EASY · THE SIMPLE PENDULUM
PERIOD FROM LENGTH
A simple pendulum hangs from a fixed pivot. The string has length L = 1.2 m and the bob swings through small angles on Earth, where g = 9.80665 m/s². Assuming the small-angle approximation holds, find the angular frequency ω and the period T of the pendulum.
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Step-by-step solution
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Step 1
Find the angular frequency ω = √(g / L). What is ω in rad/s?
Hint
The restoring force is F = −(mg/L)x for small angles, which gives ω² = g/L. Take the positive square root.
Step 2
Use T = 2π / ω (or equivalently 2π√(L/g)) to find the period in seconds.
Solution walkthrough
The simple pendulum is the prototype oscillator. When the bob is displaced by a small angle θ, gravity exerts a restoring force component F = −mg sin θ along the arc. For small angles, sin θ ≈ θ, so F ≈ −(mg/L) × arc-length. This is Hooke's law in disguise, with effective spring constant k = mg/L and mass m, giving ω² = k/m = g/L.
Step 1: ω = √(g/L) = √(9.80665/1.2) = √8.1722 ≈ 2.859 rad/s.
Step 2: the period is the time to complete 2π radians of oscillation, so T = 2π/ω = 2π√(L/g) = 2π × √(1.2/9.80665) = 2π × 0.34986 ≈ 2.198 s.
Two features are worth memorising. First, mass does not appear: a heavier bob on the same string has the same period. This is isochronism — Galileo observed it with a swinging chandelier in 1583. Second, the period scales as √L: doubling the length multiplies T by √2 ≈ 1.41, not by 2. To double the period you must quadruple the length. That square-root relationship is the architectural reason grandfather clocks are tall.
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