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 264: 69

Answer

The proof is below.

Work Step by Step

We know that the second partial derivative of equation 14.3 with respect to x is: $-Ak^2cos(kx+\omega t)$ We know that the second partial derivative of equation 14.3 with respect to t is: $-A \omega^2cos(kx+\omega t)$ We plug these into equation 14.5 to find: $-A \omega^2cos(kx+\omega t) =\frac{1}{v^2}(-Ak^2cos(kx+\omega t))$ $-A \omega^2=\frac{1}{v^2}(-Ak^2)$ We substitute the given value of v to find: $-A\omega^2 = -A\omega^2$ Thus, the proof is complete.
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