Simple Pendulum - Ticking Seconds
Example
Question:
What is the length of a simple pendulum, which ticks seconds?
Solution:
The time period of a simple pendulum is given by \[ T = 2\pi \sqrt{\frac{L}{g}} \] Rearranging, \[ L = \frac{gT^2}{4\pi^2} \] For a seconds pendulum, \(T = 2\,\mathrm{s}\), \(g = 9.8\,\mathrm{m\,s^{-2}}\): \[ L = \frac{9.8 \times 4}{4\pi^2} = 1\,\mathrm{m} \]
Pendulum Physics
A simple pendulum with length L and gravitational acceleration g has a period:
T = 2π√(L/g)
For a pendulum that "ticks" seconds (T = 2s) at Earth's gravity (g = 9.8 m/s²):
L = gT²/(4π²) ≈ 1.00 m
Current period: 2.00 s



