Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 11 - Oscillations and Waves - Problems - Page 324: 38

Answer

AM $\lambda_{min}\approx 190 m$ $\lambda_{max}\approx 550 m$ FM $\lambda_{min}\approx 2.8 m$ $\lambda_{max}\approx 3.4 m$

Work Step by Step

The wave speed is given by $v=\lambda f$. The minimum wavelength corresponds to the maximum frequency, and vice versa. First, AM radio. $$\lambda_{min}=\frac{v}{f_{max}}=\frac{3.0 \times 10^8 m/s}{1600 \times 10^3 Hz}\approx 190 m$$ $$\lambda_{max}=\frac{v}{f_{min}}=\frac{3.0 \times 10^8 m/s}{550 \times 10^3 Hz}\approx 550 m$$ Next, FM radio. $$\lambda_{min}=\frac{v}{f_{max}}=\frac{3.00 \times 10^8 m/s}{108 \times 10^6 Hz}\approx 2.8 m$$ $$\lambda_{max}=\frac{v}{f_{min}}=\frac{3.00 \times 10^8 m/s}{88 \times 10^6 Hz}\approx 3.4 m$$
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