Question

The free-fall acceleration on the moon is 1.62 m/s2

m
/
s
2
. What is the length of a pendulum whose period on the moon matches the period of a 1.70-m-long pendulum on the earth?

Answer 1

To find the length of a pendulum whose period on the moon matches the period of a 1.70-m-long pendulum on Earth, you will need to use the formula for the period of a pendulum:

T = 2π√(L/g)

where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

First, you need to find the period of the 1.70-m-long pendulum on Earth:

T = 2π√(L/g) = 2π√(1.70 m/9.81 m/s^2) = 3.18 s

Since the period of the pendulum on the moon is the same as the period of the pendulum on Earth, you can use the same period value to solve for the length of the pendulum on the moon:

T = 2π√(L/g) = 3.18 s
g = 1.62 m/s^2

Solve for L:

L = (T^2 g) / (4π^2) = (3.18 s)^2 (1.62 m/s^2) / (4π^2) = 0.31 m

Therefore, the length of the pendulum on the moon is 0.31 meters.