# Chapter 10 - Problems - Page 399: 39

(a) The period of vibration is $1.0\times 10^{-6}~s$ (b) The angular frequency is $6.3\times 10^6~rad/s$

#### Work Step by Step

(a) We can find the period of vibration: $T = \frac{1}{f} = \frac{1}{1.0\times 10^6~Hz} = 1.0\times 10^{-6}~s$ The period of vibration is $1.0\times 10^{-6}~s$ (b) We can find the angular frequency: $\omega = 2\pi~f = (2\pi)(1.0\times 10^6~Hz) = 6.3\times 10^6~rad/s$ The angular frequency is $6.3\times 10^6~rad/s$

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