oscillation of earth and another planet?

Question

What is the ratio of the periods of two identical pendulums oscillating on a planet X and on the Earth? (The acceleration of free fall on the planet X is 1.62 m/s2)

I have no idea...

Answer 1

Very easy. Period of a pendulum depends on two things only: length of the pendulum, and gravitational acceleration.
Period = 2π * sqrt[ length / g ]

Thankfully, there is no need to do any gory math because all that is wanted is the ratios.
g Earth = 9.81 m/s^2
g X = 1.62 m/s^2

So take square root of 9.81/1.62 to get 2.46. So ratio of period on Earth to X is 2.46 to 1

Answer 2

A pendulum of length L subject to acceleration due to gravity g has a period T of approximately 2π √(L/g)

So

(T_X) / (T_earth) = [2π √(L / g_X)] / [2π√(L / g_earth)] = √[(g_earth) / (g_X)]