Simple Pendulum Period Calculator
The Simple Pendulum Period Calculator computes the oscillation period (T) and frequency (f) of an ideal simple pendulum using its length and local gravitational acceleration. The governing formula is T = 2π × √(L/g), where L is the pendulum length in meters and g is gravitational acceleration (9.81 m/s² at sea level on Earth). This tool applies to small-angle oscillations (θ < 15°), where the restoring force is approximately linear. Used in physics education, clock design, seismology, and gravitational field surveys, the period depends solely on length and gravity — not on mass or initial displacement (for small angles).
When to use this calculator
- Designing a grandfather clock escapement: a 0.994 m pendulum (the 'seconds pendulum') gives exactly T = 2.000 s, one tick per second.
- Physics lab experiment: measuring local gravitational acceleration g by timing 50 oscillations of a known-length pendulum and solving g = 4π²L/T².
- Seismic station calibration: long-period pendulums (L = 10–100 m equivalent) are used as reference oscillators in gravimeters to detect g variations across geological formations.
- Amusement park ride engineering: swing rides and pendulum-style attractions are modeled as simple pendulums to set safe oscillation periods and predict maximum angular velocity at the bottom of the arc.
Calculation Example
- L=1m, g=9.81 m/s²
- T = 2π × √(1/9.81) = 2.007 s
How it works
3 min readHow It's Calculated
The period of a simple pendulum under the small-angle approximation is:
T = 2π × √(L / g)Where:
Frequency and period are reciprocals:
f = 1 / T = (1 / 2π) × √(g / L)Angular frequency (rad/s):
ω = 2π / T = √(g / L)Validity condition: This formula holds when the maximum angle θ₀ < ~15°. For larger angles, the true period is:
T_exact = 2π × √(L/g) × [1 + (1/16)θ₀² + (11/3072)θ₀⁴ + ...]At θ₀ = 15°, the error from using the small-angle formula is only ~0.5%.
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Reference Table
| Length L (m) | g = 9.81 m/s² — Period T (s) | Frequency f (Hz) | Equivalent Clock Tick |
|---|---|---|---|
| 0.248 m | 1.000 s | 1.000 Hz | 1 tick/s (frequency standard) |
| 0.994 m | 2.000 s | 0.500 Hz | Seconds pendulum (grandfather clock) |
| 1.000 m | 2.006 s | 0.499 Hz | Near-seconds pendulum |
| 2.000 m | 2.837 s | 0.352 Hz | Slow wall clock |
| 4.000 m | 4.013 s | 0.249 Hz | Large hall pendulum |
| 9.810 m | 6.283 s | 0.159 Hz | Period = 2π seconds |
| 25.000 m | 10.037 s | 0.100 Hz | Long structural pendulum |
| 100.000 m | 20.074 s | 0.050 Hz | Seismic/geophysical scale |
All values use g = 9.81 m/s² (standard gravity, NIST)
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Typical Cases
Case 1 — The Classic 1-Meter Lab Pendulum
Case 2 — The Seconds Pendulum (Clockmaking)
Case 3 — Measuring g in the Field
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Common Errors
1. Using diameter instead of effective length: L must be measured from the pivot point to the center of mass of the bob, not to its bottom edge. For a 3 cm steel ball, this error introduces ~1.5 cm offset, adding ~7 ms to T at L = 1 m.
2. Applying the formula at large angles: At θ₀ = 30°, the true period is ~1.7% longer than the small-angle prediction. At θ₀ = 90°, the error exceeds 18%. Always keep swings below 15° for formula accuracy within 0.5%.
3. Assuming g = 9.81 m/s² everywhere: Standard gravity (9.80665 m/s²) is a defined NIST constant, but real g ranges from 9.764 m/s² (equatorial mountain tops) to 9.834 m/s² (polar sea level). Using the wrong g introduces proportional T error via √g.
