Example 1
l = 1 m, blank g.
Period of a Simple Pendulum calculator
This is the school model: a point mass on a massless string, small angle, no drag. The formula matches the practical pendulum page, T = 2π√(l/g). l = 1 m and blank g give T ≈ 2.01 s.
Swing and clock version: pendulum period. From T you get f on frequency from period and ω on angular velocity.
Result
Length l (m) and g (m/s², optional). The result shows up here.
How it works
A simple pendulum is a school contract: mass at a point, string or rod with no mass of its own, motion in one plane, no friction, small angle. Under those assumptions T = 2π√(l/g). l = 1 m and g = 9.81 m/s² give T ≈ 2.01 s, the same numbers as the practical page.
The l field is string length in meters. g is optional, and blank means 9.81. There is no mass field, because T does not depend on m in this model. Amplitude stays out too, as long as the angle is small and sin θ ≈ θ.
At a large angle the harmonic model breaks and the period grows. A real string with mass, or a bob with a visible radius, is a physical pendulum: moment of inertia and center of mass enter, a different formula, a different page outside this set.
l = 0.5 m at 9.81 shortens T to about 1.42 s. l = 4 m stretches it to about 4.01 s, twice the 1 m period. On the Moon, l = 1 m and g = 1.62 give about 4.94 s. A comma and a period both parse.
Type l, leave g blank or enter your own, then Calculate. The calculator example is 1 m and blank g. For a swing or a clock without the assumption list, go back to the practical pendulum. f = 1/T is on the neighbouring page, with the same T.
A spring computes f from k and m, not from length and g. Swapping l for k does not work. ω = 2π/T from this period sits on angular velocity.
How to use
- Type length l in meters. That is the string from the support to the point mass, for example 1 m.
- You can leave g blank (9.81 m/s²) or type your own, for example 1.62. Both fields, when filled, must be positive.
- Click Calculate. 1 m and blank g give T ≈ 2.01 s. l = 0.5 m is about 1.42 s. You do not need mass.
- Keep the assumptions in mind: small angle, point mass, massless string, no drag. A large swing stretches the real period.
- Open the practical pendulum for a swing or a clock. Get f from this T on frequency from period.
Formula
T = 2π √(l/g)
Assumptions: small angle, point mass, massless string, no drag.
Letters in T = 2π√(l/g)
School model: point mass, massless string, small angle. T = 2π√(l/g) at l = 1 m and blank g is again ≈ 2.01 s, but the length letter is l, not L.
- T
- Period of the ideal model. l = 1 m and g = 9.81 m/s² is the same ≈ 2.01 s, with stricter assumptions than the L card.
- l
- String length in meters, lowercase l from the school convention. The point mass has no field; T does not use it.
- g
- Optional g. Blank means 9.81 m/s² and with l = 1 m builds those 2.01 s with no drag and no string mass.
Real-life examples
Example 2
l = 0.5 m.
Example 3
l = 2 m.
Example 4
l = 0.8 m.
Example 5
l = 1 m, g = 1.62.
Example 6
l = 0.25 m.
Example 7
l = 1 m, g = 9.81.
Example 8
l = 1.2 m.
Example 9
l = 0.1 m.
Example 10
l = 3 m.
Ways to use this calculator
- You work a school problem where the teacher listed the model assumptions.
- You compare this T with the practical pendulum and check that the numbers match.
Frequently asked questions
What does “simple” mean here?
An ideal model: point mass, massless string, small angle, no drag. The formula is T = 2π√(l/g).
What T at l = 1 m and blank g?
About 2.01 s. l = 0.5 m shortens T to about 1.42 s. Mass does not enter.
How is this different from the pendulum page?
Same formula. That page has playground and clock examples. Here you get the assumption list a teacher wants to hear.
Why the small-angle limit?
At small θ, sin θ ≈ θ and the motion is harmonic. At large angles the period grows.
Does mass enter T?
No. In the simple model T depends on l and g, not on m.
How do I get f and ω?
f = 1/T on frequency from period. ω = 2π/T on angular velocity. From 2.01 s that is about 0.50 Hz and 3.13 rad/s.
Can the string have mass?
Not in this model. A physical pendulum uses the center of mass and the moment of inertia, a different formula.
How does T depend on l at g = 9.81?
Like the square root of l. l = 4 m is about 4.01 s, twice the period at 1 m.
How is a pendulum different from a spring with k and m?
Here T comes from length and g. A spring gets f from k and m. Putting l where k belongs does nothing useful.
Can I type 1.2 m with a comma?
Yes. 1,2 and 1.2 are the same l. At 9.81, T ≈ 2.20 s.
Knowledge sources
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.