How long is a simple pendulum with a period of 5 seconds? How long would this pendulum have to be if it were to operate on the moon

Answer 1

The relevant equation is

T = 2π√(L/g)

On Earth, g = 9.8 m/s². We can use this info and T = 5 s to get the length:

5 = 2π√(L/9.8)

L = 6.2 m

Then use this length the fact that on the moon g is 1.62 m/s² , one sixth that of earth:

T = 2π√(6.2/1.62)

T = 12.3 s

We can check our answer/solve this same problem using a shortcut. Note that

T_moon = 2π√(L/g_moon)

and

T_earth = 2π√(L/g_earth)

dividing the top equation by the bottom one, the pi's and L's cancel:

T_moon/T_earth = √(g_earth/g_moon)

Then plug in:



T_moon/(5 s) = √6

T_moon = 12.3 s