1
$\begingroup$

Given :

$\phi_1(x,t)=A\cdot \cos(k_1x-\omega_1t)\hat{x},\phi_2(x,t)=A\cdot \cos(k_2x-\omega_2t)\hat{x}$ with no difference phase,$k_1-k_2 \ll k_1,k_2,\omega_1-\omega_2 \ll \omega_1,\omega_2 .$

I have to find group velocity of $\phi_{total}.$

$\phi_{total}=\phi_1(x,t)+\phi_2(x,t)=A\cdot \cos(k_1x-\omega_1t)+A\cdot \cos(k_2x-\omega_2t)=2A\Big(\cos(\frac{x(k_1+k_2)-t(\omega_1+\omega_2)}{2})\cos(\frac{x(k_1-k_2)-t(\omega_1-\omega_2)}{2})\Big)$

I know that while the wave equation is $\phi(x,t)=A\cos(kx-\omega t)$ the group velocity is $v_g=\frac{d\omega}{dk}$.

$\phi_{total}$ is a multiplication of 2 "cos" function , so I get a little bit confused how to find the group velocity.

Any help is welcome, thanks !

$\endgroup$

1 Answer 1

0
$\begingroup$

You are being asked for the velocity of the ``slowest'' of the two cosines --- as that is factor that determines how fast the amplitude modulation is moving. Look at the factors. Does not one of them look like an approxiamtion to $d\omega/dk$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.