Example 1
Length 1 m and a blank g field. The period is 2.0061 s, and there are 29.91 full swings in a minute.
T = 2.0061 s
Pendulum Period Calculator
Enter the pendulum length L in meters and, if you want, the acceleration g (a blank field means 9.81 m/s²). The calculator finds the period T = 2π√(L/g) in the small-angle limit and also reports the time of one swing, the frequency, and the angular frequency. At 1 m the period is 2.006 s; the seconds pendulum (0.994 m) is exactly 2 s. The model does not cover a physical pendulum or the elliptic correction for large amplitudes.
Result
Enter the length L. The period, the frequency, and the T(L) chart will appear here.
The model assumptions (a point mass, a massless string, and the derivation) are on the simple pendulum calculator. From the period you can also find f on the frequency-from-period calculator and ω on the angular velocity calculator.
How it works
L1 m
g9.81 m/s²
L/g0.1019 s²
T = 2π√(L/g)2.0061 s
At small angles the pendulum period is T = 2π√(L/g). The calculator divides the length L by the acceleration g, takes the square root, and multiplies by 2π. The bob’s mass does not enter the formula, and in this limit the swing angle does not enter either. Christiaan Huygens gave the formula in Horologium oscillatorium (1673). It is the textbook model of a simple pendulum when sin θ ≈ θ.
For L = 1 m and the default g = 9.81 m/s² the calculator shows a period of 2.0061 s. The ratio L/g, before the panel rounds the period, is about 0.1019 s². One swing to the other side takes 1.0030 s, the frequency is 0.4985 Hz, and there are 29.91 full swings in a minute. The seconds pendulum, with a period of 2 s, is about 0.994 m long at this g. Entering 0.994 returns 2.0000 s on the panel, which is not the same result as a 1 m string.
The L field is the distance from the pivot to the center of mass, in meters. The g field is optional: blank means 9.81 m/s². For the Moon, enter 1.62. The same length of 1 m then gives 4.9365 s, about 2.46 times longer than on Earth. A comma and a period in the field are the same decimal separator.
Small-angle limit. The formula is exact as the swing angle tends to zero. Near 20° the real period is about 0.8% longer, and near 45° about 4% longer. The calculator does not add that elliptic-integral correction, and it does not include string mass, damping, or a physical pendulum. A playground swing and a clock pendulum stay inside this model as long as the angle is not large.
The period grows like the square root of the length. L = 2 m gives 2.8370 s, not twice 2.0061 s. L = 4 m gives 4.0121 s, twice the period of a one-meter pendulum. A length of 0.25 m gives 1.0030 s, and a 3 m swing gives 3.4746 s. The rider’s mass does not change that swing.
The derivation from the equation of motion is on the simple pendulum page. This page gives the period for a swing or a clock, and the number of full swings per minute. The frequency f = 1/T is on the panel; a separate calculator finds f alone when you already know T. The angular speed ω = 2π/T is on the angular velocity calculator.
Formula
T = 2π √(L / g)
The default g is 9.81 m/s2. Model: a simple pendulum in the small-angle limit (C. Huygens, 1673). The calculator finds T, then f = 1/T and the number of full swings per minute, 60/T. It does not apply an amplitude correction.
How to use it
- Enter the length L in meters, from the pivot to the center of mass. The reference example on this page is 1 m.
- Leave g blank for Earth (9.81 m/s²). For the Moon, enter 1.62. The entered g must be greater than zero.
- Click Calculate, or wait for the live result. For 1 m and a blank g the period is 2.0061 s.
- On the panel, check the time of one swing, the number of full swings per minute, the frequency, and the ratio L/g.
- The textbook assumptions are on the simple pendulum calculator. A large angle makes the real period longer than this limit.
Symbols in T = 2π√(L/g)
A practical pendulum in the small-angle limit. For L = 1 m and g = 9.81 m/s² the period is 2.0061 s.
- T
- Period of one full swing, in seconds. At 1 m and 9.81 m/s² the result is 2.0061 s. One swing takes T/2.
- L
- Length from the pivot to the center of mass, in meters. L = 0.25 m gives a period of 1.0030 s.
- g
- Acceleration in m/s². A blank field means 9.81. There is no mass field, because T does not depend on mass in this formula.
