Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems - Page 239: 40

Answer

1680 N/m

Work Step by Step

Using the equation that equates omega to the spring constant and the mass and using the equation $\omega= 2\pi f$, we find: $k = 4\pi^2f^2m=4\pi^2(4\times10^{13})^2(2.66\times10^{-26})=\fbox{1680 N/m}$
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