Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 15 - Oscillations - Problems - Page 439: 57


$\Delta E=6.0$ percent

Work Step by Step

The mechanical energy is expressed as the maximum elastic potential energy, or $$E=U_e=\frac{1}{2}kA^2$$ Therefore, if the initial amplitude is $A$ and the next amplitude is $0.97A$ (3% lower than the initial amplitude), the new potential energy at the first oscillation is $$E_f=\frac{1}{2}k(0.97A)^2=0.47kA^2$$ Remember that the initial energy is $E_o=0.50kA^2$. Therefore, the percentage of energy lost can be expressed as $$=\frac{\Delta E}{E_o}=\frac{0.50kA^2-0.47kA^2}{0.50kA^2}=6.0 percent$$
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.