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 !


1 Answer 1


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 $d\omega/dk$?


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