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Find the ratio of the periods of a pendulum if the pendulum is first used on Earth and then on the Moon.
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Use a value of 1.63 meters per second squared for the acceleration due to gravity on the Moon.
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In this problem, weβll assume that the acceleration due to gravity on Earth, π sub e, is exactly 9.8 meters per second squared.
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In the problem statement, weβre told the value to use for the acceleration due to gravity on the Moon 1.63 meters per second squared.
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Weβll call that π sub m.
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And weβll call the acceleration due to gravity on Earth π sub e.
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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.
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Even though itβs the same pendulum with the same mass and length in these different gravitational environments, the period of the pendulum will change.
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So π sub m divided by π sub e is not one.
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Letβs find out what it is.
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To begin, letβs recall a relationship for the period of an oscillating pendulum.
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A pendulumβs period π is equal to two times pi times the square root of the length of the pendulum, capital πΏ, divided by the acceleration due to gravity.
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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.
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First, we can cancel out the two pie that appear in both the numerator and denominator.
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We can also cancel out the πΏ that appears in both the numerator and denominator.
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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.
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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.
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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.
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Entering these numbers on our calculator, we find that the ratio of the periods of this pendulum equals 2.45.
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This means that on the Moon, the pendulum would take 2.45 times longer to rock back and forth than on the Earth.