Interferensi gelombang dapat terjadi untuk gelombang-gelombang dengan frekuensi berbeda. (a) Tunjukkan bahwa resultan dari dua gelombang: y₁(x, t) = A sin (k₁x

[tex]\displaystyle y_1+y_2=A\sin(k_1x-\omega_1t)+A\sin(k_2x-\omega_2t)\\y=A(\sin(k_1x-\omega_1t)+\sin(k_2x-\omega_2t))\\y=A(2\sin\frac12(k_1x-\omega_1t+k_2x-\omega_2t)\cos\frac12(k_1x-\omega_1t-k_2x+\omega_2t))\\y=2A\sin\frac12((k_1+k_2)x-(\omega_1+\omega_2)t)\cos\frac12((k_1-k_2)x-(\omega_1-\omega_2)t))\\\\\text{misal :}\\k'=\frac12(k_1+k_2)\wedge\omega'=\frac12(\omega_1+\omega_2)\\\Delta k=k_1-k_2\wedge\Delta\omega=\omega_1-\omega_2\\\\y=2A\sin(k'x-\omega't)\cos\frac12(\Delta kx-\Delta\omega t)\\\boxed{\boxed{y=2A\cos\frac12(\Delta kx-\Delta\omega t)\sin(k'x-\omega't)}}[/tex]

[tex]\displaystyle \frac{\omega'}{k'}=\frac{\frac12(\omega_1+\omega_2)}{\frac12(k_1+k_2)}\\\frac{\omega'}{k'}=\frac{2\pi f_1+2\pi f_2}{\frac{2\pi}{\lambda_1}+\frac{2\pi}{\lambda_2}}\\\frac{\omega'}{k'}=\frac{\frac{v_1}{\lambda_1}+\frac{v_2}{\lambda_2}}{\frac{1}{\lambda_1}+\frac{1}{\lambda_2}}\\\frac{\omega'}{k'}=\frac{\frac{\lambda_2v_1+\lambda_1v_2}{\lambda_1\lambda_2}}{\frac{\lambda_1+\lambda_2}{\lambda_1\lambda_2}}}}[/tex]
[tex]\displaystyle \boxed{\boxed{\frac{\omega'}{k'}=\frac{\lambda_2v_1+\lambda_1v_2}{\lambda_1+\lambda_2}}}[/tex]