Simple Pendulum Period Calculator

Enter the pendulum length l in meters and, optionally, the acceleration g (an empty field means 9.81 m/s²). The calculator finds the period of a simple pendulum with T = 2π√(l/g) and also reports the time of one swing, the frequency, and the angular frequency. A 1 m (3.28 ft) pendulum has a period of 2.006 s, and a seconds pendulum 0.994 m long exactly 2 s.

Inputs

Result

Enter the pendulum length and, optionally, g. The period, the frequency, and the T(l) chart will appear here.

The pendulum calculator covers the same formula with swing and clock examples. From the period T, the f = 1/T calculator gives the frequency and the ω = 2π/T calculator gives the angular speed.

How it works

How T is computed for l = 1 m and g = 9.81 m/s²
l/g0.1019 s²
√0.3193 s
× 2π6.283
T2.006 s
θ≪ l T = 2π√(l/g) punkt + nić
Simple pendulum: a point mass on a massless string, small displacement. The period T = 2π√(l/g) depends only on the length and g.

The simple pendulum is an idealization: a point mass hanging on a massless, inextensible string of length l, displaced through a small angle. For small displacements the motion is harmonic, and the period of one full swing is T = 2π√(l/g), where g is the gravitational acceleration. The period depends neither on the mass of the bob nor — in this approximation — on the amplitude. This is Christiaan Huygens’s result of 1673, standard in textbooks to this day (Halliday, Resnick, Walker; Young, Freedman).

The calculator works with two numbers: the length l in meters and the acceleration g; an empty g field means 9.81 m/s². It first evaluates the quotient l/g, then its square root, and finally multiplies by 2π. For l = 1 m: l/g = 0.1019 s², √0.1019 = 0.3193 s, T = 2π·0.3193 = 2.006 s. For a seconds pendulum 0.994 m long the period is exactly 2.000 s — one swing takes one second.

What the result panel shows. Besides the period, the calculator reports the time of one swing T/2 (about 1.003 s for 1 m), the frequency f = 1/T (0.4985 Hz), the angular frequency ω = 2π/T (3.132 rad/s), the number under the root, the entered l and g, and the period of a pendulum four times longer — exactly twice as large, because the length sits under the square root. The chart compares periods for lengths from 0.25 m to 8 m at the entered g.

Limits of the model. The formula is exact only in the small-angle limit. At an amplitude of 20° the real period is about 0.8 % longer, at 45° about 4 %, and at 90° 18 %. The calculator ignores the mass of the string and the size of the bob (a physical pendulum needs the moment of inertia), does not model damping by air resistance, and does not compute the period of a spring — that is the oscillation period calculator.

The unit switch in the header changes the fields to feet and ft/s²; the values in the text stay in SI units as a reference. A comma and a period in the fields mean the same decimal separator. The length must be greater than zero and g positive; in weightlessness (g = 0) a pendulum does not oscillate, and the period is undefined.

Examples on this page: 1 m → 2.006 s; 0.994 m → 2.000 s; 0.25 m → 1.003 s; a 3 m playground swing → 3.475 s; the 67 m Foucault pendulum → 16.42 s; 0.5 m on the Moon at g = 1.62 m/s² → 3.491 s. The “Fill this example” button enters the same numbers into the fields and refreshes the result panel and the chart.

Formula

T = 2π · √(l / g)

[T] = √(m / (m/s²)) = √(s²) = s; f = 1/T, ω = 2π/T = √(g/l), l = g·T²/(4π²)

Model: simple pendulum in the small-angle approximation (C. Huygens, Horologium oscillatorium, 1673; equation of motion θ″ + (g/l)·sin θ = 0 with sin θ ≈ θ). The rearrangements l = gT²/(4π²) and g = 4π²l/T² follow from the same formula, but this calculator solves only for T. Order of computation: l/g in SI units, square root, multiplication by 2π, then f = 1/T and ω = 2π/T.

Theory, derivation, and domain

Definition of the simple pendulum and the period formula

The simple pendulum is a model: a point mass m hanging on a massless, inextensible string of length l in a uniform gravitational field. When displaced through an angle θ it feels a restoring force −m·g·sin θ tangent to its path. For small angles sin θ ≈ θ (in radians), so the motion becomes harmonic with angular frequency ω = √(g/l), and the period is T = 2π/ω = 2π√(l/g).

