# Video: Finding the Variation of Period of an Oscillating Pendulum with Acceleration Due to Gravity

Ed Burdette

Find the ratio of the periods of a pendulum if the pendulum is first used on Earth and then on the Moon. Use a value of 1.63 m/s² for the acceleration due to gravity on the Moon.

03:10

### Video Transcript

Find the ratio of the periods of a pendulum if the pendulum is first used on Earth and then on the Moon. Use a value of 1.63 meters per second squared for the acceleration due to gravity on the Moon.

In this problem, weβll assume that the acceleration due to gravity on Earth, π sub e, is exactly 9.8 meters per second squared. In the problem statement, weβre told the value to use for the acceleration due to gravity on the Moon 1.63 meters per second squared. Weβll call that π sub m. And weβll call the acceleration due to gravity on Earth π sub e.

If the period of the pendulum on Earth is capital π sub e and we call the period of the pendulum on the Moon capital π sub m, then the problem is asking for the ratio of π sub m to π sub e. Even though itβs the same pendulum with the same mass and length in these different gravitational environments, the period of the pendulum will change. So π sub m divided by π sub e is not one. Letβs find out what it is.

To begin, letβs recall a relationship for the period of an oscillating pendulum. A pendulumβs period π is equal to two times pi times the square root of the length of the pendulum, capital πΏ, divided by the acceleration due to gravity. If we apply this relationship to our scenario, solving for π sub m and π sub e, then as we look at this equation, we see that some cancellation occurs. First, we can cancel out the two pie that appear in both the numerator and denominator. We can also cancel out the πΏ that appears in both the numerator and denominator.

Algebraically, this simplified fraction is equal to the square root of π sub e, the acceleration due to gravity on the Earth divided by π sub m, the acceleration due to gravity on the Moon. So π sub m, the period of the pendulum on the Moon divided by π sub e, the period of the pendulum on Earth, is equal to this fraction. When we enter in the values for π sub e and π sub m, we see that π sub m divided by π sub e is equal to the square root of 9.8 meters per second squared divided by 1.63 meters per second squared.

Entering these numbers on our calculator, we find that the ratio of the periods of this pendulum equals 2.45. This means that on the Moon, the pendulum would take 2.45 times longer to rock back and forth than on the Earth.