Duration of one complete cycle.
Physics · Mechanics · Motion & Energy
Simple pendulum calculator
Estimate period and frequency of a simple pendulum at small angular amplitude. Measure length from the pivot to the bob’s centre of mass.
Created v1.0
By Thibaut Grzelak, Mechanical Analysis Engineer
Results
Number of complete cycles per second.
T = 2π × √(L / g)—
Small angular oscillations; L,g > 0; no damping; point mass.
01Formulas and symbols
Formulas used
T = 2π × √(L / g)Duration of one complete cycle.f = 1 / TNumber of complete cycles per second.02Assumptions and limits
Scope of validity
- Small angular oscillations; L,g > 0; no damping; point mass.
03Validation example
Reference numerical case
- L = 1 m, g = 9.80665 m/s²: T = 2π√(1/9.80665) ≈ 2.00640929 s; f ≈ 0.49840280 Hz. Standard display: T ≈ 2.0 s, f ≈ 0.50 Hz.
04References
Technical references
FAQ
Why is the pendulum period approximately independent of amplitude?
For small angles, sinθ ≈ θ, which makes the equation of motion approximately linear. In that approximation the period depends on length and gravity, not on the oscillation amplitude.
When does the small-angle approximation stop being accurate?
There is no sharp cutoff: the error grows progressively as the maximum angle increases. For larger amplitudes, use a finite-amplitude pendulum model when period accuracy matters.
05How it works
For small angular amplitudes, the restoring torque gives approximately harmonic motion. The period grows with the square root of length: quadrupling L doubles T. Mass does not appear in the ideal period because it cancels between inertia and gravitational restoring torque.
06Limits of the model
Point bob, massless inextensible string, fixed pivot, uniform gravity, negligible friction and small amplitude. At large angles the actual period is longer; this tool does not apply a finite-amplitude correction. A distributed rigid body is a physical pendulum, not this model.
07Common mistakes
Use pivot-to-centre-of-mass length, not only exposed string length. One full period returns to the same position and motion direction; a one-way crossing is not a full period. The drawing’s arc is only illustrative.