Three properties follow from the form of the formula. The period does not depend on mass, because the mass cancels in the equation of motion (the same reason all bodies fall alike). The period does not depend on amplitude — the isochronism noticed by Galileo. The period grows with the square root of length: a pendulum four times longer swings twice as slowly, and one a hundred times longer ten times as slowly.

Historical note: Galileo, Huygens, and the pendulum clock

According to Vincenzo Viviani, Galileo as a student in Pisa (about 1583) watched a swinging chandelier in the cathedral and, timing it with his pulse, noticed that the period did not depend on the amplitude. He studied the pendulum throughout his life and near its end designed a pendulum clock he did not live to build. He demonstrated the independence of the period from mass experimentally by comparing pendulums with lead and cork bobs.

Christiaan Huygens built the first working pendulum clock in 1656, and in the treatise Horologium oscillatorium (1673) he gave the exact theory: he derived T = 2π√(l/g), showed that Galileo’s isochronism is only approximate and that motion along a cycloid is exactly isochronous, and introduced the center of oscillation for the physical pendulum. The seconds pendulum, about 0.994 m long, was for a time considered as the standard of the meter before a fraction of the meridian was chosen in 1791.

Deriving the formula T = 2π√(l/g)

Step 1. A bob of mass m displaced through an angle θ moves along an arc of length s = l·θ. The bob feels its weight m·g and the string tension; the component of the weight along the path is −m·g·sin θ and acts as the restoring force. Step 2. Newton’s second law along the arc: m·l·θ″ = −m·g·sin θ, that is θ″ + (g/l)·sin θ = 0. The mass cancels already at this stage.

Step 3. For small angles sin θ ≈ θ and the equation takes the harmonic-oscillator form θ″ + ω²θ = 0 with ω² = g/l. Its solution is θ(t) = θ₀·cos(ωt + φ) with period T = 2π/ω = 2π√(l/g). Step 4. For l = 1 m and g = 9.81 m/s²: l/g = 0.10194 s², √0.10194 = 0.31928 s, T = 6.2832·0.31928 = 2.006 s. One swing takes T/2 = 1.003 s, and the frequency is f = 1/T = 0.4985 Hz.

Model assumptions and the amplitude correction

The formula assumes a point mass, a massless and inextensible string, no air resistance, constant g, and a small displacement angle. For a real bob of finite size and a string with mass one must use the physical pendulum: T = 2π√(I/(m·g·d)), where I is the moment of inertia about the pivot and d the distance of the center of mass from the pivot; for a ball of radius R on a string l the relative correction is about ⅕(R/l)². For a ball of radius 2 cm on a 1 m string that is 0.008 %.

The amplitude matters most in practice. The exact period is given by an elliptic integral, expanded as a series: T = T₀·(1 + θ₀²/16 + 11θ₀⁴/3072 + …), where θ₀ is the amplitude in radians. At 10° (0.1745 rad) the correction is 0.19 %, at 20° 0.77 %, at 45° 4.0 %, at 90° 18 %. Pendulum clocks run at amplitudes of a few degrees precisely so that the error is negligible; the calculator’s result corresponds to the limit θ₀ → 0.

The pendulum as a gravimeter and a clock

Inverting the formula, g = 4π²l/T², turns the pendulum into an instrument for measuring the gravitational acceleration. In 1672 Jean Richer found that a pendulum clock regulated in Paris lost about 2.5 minutes a day in Cayenne — the first evidence that g decreases toward the equator and that the Earth is flattened. Kater’s reversible pendulum (1817) measured g to 10⁻⁵ and for more than a century was the basic tool of gravimetry.

A pendulum clock is regulated by length, not by mass: shortening a 0.994 m pendulum by 1 mm shortens the period by 0.05 %, making the clock gain about 43 s per day. That is why clocks have an adjusting nut under the pendulum bob, and precision clocks have temperature compensation (Graham’s mercury pendulum of 1721, Harrison’s gridiron pendulum of 1726), because a steel rod lengthens by 12 μm per meter for every kelvin.

The value of g, other planets, and the Foucault pendulum

The period scales as 1/√g. On the Moon (g = 1.62 m/s²) a 0.5 m pendulum has a period of 3.49 s instead of 1.42 s on Earth, and on Mars (3.71 m/s²) 2.31 s. On Earth itself g varies from 9.780 m/s² at the equator to 9.832 m/s² at the poles, which for a one-meter pendulum gives a period difference of 5 ms — measurable already in the 17th century. The g for any planet comes from the g = GM/r² calculator; the result can be entered in the g field.

