Question

How long is a pendulum with a period of 1.0 S on the moon which has 1/6 of the earths gravity

Answer 1

For small deflections, T = 2*pi*sqrt(L / g) where T is period, L is length and g is gravity. Setting the equations to the same period, 2*pi*sqrt(3.85 / g) = 2*pi*(L / (1/6 * g)) The equation reduces to 3.85 m = 6 * L so L = 0.642 m chrsclrk · 7 years ago