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: 55

Answer

(a) $L = 7.5~m$ (b) $v_{max} = 0.45~m/s$

Work Step by Step

(a) We can find the length of the chain as: $T = 2\pi~\sqrt{\frac{L}{g}}$ $L = \frac{T^2~g}{(2\pi)^2}$ $L = \frac{(5.5~s)^2(9.80~m/s^2)}{(2\pi)^2}$ $L = 7.5~m$ (b) We can find the amplitude of the motion as: $\frac{A}{L} = sin(\theta)$ $A = L~sin(\theta)$ $A = (7.5~m)~sin(3.0^{\circ})$ $A = 0.39~m$ We can find the maximum speed as: $v_{max} = A~\omega$ $v_{max} = A~\sqrt{\frac{g}{L}}$ $v_{max} = (0.39~m)~\sqrt{\frac{9.80~m/s^2}{7.5~m}}$ $v_{max} = 0.45~m/s$
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