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

Published by Pearson

# Chapter 15 - Oscillations - Exercises and Problems - Page 416: 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 $3.67~m/s^2$.

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.