Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 13 - Exercises and Problems: 44

Answer

The proof is below.

Work Step by Step

Equation 13.3 is: $m\frac{d^2x}{dt^2}=-kx$ If the given equation is a solution to equation 13.3, it will result in a true statement when plugged into equation 13.3. Thus, we find: $m\frac{d^2x}{dt^2}=-kx$ $m\frac{d^2(Asin\omega t)}{dt^2}=-kAsin\omega t$ $-mA\omega^2=-kA$ We know that $\omega^2 = \frac{k}{m}$, so we find: $-mA \frac{k}{m}=-kA$ $-kA=-kA$ This is true, so we see that the given equation is a solution to equation 13.3.
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