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Pendulum Calculator | Period & Frequency

Calculate the period, frequency, and length of a simple pendulum.

GRAVITY PRESETS

What Is the Pendulum Calculator | Period & Frequency?

This pendulum calculator solves for any of the four key quantities, period T, frequency f, length L, or gravitational acceleration g, given the other two. It includes gravity presets for six planets and derives angular frequency, ticks per minute, and the seconds-pendulum length.

  • Four solve modes: Find Period, Frequency, Length, or Gravity, enter any two to solve for the rest.
  • Planet gravity presets: Earth, Moon, Mars, Venus, Jupiter, Saturn, one click sets g.
  • Angular frequency ω: Displayed in rad/s alongside T and f.
  • Ticks per minute: Useful for clock-making, counts half-swings per minute.
  • Seconds-pendulum length: The length needed for T = 2s at the selected gravity.

Formula

T = 2π√(L/g)  ·  f = 1/T  ·  ω = 2π/T = √(g/L)
Valid for small angles (θ ≤ 15°). For large angles: T ≈ T₀(1 + θ²/16 + 11θ⁴/3072 + …)
SymbolQuantityFormula
TPeriod (s)2π√(L/g)
fFrequency (Hz)1/T = (1/2π)√(g/L)
ωAngular frequency (rad/s)2π/T = √(g/L)
LLength (m)g·(T/2π)²
gGravity (m/s²)L·(2π/T)²

How to Use

  1. 1Select a solve mode: Find Period, Find Frequency, Find Length, or Find Gravity.
  2. 2Optionally click a planet preset to set gravitational acceleration g.
  3. 3Enter the known quantities in the input fields.
  4. 4Press Enter or click Calculate.
  5. 5Read the result grid and derived quantities (ω, ticks/min, seconds-pendulum length).
  6. 6Click Clear to reset.

Example Calculation

Pendulum length for a grandfather clock (T = 2 s)

Mode: Find Length Period T = 2 s Gravity g = 9.80665 m/s² L = g·(T/2π)² = 9.80665 × (2/6.2832)² = 9.80665 × 0.10132 = 0.9933 m f = 0.5 Hz ω = π rad/s ≈ 3.14159 rad/s Ticks/minute: 60 (one tick per second)

The seconds pendulum

A seconds pendulum has a period of exactly 2 seconds (one second each way). On Earth it is ≈ 0.9933 m long. Its half-swing ticks were the original definition of the metre, the metre was briefly defined as 1/10 of the length of a seconds pendulum. The Moon's weaker gravity requires a longer pendulum (2.39 m) for the same period.

Understanding Pendulum | Period & Frequency

Simple Pendulum Physics

A simple pendulum consists of a mass (bob) suspended from a fixed point by a massless string of length L. When displaced by a small angle and released, it oscillates with period T = 2π√(L/g). This result follows from the small-angle approximation sin(θ) ≈ θ, which holds for θ ≤ 15° (about 26 cm of horizontal displacement per metre of length).

The remarkable feature is that the period depends only on length and gravity, not on the mass of the bob or the amplitude of swing (for small angles). This isochronism property was exploited by Galileo and later by Huygens to build the first accurate pendulum clocks.

Gravity on Different Planets

Bodyg (m/s²)Relative to EarthT for L = 1 m
Earth9.8071.0002.006 s
Moon1.6200.1654.930 s
Mars3.7210.3793.253 s
Venus8.8700.9052.110 s
Jupiter24.792.5281.263 s
Saturn10.441.0651.948 s

Large-Angle Corrections

For amplitudes above 15°, the small-angle approximation breaks down and the true period is longer. The correction uses an elliptic integral, but a useful approximation for moderately large angles is: T ≈ T₀(1 + θ²/16) where θ is in radians. At 30° (π/6 rad), the correction is about 1.5%, a 1 m pendulum takes 2.036 s instead of 2.006 s.

Frequently Asked Questions

Why doesn't the mass of the bob affect the period?

From Newton's second law, the restoring force is F = −mg·sin(θ) ≈ −mgθ for small angles. The equation of motion becomes θ̈ = −(g/L)θ. The mass m cancels completely, it appears on both sides of the equation. This is the same principle that makes all objects fall at the same rate in vacuum regardless of mass.

What is angular frequency and how does it differ from frequency?

Frequency f (in Hz) counts complete oscillations per second. Angular frequency ω (in rad/s) measures the rate in radians per second:

ω = 2π × f = 2π / T Example: T = 2s → f = 0.5 Hz → ω = π ≈ 3.14159 rad/s

Angular frequency appears naturally in the differential equation of motion (θ̈ = −ω²θ) and in physics calculations involving energy, phase, and resonance.

How can I use this to measure gravitational acceleration?

Use Find Gravity mode. Set up a pendulum of known length L, time several complete oscillations, and divide by the number of swings to get T. Enter L and T, the calculator computes g = L·(2π/T)².

Experiment: L = 1.000 m, 10 swings in 20.06 s T = 20.06 / 10 = 2.006 s g = 1.000 × (2π/2.006)² = 9.806 m/s²

To minimise error: keep amplitude below 10°, use a string long enough that air resistance is negligible, and time at least 20 full oscillations.

What is a seconds pendulum?

A seconds pendulum has a period of exactly 2 seconds, it swings from one side to the other in exactly one second (a tick), then returns in one second (a tock). On Earth (g = 9.807 m/s²), its length is approximately 0.9933 m (just under 1 metre). It was historically significant: in the early 1790s, one proposed definition of the metre was 1/10 the length of the seconds pendulum, which would have given 0.993 m, very close to the modern metre.

How does gravity vary across Earth?

Gravitational acceleration varies from about 9.764 m/s² at the equator to 9.832 m/s² at the poles, due to Earth's oblate shape and centrifugal effect. Altitude also matters: g decreases by about 0.003 m/s² per kilometre of height. The standard reference value 9.80665 m/s² is the defined standard gravity.

  • Equator: ~9.764 m/s² (fastest rotation, most centrifugal)
  • Sea level, 45°N/S: ~9.806 m/s²
  • North/South Poles: ~9.832 m/s² (closer to Earth's centre)
  • Everest summit: ~9.764 m/s²

What happens for large amplitude swings?

For small angles (θ ≤ 15°), T = 2π√(L/g) is accurate to 0.5%. For larger amplitudes, the true period is longer:

T_actual ≈ T₀ × (1 + θ²/16 + 11θ⁴/3072 + …) At θ = 30°: T ≈ T₀ × 1.017 (+1.7%) At θ = 60°: T ≈ T₀ × 1.073 (+7.3%) At θ = 90°: T ≈ T₀ × 1.180 (+18%) At θ = 180°: T → ∞ (balancing point)

For θ = 90° (horizontal release), the exact period uses an elliptic integral K(sin(45°)) and is approximately 1.18 × T₀.

What are ticks per minute and why do they matter?

Each half-swing (one side to the other) is one tick. A pendulum with period T makes 2/T ticks per second, or 120/T ticks per minute. Grandfather clocks use a seconds pendulum (T = 2s) producing exactly 60 ticks per minute, one per second. Watchmakers and clockmakers use this relationship to design pendulums for specific tick rates and gear train requirements.

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