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Simple Pendulum Calculator – Online Period & Length

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Simple Pendulum Calculator

Calculate the period, frequency, and angular frequency of a simple pendulum. Switch modes to find the required pendulum length for a desired period. Adjust gravity for different celestial bodies.

Calculate Period (T) Calculate Length (L)
Input Parameters
Length from pivot point to center of mass
Quick:
Animation synced to period
Results
Period (T)
2.0061 s
Time for one full oscillation
Frequency (f)
0.4985 Hz
Oscillations per second (f = 1/T)
Angular Frequency (ω)
3.1321 rad/s
ω = 2πf = √(g/L)
Formula

Period from Length:

T = 2π √(L / g)

T = period (s)  |  L = length (m)  |  g = gravity (m/s²)

Length from Period:

L = g · T² / (4π²)

Rearranged formula  |  Valid for small angles (< 15°)

Frequently Asked Questions

A simple pendulum consists of a point mass (called the bob) suspended from a fixed pivot by a massless, inextensible string or rod. When displaced from its equilibrium position and released, it swings back and forth under the influence of gravity, exhibiting simple harmonic motion for small angular displacements (typically less than 15°).

The period T of a simple pendulum (for small angles) is given by: T = 2π √(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. This formula shows that the period is independent of the mass of the bob and depends only on the length and gravity. This is known as isochronism.

A seconds pendulum is a pendulum with a period of exactly 2 seconds—it takes 1 second to swing in one direction and 1 second to return. On Earth (g = 9.81 m/s²), the required length is approximately 0.994 meters. Historically, seconds pendulums were used in precision clocks and were even proposed as a standard for defining the meter.

The simple pendulum formula T = 2π√(L/g) is derived using the small-angle approximation (sin θ ≈ θ). This holds well for angles up to about 15° (0.26 radians), where the error is less than 0.5%. For larger angles, the period becomes amplitude-dependent and requires an elliptic integral solution. The exact period for any angle θ₀ is: T = 2π√(L/g) × [1 + (1/16)θ₀² + (11/3072)θ₀⁴ + ...].

The period is inversely proportional to the square root of gravity: T ∝ 1/√g. This means that on the Moon (g ≈ 1.62 m/s²), the same pendulum swings about 2.46 times slower than on Earth. A 1-meter pendulum on Earth has a period of ~2.01 seconds, but on the Moon it would have a period of ~4.94 seconds. This principle can be used to measure local gravity variations.

Pendulums have many applications: grandfather clocks use pendulums for timekeeping; seismometers use inverted pendulums to detect ground motion; Foucault pendulums demonstrate Earth's rotation; pendulum wave machines create mesmerizing visual patterns; and gravimeters use precise pendulum measurements to map gravitational anomalies for geological surveys.

You can determine local gravity by measuring the period T of a pendulum of known length L. Rearranging the formula: g = 4π²L / T². For accurate results: use a long pendulum (1+ meter), measure the time for many oscillations (50-100) and divide to get the average period, keep the swing angle small (<10°), and ensure the pivot has minimal friction.

No. The mass of the bob cancels out in the equation of motion. This is because both the gravitational force (mg) and the inertia (m) are proportional to mass. As a result, the period depends only on length and gravity. This was one of Galileo's key observations: pendulums of equal length swing with the same period regardless of bob weight.