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: 81

Answer

The proof is below.

Work Step by Step

We first use the distance formula to find: $ A = \sqrt{a^2 + (-b)^2} \\ A = \sqrt{a^2+b^2}$ We now are asked to find the phase constant. To do so, we set the derivative of the equation equal to zero: $0=-asin\omega t - b cos\omega t $ This becomes: $asin \phi = b cos \phi $ $atan \phi = b $ $\phi = tan^{-1}(\frac{b}{a})$
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