Simple Pendulum Calculator

Enter the pendulum's length, local gravitational acceleration, and initial swing angle to get its period, frequency, and angular frequency using T = 2π√(L/g).

Quick Facts

Period formula
T = 2π√(L/g)
Valid for small swing angles; depends only on length and local gravity.
Standard gravity
g = 9.80665 m/s²
Internationally adopted average for Earth's surface; local values range roughly 9.78-9.83 m/s².
Small-angle rule
θ₀ < 15°
Keeps the simple-harmonic period accurate to within about 0.2%.
Mass independence
Period ∝ √L, not mass
A heavier bob swings with the same period as a lighter one of equal length.

Your Results

Calculated
Period (small-angle)
-
T = 2π√(L/g)
Frequency
-
f = 1/T
Angular Frequency
-
ω = √(g/L)
Corrected Period (large-amplitude)
-
Series correction for swing angle θ₀

Ready

Enter the pendulum length, gravity, and swing angle, then press Calculate.

How the Simple Pendulum Period Formula Works

A simple pendulum is an idealized model: a point mass (the "bob") swinging from a fixed pivot on a massless, inextensible string or rod, with no friction at the pivot and no air resistance. For swings that stay within a small angle from vertical, the bob's motion is simple harmonic, and its period — the time for one complete back-and-forth swing — depends only on the pendulum's length and the local gravitational acceleration: T = 2π√(L/g), where L is the length from the pivot to the bob's center of mass (in meters) and g is the gravitational acceleration (in m/s²). This calculator also reports the frequency (f = 1/T), the angular frequency (ω = √(g/L)), and a corrected period that accounts for a larger swing angle.

Deriving the period from Newton's second law

For a bob displaced by angle θ, gravity supplies a restoring torque proportional to sin θ. Newton's second law for rotation gives the equation of motion θ'' + (g/L) sin θ = 0. For small angles, sin θ ≈ θ (in radians), which reduces this to the classic simple-harmonic-motion equation θ'' + (g/L)θ = 0, whose angular frequency is ω = √(g/L) and whose period is T = 2π/ω = 2π√(L/g). Notice that the bob's mass never appears — it cancels out of the torque and moment-of-inertia terms, which is why a heavier bob and a lighter one of the same length keep the same beat.

Working with units

  • Length and gravitational acceleration must use the same unit system before you take the square root — this calculator converts the length you enter (m, cm, mm, ft, or in) to meters internally so it stays consistent with g in m/s².
  • Exact length conversions used: 1 ft = 0.3048 m, 1 in = 0.0254 m, 1 cm = 0.01 m, 1 mm = 0.001 m.
  • Standard gravity is 9.80665 m/s² by international convention; use a locally measured value (typically 9.78-9.83 m/s²) for precision timing or lab work.

Large-angle corrections and limits

The small-angle approximation sin θ ≈ θ is what makes T = 2π√(L/g) simple, but it is only an approximation. For larger initial swing angles θ₀, the true period lengthens according to the series T ≈ T₀[1 + θ₀²/16 + 11θ₀⁴/3072 + 173θ₀⁶/737280 + ...] (θ₀ in radians), which this calculator uses for the "corrected period" result. This model still assumes an ideal point mass, a rigid massless support, no air drag, and no damping — a real pendulum's amplitude decays over time, and a physical (extended) pendulum needs its moment of inertia and center of mass, not just a point-mass length, for exact results.

Real-world uses of the simple pendulum

  • Pendulum clocks use a fixed length and a small, near-constant swing angle to keep a stable time reference.
  • Physics labs use timed swings of a known length to measure the local value of g.
  • Metronomes apply the same length-period relationship to set a musical tempo.
  • Seismometers and the Foucault pendulum use long, slow-swinging pendulums to detect ground motion and demonstrate Earth's rotation.

Frequently Asked Questions

What is the formula for the period of a simple pendulum?
The period of a simple pendulum swinging through a small angle is T = 2π√(L/g), where L is the length of the pendulum (pivot to center of mass) and g is the local gravitational acceleration. For example, a 1 m pendulum on Earth (g = 9.80665 m/s²) has a period of about 2.006 seconds.
Does the mass of the pendulum bob affect the period?
No. For an idealized simple pendulum (a point mass on a massless, inextensible string or rod), the mass cancels out of the equation of motion, so a heavier bob and a lighter bob of the same length swing with the same period. Only length and gravitational acceleration matter.
How accurate is the small-angle approximation for larger swings?
T = 2π√(L/g) assumes sin θ ≈ θ. At a 10° swing amplitude the resulting period is about 0.19% short of the true nonlinear value, which is negligible for most purposes. Past roughly 15-20° the error grows quickly — around 4% at 45° and about 18% at 90° — so this calculator applies a series correction, T ≈ T₀[1 + θ₀²/16 + 11θ₀⁴/3072 + ...], for larger swing angles.
What value of g should I use?
9.80665 m/s² is standard gravity, the internationally adopted average for Earth's surface. Actual local gravity ranges from about 9.78 m/s² near the equator to about 9.83 m/s² near the poles, and decreases slightly with altitude, so use a measured local value for precise timing applications.