University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 14 - Periodic Motion - Problems - Discussion Questions - Page 459: Q14.11

Answer

See explanation.

Work Step by Step

The period of a simple pendulum is $T=2 \pi\sqrt{\frac{L}{g}}$. a. Accelerating the elevator upward effectively increases the perceived g, the acceleration due to gravity, inside the elevator. This decreases the period T. b. With zero acceleration, the perceived g, the acceleration due to gravity, is unchanged inside the elevator. This doesn’t affect the period T. c. Accelerating the elevator downward effectively decreases the perceived g, the acceleration due to gravity, inside the elevator. This increases the period T. d. Accelerating the elevator downward at g effectively decreases the perceived g to zero inside the elevator. Under these “weightless” conditions a pendulum would not work.
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