pi = 3.14159
Formula
The period of a simple pendulum depends only on the string length and the free-fall acceleration: T = 2π√(l/g).
ENT-level problem
A simple pendulum has a length $l = 2.45$ m. Find its period of oscillation. The acceleration due to gravity is $g = 9.8$ m/s².
The period of a simple pendulum:
$$T = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{2.45}{9.8}} = 2\pi \sqrt{0.25} = 2\pi \cdot 0.5 = \pi \approx 3.14 \text{ s}$$
A simple pendulum is a mass on a weightless string. Its period depends only on the string length l and the free-fall acceleration g: neither the mass of the bob nor the amplitude (for small angles) affects the period.
With l = 1 m and g = 9.81 m/s² the period is about 2 s — a fact Galileo used to measure time. On another planet with a different g the period changes.
Only on the string length and the free-fall acceleration: T = 2π√(l/g). Mass and amplitude do not matter (for small angles).
Solve for l: l = g · T² / (4π²). For T = 2 s and g = 9.81 m/s² the string length is ≈ 1 m.
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