= 2π√(l/g).
And if you try you will find that this is so. For instance, take a thread 39-1/7 inches long, that is 3·25 feet. Hang anything heavy from one end of it, and cause it to swing round and round in a small circle. Now g the acceleration of gravity = 32·2 feet per second. π the ratio of the circumference of a circle to its diameter = 3·14. From which it follows that the time of rotation = 2 × 3·14√(3·25/32·2) seconds = 2 seconds. But if we look at the rotating body sideways, it appears to act as a pendulum; it matters nothing whether we swing it round and round or to and fro. For in any case the accelerative force tending to bring it back to a position of rest is always proportional to the distance of displacement, and, therefore, its time of motion must always be 2π√(l/g) and its motion harmonic.
The length of a seconds pendulum, that is a pendulum that makes its double swing in two seconds, will therefore be
l = 4/((2π)²) × g feet
= (g × 12)/π² inches
= 39·14 inches.