Wavelength and Frequency in a Waveguide

[James Brady]

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]

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.

 

[James Brady]

@rude man I see this now... Thank you.