0.3%), but the standard textbook value remains ≈ 99.39 cm.



Why Other Options Are Wrong:

  • 94.9 cm, 100 cm, 101 cm: outside the accepted range for standard gravity with small-angle motion.


Common Pitfalls:
Using g = 10 m/s^2 for estimates gives L ≈ 1.013 m (≈ 101.3 cm), which is a rough approximation and not the standard precise value.



Final Answer:
99.4 cm

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