Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
11
Vibration and Waves
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11-1 to 11-3: Simple Harmonic Motion
11-4: Simple Pendulum
11-7 and 11-8: Waves
11-9: Energy Transported by Waves
11-11: Interference
11-12: Standing Waves; Resonance
11-13: Refraction
11-14: Diffraction

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 38
Q

AM radio signals have frequencies between 550 kHz and 1600 kHz (kilohertz) and travel with a speed of 3.0×108 m/s3.0 \times 10^8 \textrm{ m/s}. What are the wavelengths of these signals? On FM the frequencies range from 88 MHz to 108 MHz (megahertz) and travel at the same speed. What are their wavelengths?

A
AM: λmin=190 m, \textrm{AM: } \lambda_{min} = 190 \textrm{ m, } λmax=550 m, FM: λmin=2.8 m, \lambda_{max} = 550 \textrm{ m, FM: } \lambda_{min} = 2.8 \textrm{ m, } λmax=3.4 m \lambda_{max} = 3.4 \textrm{ m}
Giancoli 7th Edition, Chapter 11, Problem 38 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The speed of a wave is its frequency times its wavelength and we can divide this formula by f on both sides. And we get wavelength is the wave speed divided by its frequency. So, for AM signals, the minimum wavelength is going to be 3 times 10 to the 8 meters per second speed divided by its maximum frequency of 1600 times 10 to the 3 hertz. We have to convert the kilohertz into hertz and that is 190 meters for the minimum wavelength. And then for the maximum wavelength we get the same speed divided by 550 kilohertz or 550 times 10 to the 3 hertz which gives 550 meters. For FM signals we have that the maximum wavelength is going to be 3 times 10 to the 8 divided by 88 times 10 to the 6 hertz, it's megahertz, so that means times 10 to the 6. That gives 3.4 meters. And then divide that speed by 108 times 10 to the 6 to get to 0.8 as the minimum wavelength for FM signals.

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