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Wavelength and Frequency in a Waveguide

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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?
 

rude man

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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}##
Just algebra! Take λ = c/√(f2-f02), multiply both sides, solve for f. Then ω(k) = 2πf.
 
103
4
@rude man I see this now... Thank you.
 

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