- L/g
- The number under the square root, in s². For 1 m and 9.81 m/s² it is about 0.1019 s².
- f
- Frequency in hertz, f = 1/T. For 1 m it is 0.4985 Hz, or 29.91 full swings per minute.
- θ
- Swing angle. The formula is the limit θ → 0. Near 20° the real period is about 0.8% longer.
Worked examples
Example 2
Length 2 m at Earth g. The period is 2.8370 s, not twice 2.0061 s, because the length is under the square root.
T = 2.8370 s
Example 3
A short string, 0.25 m. The period is 1.0030 s.
T = 1.0030 s
Example 4
Length 0.5 m on Earth. The period is 1.4185 s.
T = 1.4185 s
Example 5
Length 1 m at g = 1.62 m/s² (the Moon). The period is 4.9365 s.
T = 4.9365 s
Example 6
80 cm entered as 0.8 m. The period is 1.7943 s.
T = 1.7943 s
Example 7
Length 1.5 m. The period is 2.4569 s.
T = 2.4569 s
Example 8
Length 1 m and g = 9.8 m/s². The period is 2.0071 s, which is 0.0010 s more than at 9.81.
T = 2.0071 s
Example 9
A short length, 0.1 m. The period is 0.6344 s, so the frequency is 1.5764 Hz.
T = 0.6344 s
Example 10
A swing 3 m from the bar to the seat. The period is 3.4746 s and does not depend on the rider’s mass.
T = 3.4746 s
Ways to use this calculator
- Estimate a swing’s period: 3 m on Earth gives 3.4746 s.
- Compare Earth and the Moon at L = 1 m: 2.0061 s and 4.9365 s.
- Read how many full swings fall in a minute before you move on to f = 1/T or ω = 2π/T.
Frequently asked questions
What does this calculator compute, and from which formula?
The calculator finds the pendulum period from T = 2π√(L/g): it divides the length by g, takes the square root, and multiplies by 2π. For 1 m and 9.81 m/s² the result is 2.0061 s. It also shows the time of one swing, the frequency, and the number of full swings per minute.
Which units should I use for the length and g?
By default, the length is in meters and g is in meters per second squared. Enter 80 cm as 0.8 m; the period is then 1.7943 s. A comma and a period are the same decimal separator. After you switch the header to US units, the fields take feet and ft/s², and the period stays in seconds.
What happens if I leave g blank?
The calculator uses 9.81 m/s². Enter another value yourself, for example 1.62 on the Moon. At L = 1 m the period on the Moon is 4.9365 s.
Does the bob’s mass change the period?
Not in this formula. A 1 kg bob and a 200 g bob on the same length have the same period, because the mass cancels in the equation of motion.
Does the formula hold at a large swing angle?
Not outside the small-angle limit. The calculator reports T for sin θ ≈ θ. Near 20° the real period is about 0.8% longer, and near 45° about 4% longer. The calculator does not add that correction.
How does the period grow with length?
Like the square root of L. Four meters gives 4.0121 s, which is twice the period of a 1 m pendulum (2.0061 s), not four times that period.
How do I get the frequency from the period?
The frequency is f = 1/T. For 2.0061 s it is 0.4985 Hz. The same quotient is on the result panel. A separate frequency-from-period calculator finds f alone when you already know T.
How is this page different from the simple pendulum page?
The formula is the same. That page states the model assumptions: a point mass, a massless string, and the derivation. Here the examples are a swing, a clock, and your own g, and the panel reports swings per minute.
Does the calculator include damping?
No. Air resistance reduces the swing, while the period in the small-angle limit hardly changes. The calculator does not compute the decay of the motion.
Why is there no result when g is zero?
Without acceleration there is no restoring force, and the pendulum does not oscillate. The calculator requires L > 0 and g > 0, and until then it shows a message instead of a period.
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.
Sources
The formula, the small-angle limit, and the units come from textbooks and from the SI. The calculator does not replace a laboratory measurement.
- Wikipedia — Pendulum (amplitude correction)
- Wikipedia — Seconds pendulum
- BIPM — SI Brochure
- NIST SP 330 — SI units
- Huygens, C., Horologium oscillatorium, Paris, 1673.
- Young, H. D.; Freedman, R. A., University Physics — Periodic Motion: The Simple Pendulum.
Page updated November 3, 2026.