Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 15 - Oscillations - Exercises and Problems: 29

Answer

The free-fall acceleration on Mars is $3.67~m/s^2$

Work Step by Step

We can write an expression for the period of the pendulum on the earth as: $T_e = 2\pi~\sqrt{\frac{L}{g}}$ We can write an expression for the period of the pendulum on Mars as: $T_m = 2\pi~\sqrt{\frac{L}{g_m}}$ We then divide the first expression by the second expression; $\frac{T_e}{T_m} = \frac{2\pi~\sqrt{\frac{L}{g}}}{2\pi~\sqrt{\frac{L}{g_m}}}$ $\frac{T_e}{T_m} = \sqrt{\frac{g_m}{g}}$ $g_m = (\frac{T_e}{T_m})^2~g$ $g_m = (\frac{1.50~s}{2.45~s})^2~(9.80~m/s^2)$ $g_m = 3.67~m/s^2$ The free-fall acceleration on Mars is $3.67~m/s^2$.
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.