College Physics (4th Edition)

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

Chapter 12 - Problems - Page 464: 47

Answer

The speed of the child at the bottom of the swing is $8.37~m/s$

Work Step by Step

Let $v_c$ be the child's speed at the bottom of the swing. We can use the equation for the Doppler effect when the observer is approaching to find an expression for the higher frequency $f_h$: $f_h = \left(\frac{v+v_c}{v}\right)~f_s$ We can use the equation for the Doppler effect when the observer is moving away to find an expression for the lower frequency $f_l$: $f_l = \left(\frac{v-v_c}{v}\right)~f_s$ Note that $f_h = 1.05~f_l$: $f_h = 1.05~f_l$ $\left(\frac{v+v_c}{v}\right)~f_s = 1.05\left(\frac{v-v_c}{v}\right)~f_s$ $v+v_c = 1.05~(v-v_c)$ $2.05~v_c = 0.05~v$ $v_c = \frac{0.05~v}{2.05}$ $v_c = \frac{(0.05)~(343~m/s)}{2.05}$ $v_c = 8.37~m/s$ The speed of the child at the bottom of the swing is $8.37~m/s$.
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