# Group velocity of a wave

• Jul 4th 2011, 10:57 AM
techmath
Group velocity of a wave
The phase velocity for waves in a determined material is given by: v= $v=C_1 + C_2 \lambda$

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

My calculation:

$\omega=v \cdot k$

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

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

This is physics, but purely a mathematical difficulty.
• Jul 4th 2011, 11:36 AM
skeeter
Re: Group velocity of a wave
Quote:

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

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

My calculation:

$\omega=v \cdot k$

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

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

This is physics, but purely a mathematical difficulty.

$\omega = v \cdot k = (C_1+C_2\lambda) \cdot \frac{2\pi}{\lambda} = \frac{2\pi \cdot C_1}{\lambda} + 2\pi \cdot C_2$

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

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

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

note that ...

$\frac{d\omega}{dk} = \frac{d\omega}{d\lambda} \cdot \frac{d\lambda}{dk}$
• Jul 4th 2011, 01:32 PM
techmath
Re: Group velocity of a wave
Thank you very much.

(Evilgrin)