Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 13 - Oscillations About Equilibrium - Problems and Conceptual Exercises - Page 446: 28

Answer

(a) $1.877s$ (b) $0.23m$ (c) $0.77m/s$ (d) $2.58m/s^2$

Work Step by Step

(a) We can find the time period as follows: $T=\frac{2\pi r}{v}$ We plug in the known values to obtain: $T=\frac{2\pi (0.23m)}{0.77m/s}$ $T=1.877s$ (b) We know that amplitude is the maximum distance from mean position. Thus, $A=0.23m$ (c) We can find the maximum speed as follows: $v_{max}=A(\frac{v}{r})$ as $A=r$ $v_{max}=\frac{rv}{r}=v$ $v_{max}=0.77m/s$ (d) We can find the magnitude of the maximum acceleration as follows: $a_{max}=\frac{v^2}{r}$ We plug in the known values to obtain: $a_{max}=\frac{(0.77m/s)^2}{0.23m}$ $a_{max}=2.58m/s^2$
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