Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 14 - Exercises and Problems - Page 263: 49


$E =\frac{4\pi^2FA^2}{\lambda}$

Work Step by Step

We know that the energy is given by: $E=\mu \omega^2 A^2 vT$ $E=\frac{F}{v^2} \omega^2 A^2 vT$ $E=\frac{F}{v} \omega^2 A^2 T$ $E=\frac{F}{v} (\frac{2\pi}{T})^2 A^2 T$ $E=\frac{F}{v} (\frac{4\pi^2}{T}) A^2 $ $E =\frac{4\pi^2FA^2}{\lambda}$
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