Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 14 - Oscillations - Exercises and Problems: 24

Answer

$L = 0.54~m$

Work Step by Step

We can use the following equation for the period of a physical pendulum: $T = 2\pi~\sqrt{\frac{I}{mgl}}$ $T = 2\pi~\sqrt{\frac{\frac{1}{3}mL^2}{mg~(\frac{L}{2})}}$ $T = 2\pi~\sqrt{\frac{2L}{3g}}$ $L = \frac{T^2~(3g)}{2~(2\pi)^2}$ $L = \frac{(1.2~s)^2~(3)(9.80~m/s^2)}{2~(2\pi)^2}$ $L = 0.54~m$
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