Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 11 - Oscillations and Waves - General Problems - Page 325: 69

Answer

$2.6 \times 10^{13}Hz$

Work Step by Step

The spring constant of the bond stays the same. Use equation 11-6b to relate frequency, mass, and spring constant. $$f=\frac{1}{2 \pi}\sqrt{\frac{k}{m}}$$ $$k=4 \pi^2f^2m $$ $$ (3.7 \times 10^{13}Hz)^2(16amu) = (f)^2(32amu)$$ Solve for f = $2.6 \times 10^{13}Hz$.
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