Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 14 - Oscillations - Exercises and Problems: 22

Answer

The length of the pendulum is 0.33 meters.

Work Step by Step

We can find the period of a 2.0-meter long pendulum on the earth as: $T = 2\pi~\sqrt{\frac{L}{g}}$ $T = 2\pi~\sqrt{\frac{2.0~m}{9.80~m/s^2}}$ $T = 2.84~s$ We can find the length $L_m$ of a pendulum that would have the same period on the moon, which has an acceleration due to gravity of $g_m = 1.63~m/s^2$. $T = 2\pi~\sqrt{\frac{L_m}{g_m}}$ $L_m = \frac{T^2~g_m}{(2\pi)^2}$ $L_m = \frac{(2.84~s)^2(1.63~m/s^2)}{(2\pi)^2}$ $L_m = 0.33~m$ The length of the pendulum is 0.33 meters.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.