4. Ignoring air resistance and pivot friction: In a real pendulum, damping gradually reduces amplitude. The formula gives the undamped natural period; real clocks use an escapement mechanism precisely to compensate for this energy loss and maintain constant amplitude.
5. Confusing period with half-period: A "tick" on a grandfather clock is a half-period (one directional swing). T = 2.00 s means 1 second per tick, not 2 seconds. Misreading this doubles or halves all timing calculations.
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Related Calculators
Currently no related calculators are linked for this tool. Check back as the physics section of Hacé Cuentas grows.
Frequently asked questions
What is the period of a 1-meter simple pendulum?
Using T = 2π × √(L/g) with L = 1 m and g = 9.81 m/s², the period is T = 2π × √(0.10194) = 2π × 0.31928 ≈ 2.006 s. Its frequency is f = 1/2.006 ≈ 0.498 Hz. This is one of the most common textbook examples in introductory physics.
Does the mass of the pendulum bob affect the period?
No. Under the simple pendulum model, mass cancels out of the equation of motion entirely. The net restoring force is proportional to mass (F = mg sinθ), but so is the inertia (F = ma), so mass divides out. This was experimentally confirmed by Galileo around 1602 and later formalized by Newton's second law.
What value of g should I use?
NIST defines standard gravity as exactly g = 9.80665 m/s². For most calculations, 9.81 m/s² is sufficient (error < 0.004%). For precise work — geodesy, metrology, or gravimeter calibration — use your local g value. It ranges from ~9.764 m/s² near the equator at altitude to ~9.834 m/s² at sea level near the poles. NOAA provides local g measurements via the National Geodetic Survey.
What is the 'seconds pendulum' and why does it matter?
A seconds pendulum has a period of exactly T = 2.000 s, so each half-swing (one 'tick') lasts 1 second. Its required length is L = g × T² / (4π²) = 9.81 × 4 / 39.478 ≈ 0.994 m at standard gravity. It was historically proposed as a basis for defining the meter — in fact, 1 meter was very nearly defined as 1/10 of this length before France adopted the meridian definition in 1793.
At what angle does the small-angle approximation break down?
The formula T = 2π√(L/g) uses the approximation sinθ ≈ θ (in radians), which holds within 1% error for θ < ~14° (~0.245 rad). At 15° the error is ~0.5%; at 30° it's ~1.7%; at 45° it's ~4%; at 90° it exceeds 18%. For angles above 15°, use the full elliptic integral solution or the series correction: T_corrected = T₀ × [1 + θ²/16 + ...].
How can I measure g experimentally with a pendulum?
Rearranging the period formula gives g = 4π² × L / T². Measure L carefully (pivot to bob center of mass), time at least 20–50 full oscillations to reduce human reaction-time error, then divide total time by the number of cycles to get T. Example: L = 1.000 m, 50 swings in 100.3 s → T = 2.006 s → g = 4π² × 1.000 / 4.024 = 9.810 m/s², matching NIST standard gravity.
How does pendulum length affect the period on the Moon vs. Earth?
The Moon's surface gravity is approximately g_Moon = 1.625 m/s² (~1/6 of Earth's). For L = 1 m: T_Moon = 2π × √(1/1.625) = 2π × 0.7845 ≈ 4.93 s — about 2.46× longer than on Earth. Astronauts on the Moon would perceive a clock pendulum as swinging in 'slow motion'. This gravitational dependence is why pendulum clocks must be recalibrated when moved to different latitudes or altitudes.
Why do grandfather clocks use a pendulum instead of a spring?
A pendulum's period is highly stable and depends only on its length and local gravity — both nearly constant in a fixed location. The isochronous property (period independent of amplitude for small angles) means minor disturbances don't affect timekeeping. Spring oscillators are temperature-sensitive and non-isochronous. Christiaan Huygens invented the pendulum clock in 1656, achieving accuracy of ~15 seconds per day — roughly 30× better than prior spring clocks of that era.