# Thread: Group velocity of a wave

1. ## Group velocity of a wave

The phase velocity for waves in a determined material is given by: v= $\displaystyle v=C_1 + C_2 \lambda$

Both Cs are constants. $\displaystyle \lambda$ is wavelength. Determine the group velocity $\displaystyle v_g$!

My calculation:

$\displaystyle \omega=v \cdot k$

$\displaystyle k=\frac{2 \pi}{\lambda}$

$\displaystyle v_g=\frac{d \omega}{dk} = \frac{d(vk)}{dk}= (C_1 +C_2 \lambda)'k + (C_1 +C_2 \lambda)k' = C_2k + (C_1 + C_2 \lambda)(-\frac{2 \pi}{\lambda^{2}})$

This is physics, but purely a mathematical difficulty.

2. ## Re: Group velocity of a wave

Originally Posted by techmath
The phase velocity for waves in a determined material is given by: v= $\displaystyle v=C_1 + C_2 \lambda$

Both Cs are constants. $\displaystyle \lambda$ is wavelength. Determine the group velocity $\displaystyle v_g$!

My calculation:

$\displaystyle \omega=v \cdot k$

$\displaystyle k=\frac{2 \pi}{\lambda}$

$\displaystyle v_g=\frac{d \omega}{dk} = \frac{d(vk)}{dk}= (C_1 +C_2 \lambda)'k + (C_1 +C_2 \lambda)k' = C_2k + (C_1 + C_2 \lambda)(-\frac{2 \pi}{\lambda^{2}})$

This is physics, but purely a mathematical difficulty.
$\displaystyle \omega = v \cdot k = (C_1+C_2\lambda) \cdot \frac{2\pi}{\lambda} = \frac{2\pi \cdot C_1}{\lambda} + 2\pi \cdot C_2$

$\displaystyle \frac{d\omega}{d\lambda} = -\frac{2\pi \cdot C_1}{\lambda^2}$

$\displaystyle \lambda = \frac{2\pi}{k}$

$\displaystyle \frac{d\lambda}{dk} = -\frac{2\pi}{k^2}$

note that ...

$\displaystyle \frac{d\omega}{dk} = \frac{d\omega}{d\lambda} \cdot \frac{d\lambda}{dk}$

3. ## Re: Group velocity of a wave

Thank you very much.