Case 1
- Pendulum length
- 1m
2.006409s
Open with these valuesPendulum Period Calculator
2.006409s
Result: 2.006409 sA simple pendulum's period depends on its length and on gravity — and on nothing else. A one-metre pendulum takes 2.006409 s for one full swing. The mass of the bob does not appear in the formula at all, and neither does the size of the swing, as long as it stays small.
Worked examples
Case 2
- Pendulum length
- 0.25m
1.003205s
Open with these valuesCase 3
- Pendulum length
- 0.994m
2.000381s
Open with these valuesHow it's calculated
T = 2π × √(L ÷ g)
- StepMeasure from the pivot to the centre of the bob, in metres.
- StepDivide by g = 9.80665 m/s² and take the square root.
- ResultMultiply by 2π to get one full swing, over and back.
Reference table
| Length (m) | Where you meet it | Period (s) |
|---|---|---|
| 0.25 | A short classroom pendulum | 1.003205 |
| 0.994 | The seconds pendulum of a longcase clock | 2.000381 |
| 1 | The round number most people try first | 2.006409 |
| 2 | A tall clock case | 2.837491 |
| 4 | Four times the length, twice the period | 4.012819 |
| 9.80665 | Length equal to g: the period is exactly 2π | 6.283185 |
Questions
What is the period of a pendulum and how is it calculated?
The period is the time one full swing — over and back — takes. For a simple pendulum it depends only on the length and gravity: period = 2π√(L ÷ g). With g = 9.80665 m/s², a 1 m pendulum gives 2.006409 s.
Why doesn't the mass of the bob matter?
The period formula contains no mass term — only length and gravity. A heavier bob feels a larger restoring force, but it also has more inertia, and the two effects cancel exactly, so a heavy bob and a light bob of the same length swing in step. This is the same reason all objects fall at the same rate in a vacuum.
How does length affect the period?
The period grows with the square root of the length, so the relationship is not linear. To double the period you must quadruple the length, and to halve it you cut the length to a quarter. A 1 m pendulum swings in about 2 s, a 4 m pendulum in about 4 s, a 0.25 m pendulum in about 1 s.
Does this formula work for any swing size?
It uses the small-angle approximation, which is accurate for swings up to about 15 to 20 degrees. For larger amplitudes the true period is slightly longer than the formula predicts. For clocks, metronomes and most demonstrations the swing is small, so the approximation is excellent.
Why does a length of 9.80665 m give exactly 2π seconds?
Because the length and g then carry the same number, so L ÷ g is 1 and its square root is 1. What is left is the bare factor 2π, about 6.283185 s. It is a coincidence of units, not of physics — in feet the number would be different.
Sources and last check
Information, not professional advice.