Pendulum Period Calculator

Results

How amplitude changes the period

How to Use the Pendulum Period Calculator

1. 1. Pick the unknown. Solve for the period T, the length L, or gravity g. The box for the unknown shows the answer; fill in the other two. Each has a unit menu, from millimetres to feet and from seconds to minutes.
2. 2. Set g. The default is standard gravity, 9.80665 m/s². The Set g for menu loads the surface gravity of the Moon, Mars, Venus and every other planet from NASA's Planetary Fact Sheet.
3. 3. Enter the amplitude. This is the angle from the vertical at which the bob is released. Leave Correct for the amplitude ticked to use the exact formula, or untick it for the textbook small-angle result.
4. 4. Read the results and the working. Besides T you get the frequency, the speed at the bottom of the swing, the size of the amplitude correction and a table of how the period grows with angle.

The Pendulum Period Formula

T₀ = 2π √(L / g)

T = T₀ × 2K(sin(θ₀/2)) / π = T₀ / AGM(1, cos(θ₀/2))

T ≈ T₀ (1 + θ₀²/16 + 11θ₀⁴/3072 + …)   (θ₀ in radians)

For a small swing the restoring force on the bob is almost proportional to its displacement, so the motion is simple harmonic and the period depends only on the length of the string and on g. Notice what is missing: the mass of the bob and, for small angles, the amplitude. Doubling the length multiplies the period by √2; a bob four times heavier swings at the same rate.

The small-angle formula treats sin θ as θ. As the amplitude grows that stops being true, and the real period lengthens. The exact result involves K, the complete elliptic integral of the first kind. The calculator evaluates it through the arithmetic–geometric mean (AGM), which converges in a handful of steps and agrees with the tabulated values to at least eight digits. The series form is the same correction written as a power series, and the table above shows where its first two terms stop being adequate. OpenStax University Physics derives the small-angle result in section 15.4.

Worked Examples

1. A one-metre pendulum on Earth. With L = 1 m and g = 9.80665 m/s², T₀ = 2π√(1/9.80665) = 2.006 s. Released from 10° the correction is 1.00191, so T = 2.010 s, and the bob passes the bottom at 0.5459 m/s.
2. The seconds pendulum. A pendulum with a two-second period (one second each way) needs L = 0.9936 m at standard gravity, which is why the length of a seconds pendulum was once proposed as a measure of length.
3. Measuring g. A student times a one-metre pendulum at 2.007 s. Without the amplitude correction, g = 4π²L/T² = 9.801 m/s². A 10° release makes the true small-angle period slightly shorter than the measured one, so using the correction gives g = 9.838 m/s². The difference is the systematic error from the amplitude.
4. On the Moon. With g = 1.62 m/s², the same one-metre pendulum has T₀ = 4.937 s, about 2.46 times slower than on Earth.

The calculator's table lists the exact factor for each amplitude. Below about 10° the small-angle formula is within 0.2% (T/T₀ = 1.0019 at 10°), which is why labs use it. At 20° the period is already 0.8% longer (1.0077), at 45° it is 4.0% longer, at 60° 7.3% longer, and a release from 90° takes 18% longer than the textbook value (1.1803). The first two terms of the series, 1 + θ₀²/16, hold up to about 30° and then fall behind.

Assumptions and Limits

• Simple pendulum. A point mass on a massless, inextensible string. A real bob has size: measure L to its centre, and for a heavy bob use the physical-pendulum formula with its moment of inertia instead.
• One plane, no damping. Air resistance and friction at the pivot shorten the swing over time and change the period very slightly.
• Constant g. Near Earth's surface g varies from about 9.78 m/s² at the equator to 9.83 m/s² at the poles, which changes a one-metre pendulum's period by about 0.3%.
• Rigid string. A bouncing or stretching string, or a bob that is not released from rest, is outside the model.

Frequently Asked Questions

What is the formula for the period of a pendulum?

T = 2π√(L/g), where L is the length from the pivot to the bob's centre in metres and g is the acceleration due to gravity in m/s². It holds for small swings, below about 10° from the vertical. A 1 m pendulum on Earth has a period of about 2.01 seconds.

Does the mass of the bob affect the period?

No. Gravity's pull on the bob and the bob's inertia both scale with mass, so mass cancels from the equation. A heavy and a light bob on equal strings swing together, as Galileo noticed.

How does the amplitude change the period?

For small angles it does not, to a good approximation. For larger angles the period grows: 0.19% longer at 10°, 1.74% at 30°, 7.32% at 60° and 18.03% at 90°. The calculator's correction uses the exact elliptic-integral result.

How do I find g with a pendulum?

Measure the length and time many swings, for example 20, then divide for the period. Rearranging the formula gives g = 4π²L/T². Use a small amplitude, or the correction here, and time more swings to cut the effect of your reaction time.

What is a seconds pendulum?

It is a pendulum with a period of exactly two seconds, taking one second per swing. At standard gravity its length is 0.9936 m, close to one metre.

Why is a pendulum slower on the Moon?

The period varies as 1/√g, and the Moon's surface gravity of 1.62 m/s² is about one-sixth of Earth's. A given pendulum swings √6 ≈ 2.46 times slower there.