The Foucault pendulum (1851) is an ordinary long pendulum, 67 m long in the Paris Panthéon with a period of 16.4 s. Its plane of swing rotates relative to the floor with a period of 24 h/sin φ, where φ is the latitude — about 32 hours in Paris — and so demonstrates the Earth’s rotation. The swing period is still described by T = 2π√(l/g); the great length serves to make the precession clearly visible within an hour.

Sources

The definition, derivation, amplitude correction, and historical note come from physics textbooks and historical sources: C. Huygens, Horologium oscillatorium, Paris 1673; D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics, chapter on oscillations (simple and physical pendulum); H. D. Young, R. A. Freedman, University Physics, chapter “Periodic Motion,” sections “The Simple Pendulum” and “The Physical Pendulum”; R. P. Feynman, R. B. Leighton, M. Sands, The Feynman Lectures on Physics, vol. I, chapter 21 (The Harmonic Oscillator); L. D. Landau, E. M. Lifshitz, Mechanics, § 11 (the pendulum, elliptic integral); H. Kater, Philosophical Transactions 108 (1818); L. Foucault, Comptes rendus 32 (1851); BIPM, SI Brochure, 9th edition. This page is for educational purposes.

How to use

  1. Enter the pendulum length l in meters, measured from the pivot to the center of the bob, for example 1.
  2. Leave the g field empty for Earth (9.81 m/s²). For the Moon enter 1.62, for Mars 3.71.
  3. Click Calculate or wait for the live result. For 1 m and 9.81 m/s² the calculator shows 2.006 s.
  4. Check the panel: the time of one swing, the frequency, the angular frequency, the number under the root, and the period of a pendulum four times longer.
  5. The unit switch in the header changes the fields to feet and ft/s². The Clear button removes the data and hides the panel.

Symbol glossary

T is the period, l the string length, g the gravitational acceleration. For 1 m and 9.81 m/s² the result is 2.006 s.

T
Period of a full swing [s], T = 2π√(l/g); one swing takes T/2.
l
Length from the pivot to the center of mass [m].
g
Gravitational acceleration [m/s²]; an empty field means 9.81.
l/g
The number under the root [s²]; for 1 m and 9.81 m/s² it equals 0.1019 s².
f
Frequency [Hz], f = 1/T; about 0.4985 Hz for 1 m.
ω
Angular frequency [rad/s], ω = 2π/T = √(g/l); about 3.132 rad/s for 1 m.
θ₀
Angular amplitude; the formula is exact for θ₀ → 0, with a 0.8 % error at 20°.

Real-life examples

Example 1

A one-meter pendulum on Earth: l = {{u:1:dist-m:0}}, default g. This is the reference value for this page: a full swing takes just over two seconds.

T = 2.006 s

Example 2

The seconds pendulum: l = {{u:0.994:dist-m:3}}. One swing takes exactly one second — hence the old idea of defining the meter with it.

T = 2.000 s

Example 3

A desk toy: a bob on a string {{u:0.25:dist-m:2}} long. The period is close to one second, and the bob swings out twice in that time.

T = 1.003 s

Example 4

A playground swing: from the beam to the seat with a child about {{u:3:dist-m:0}}. The parent pushes every 3.5 s, whether the child is light or heavy.

T = 3.475 s

Example 5

The Foucault pendulum in the Paris Panthéon: l = {{u:67:dist-m:0}}. One swing takes more than sixteen seconds, and the plane of swing turns with the Earth.

T = 16.42 s

Example 6

A pendulum {{u:0.5:dist-m:1}} long on the Moon at g = {{u:1.62:accel:2}}. There the pendulum swings almost two and a half times slower than on Earth.

T = 3.491 s

Example 7

The same {{u:0.5:dist-m:1}} pendulum on Earth, for comparison with the lunar example.

T = 1.418 s

Example 8

A {{u:0.5:dist-m:1}} pendulum on Mars at g = {{u:3.71:accel:2}}. The period scales as 1/√g.

T = 2.307 s

Example 9

A long pendulum in a stairwell: l = {{u:10:dist-m:0}}. Ten times longer than the one-meter pendulum, so the period is √10 ≈ 3.16 times longer.

T = 6.344 s

Example 10

A short pendulum {{u:0.1:dist-m:1}} long, such as a key fob on a finger. It completes more than one and a half full swings per second.

T = 0.634 s

Example 11

The square-root test: l = {{u:4:dist-m:0}}, four times the first example. The period is exactly twice as long.

