College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 10 - Problems - Page 400: 50

Answer

The maximum velocity is $0.157~m/s$ The maximum acceleration is $24.7~m/s^2$

Work Step by Step

We can find the maximum velocity: $v_m = A~\omega$ $v_m = A~(2\pi~f)$ $v_m = (1.00\times 10^{-3}~m)~(2\pi)~(25.0~Hz)$ $v_m = 0.157~m/s$ We can find the maximum acceleration: $a_m = A~\omega^2$ $a_m = A~(2\pi~f)^2$ $a_m = (1.00\times 10^{-3}~m)~(2\pi)^2~(25.0~Hz)^2$ $a_m = 24.7~m/s^2$ The maximum velocity is $0.157~m/s$ The maximum acceleration is $24.7~m/s^2$
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.