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: 20

Answer

$L = 0.357~m$

Work Step by Step

We can find the period as: $T = \frac{time}{oscillations}$ $T = \frac{12.0~s}{10~oscillations}$ $T = 1.20~s$ We then find the length of the string as: $T = 2\pi~\sqrt{\frac{L}{g}}$ $L = \frac{T^2~g}{(2\pi)^2}$ $L = \frac{(1.20~s)^2~(9.80~m/s^2)}{(2\pi)^2}$ $L = 0.357~m$
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