College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 22 - Problems - Page 866: 61


(a) $f_2 \lt f_1$ (b) $f_2-f_1 = -10,320~Hz$

Work Step by Step

(a) The source and observer are moving away from each other so the observed frequency is less than the emitted frequency. Therefore, $f_2 \lt f_1$ (b) Let $f_s$ be the frequency observed by the speeder. We can find an expression for $f_s$: $f_s = f_1~(1-\frac{v}{c})$ We can find the frequency $f_2$: $f_2 = f_s~(1-\frac{v}{c})$ $f_2 = [f_1~(1-\frac{v}{c})]~(1-\frac{v}{c})$ $f_2 = f_1~(1-\frac{v}{c})^2$ $f_2 = (36.0~GHz)~(1-\frac{43.0~m/s}{3.0\times 10^8~m/s})^2$ $f_2 = 35.99998968~GHz$ We can find $f_2-f_1$: $f_2-f_1 = 35.99998968~GHz-36.0~GHz$ $f_2-f_1 = -0.00001032~GHz$ $f_2-f_1 = -10,320~Hz$
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