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: 26

Answer

(a) The period is 4.00 seconds. (b) The period is 5.66 seconds. (c) The period is 2.83 seconds. (d) The period is 4.00 seconds.

Work Step by Step

We can find an expression for the original period as: $T = 2\pi~\sqrt{\frac{L}{g}} = 4.00~s$ (a) Since the period does not depend on the mass, the period is still 4.00 seconds. (b) $T' = 2\pi~\sqrt{\frac{2L}{g}}$ $T' = \sqrt{2}\times ~2\pi~\sqrt{\frac{L}{g}}$ $T' = \sqrt{2}~T$ $T' = \sqrt{2}~(4.00~s)$ $T' = 5.66~s$ (c) $T' = 2\pi~\sqrt{\frac{L/2}{g}}$ $T' = \frac{1}{\sqrt{2}}\times ~2\pi~\sqrt{\frac{L}{g}}$ $T' = \frac{1}{\sqrt{2}}~T$ $T' = \frac{1}{\sqrt{2}}~(4.00~s)$ $T' = 2.83~s$ (d) Since the period does not depend on the amplitude, the period is still 4.00 seconds.
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.