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Wavelength and Frequency in a Waveguide

  1. Mar 2, 2017 #1
    • Moved from a technical forum, so homework template missing.
    The problem states that the wavelength and frequency in a waveguide are related by:

    ##\lambda = \frac{c}{\sqrt{f^2 - f_0^2}}##

    then asks to express the group velocity ##v_g## in terms of c and the phase velocity ##v_p = \lambda f##
    Solution:

    Given that ##\omega = 2\pi f##,

    ## \omega(k) = 2\pi \sqrt{\frac{c^2}{\lambda^2} + f_0^2} = 2 \pi \sqrt{\frac{c^2 k^2}{4 \pi^2} + f_0^2}##

    So I'm trying to understand how they got omega there. I figured that by using ## f \lambda = c, f = \frac{c}{\lambda}##, we could change it from a wavelength term to a frequency term, then by multiplying both sides by 2pi, it is now in terms of radians per second.

    ##f = \frac{c}{\lambda} = c \frac{\sqrt{f^2 - f_0^2}}{c}##

    So I'm trying to understand how they got that plus sign in there. Is this some simple algebra thing that I'm completely missing or is my reasoning wrong?
     
  2. jcsd
  3. Mar 3, 2017 #2

    rude man

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    Gold Member

    Just algebra! Take λ = c/√(f2-f02), multiply both sides, solve for f. Then ω(k) = 2πf.
     
  4. Mar 3, 2017 #3
    @rude man I see this now... Thank you.
     
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