T = 4.012 s

Example 12

A lamp on a {{u:2:dist-m:0}} chain swaying in a draft — the scene from Pisa Cathedral where Galileo began his study of the pendulum.

T = 2.837 s

Frequently asked questions

What does this calculator compute, and from which formula?

The calculator finds the period of a simple pendulum from T = 2π√(l/g): it divides the length by the gravitational acceleration, takes the square root, and multiplies by 2π. For 1 m and 9.81 m/s² the result is 2.006 s.

Which units should I use for the length and g?

By default the length in meters and g in meters per second squared; the result is in seconds. Convert centimeters to meters (25 cm = 0.25 m). After switching the header to US units the fields take feet and ft/s².

What happens if I leave the g field empty?

The calculator uses 9.81 m/s², a typical value of the Earth’s gravitational acceleration. The difference between the equator (9.780) and the poles (9.832) changes the period of a one-meter pendulum by 5 ms, which is negligible for school purposes.

Why does the period not depend on the mass of the bob?

Because the restoring force m·g·sin θ and the inertia m·a are both proportional to the mass, which cancels in the equation of motion. It is the same reason all bodies fall alike. A heavy and a light bob on the same string swing with the same period.

Does the period depend on the amplitude?

For small angles practically not — the isochronism discovered by Galileo. The exact period grows with amplitude: by 0.19 % at 10°, 0.77 % at 20°, 4 % at 45°, 18 % at 90°. The calculator gives the limiting value for small displacements.

How long is a seconds pendulum?

One whose single swing takes one second, that is a full period of 2 s. From l = gT²/(4π²) with g = 9.81 m/s² that is 0.994 m. That is why many longcase clocks have a pendulum about one meter long.

What does “one swing T/2” in the panel mean?

The time to pass from one extreme position to the other, that is half the full period. Pendulum clocks “tick” every half period: a seconds pendulum ticks once a second, although its period is 2 s.

How does a simple pendulum differ from a physical pendulum?

A simple pendulum is a point on a massless string; a physical pendulum is any body swinging about an axis, with period T = 2π√(I/(m·g·d)). For a small ball on a long string both models give almost the same result — the correction is about ⅕(R/l)², that is 0.008 % for a ball of radius 2 cm on a 1 m string.

How do I measure g with a pendulum?

Measure the length l and the period T (best as the time of 20–50 swings divided by their number), then compute g = 4π²l/T². For l = 1 m and T = 2.006 s you get 9.81 m/s². This calculator works the other way: T from l and g.

How do I compute the period on the Moon or Mars?

Enter the matching g: Moon 1.62 m/s², Mars 3.71 m/s². The period scales as 1/√g, so a half-meter pendulum has 3.49 s on the Moon, 2.31 s on Mars, and 1.42 s on Earth.

Does the calculator account for damping and air resistance?

No. Air resistance reduces the amplitude but changes the period very little; the pendulum dies down yet ticks at the same rate. That is why clocks need a drive to maintain the amplitude, not a correction of the period.

Why is there no result for g = 0?

In weightlessness there is no restoring force, so the pendulum does not oscillate — the period is undefined (formally infinite). The calculator requires l > 0 and g > 0, and until then it displays a message instead of a result.

How do I calculate the period of a simple pendulum?

Use T = 2π√(l/g). At l = 1 m and g = 9.81 m/s² you get √(1/9.81)·2π ≈ 2.006 s. Bob mass does not enter the formula.

What is the period of a seconds pendulum?

A full period of 2 s (one swing = 1 s). At g = 9.81 m/s² the length is l = gT²/(4π²) ≈ 0.994 m.

What is T at l = 0.5 m on the Moon?

About 3.49 s at g = 1.62 m/s². On Earth the same pendulum has T ≈ 1.42 s — period scales as 1/√g.

Is this a physical pendulum?

No. Here T = 2π√(l/g) for a point mass on a massless string. A physical pendulum uses T = 2π√(I/(mgd)) — a separate model.

Does the calculator include large swing angles?

No. The result is the small-angle limit. Near 20° the real period is about 0.8% longer, near 45° about 4%; the elliptic correction is not in the panel.

The result is T = 2π√(l/g): for 1 m and g = 9.81 m/s² it equals 2.006 s, with a frequency of 0.4985 Hz. For the angular frequency from the period, use the ω = 2π/T calculator.

Knowledge sources

The definition, unit, history, and derivation below come from recognized physics and metrology sources.

Page updated November 3, 2026.