Physics: Principles with Applications (7th Edition)

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

Chapter 22 - Electromagnetic Waves - Problems - Page 642: 36

Answer

The wavelength of the AM wave is larger by a factor of $1.00\times10^2$.

Work Step by Step

See problem 34. The two frequencies are 980 kHz (AM) and 98.1 MHz (FM). Use equation 22–4, $c=\lambda f$. The lower frequency (AM) has the longer wavelength. $$\lambda_{AM}=\frac{c}{f}=\frac{3.00\times10^8m/s }{980\times10^3Hz}$$ $$\lambda_{FM}=\frac{c}{f}=\frac{3.00\times10^8m/s }{98.1\times10^6Hz}$$ Divide to find the ratio. $$\frac{\lambda_{AM}}{\lambda_{FM}}=\frac{98.1\times10^6Hz }{980\times10^3Hz }=1.00\times10^2$$
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