Answer
$0.13m$
Work Step by Step
$v_{max}=\sqrt (\frac{k}{m})A$
Therefore, $A=\sqrt (\frac{m}{k})v_{max}$
Now, we have
$v_{max}=0.50 m/s$
m = 1.0 kg
k=15N/m
So, $A=\sqrt (\frac{1}{15})\times0.50=0.13m$
College Physics (7th Edition)
by Wilson, Jerry D.; Buffa, Anthony J.; Lou, Bo
Published by
Pearson
ISBN 10:
0-32160-183-1
ISBN 13:
978-0-32160-183-4
Chapter 13 - Vibrations and Waves - Learning Path Questions and Exercises - Exercises - Page 484: 7
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