Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 15 - Oscillations - Problems - Page 436: 6a

Answer

$628318.5 rad/s$

Work Step by Step

As we know that angular $\omega$ and time period $T$ are mutually related as follows $\omega=\frac{2\pi}{T}$ Putting the values,we get $\omega=\frac{2\times3.1415}{1\times10^{-5}}$ $\omega= 628318.5 rad/s$
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