Question

Consider two waves $y_1=a\cos(\omega t-kz)$ and $y_2=a\cos[(\omega+\Delta \omega)t-(k+\Delta k)z]$. The group velocity of the superposed wave will be ($\Delta \omega \ll \omega$ and $\Delta k \ll k$)

• A. $\dfrac{\omega-\Delta \omega}{k-\Delta k}$
• B. $\dfrac{2\omega+\Delta \omega}{2k+\Delta k}$
• C. $\dfrac{\Delta \omega}{\Delta k}$
• D. $\dfrac{\omega+\Delta \omega}{k+\Delta k}$

Hint:

Group velocity comes from the slope of $\omega(k)$, not the ratio $\omega/k$.

Answer 1

The Correct Option is C

Step 1: Write the standard formula for group velocity.
Group velocity $v_g = \dfrac{d\omega}{dk}$.

Step 2: For two close frequencies.
When two waves differ slightly in $(\omega, k)$, the group velocity is approximated by the finite difference: $v_g = \dfrac{\Delta \omega}{\Delta k}$.

Step 3: Why the other options fail.
(A), (B), and (D) give phase velocity-like expressions, not group velocity. Only the ratio of differences gives the envelope propagation speed.

Step 4: Conclusion.
Therefore the group velocity is $\Delta \omega / \Delta k$.