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For a simple pendulum, with very small, technically infinitesimal, swings, the period T is approximately 2 pisquare root (Length / gravity). From there, it is simple algebra.
T2 = 4 pi2 length / gravity
gravity = 4 pi2 length / T2
For more information, including how to compensate for circular error introduced by non-infinitesimal swings, please see the Related Link below.
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What is the maths of period of pendulum?
For small swings, and a simple pendulum:T = 2 pi root(L/g) where T is the time for one period, L is the length of the pendulum, and g is the strength of the gravitational field.
What is the relationship between the period of a pendulum and its length?
For small angles, the formula for a pendulum's period (T) can be approximated by the formula:T = 2 * pi * sqrt(L/g), where L is the length of the pendulum length, and g is acceleration due to gravity. See related link for Simple Pendulum.
How do you solve a problem in numerical analysis?
g
What is the difference in period for a pendulum on earth and a pendulum on moon?
The period of a simple pendulum swinging at a small angle is approximately 2*pi*Sqrt(L/g), where L is the length of the pendulum, and g is acceleration due to gravity. Since gravity on the moon is approximately 1/6 of Earth's gravity, the period of a pendulum on the moon with the same length will be approximately 2.45 times of the same pendulum on the Earth (that's square root of 6).
What is the effect of acceleration due to gravity on the time period of a simple pendulum?
T=2pi(l/g)1/2
