A “seconds” pendulum has a period of exactly 2.000 s—each one-way swing takes 1.000 s

Chapter 11, Problem 82GP

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QUESTION:

Problem 82GP

A “seconds” pendulum has a period of exactly 2.000 s—each one-way swing takes 1.000 s. (a) What is the length of a seconds pendulum in Austin, Texas, where g = 9.793 m/s2? (b) If the pendulum is moved to Paris, where g = 9.809 m/s2, by how many millimeters must we lengthen the pendulum? (c) What would be the length of a seconds pendulum on the Moon, where g = 1.62 m/s2?

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QUESTION:

Problem 82GP

A “seconds” pendulum has a period of exactly 2.000 s—each one-way swing takes 1.000 s. (a) What is the length of a seconds pendulum in Austin, Texas, where g = 9.793 m/s2? (b) If the pendulum is moved to Paris, where g = 9.809 m/s2, by how many millimeters must we lengthen the pendulum? (c) What would be the length of a seconds pendulum on the Moon, where g = 1.62 m/s2?

ANSWER:

Solution 82GP

Step 1 of 3

Part a

We are required to calculate the length of the pendulum in Austin, Texas from other values given.

Given:

If the length of the pendulum in Austin is

From the relation,

The length of the pendulum in Austin is 0.9921 m.

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