Frequency of superimposed wave

Question

If two waves with frequency $f_1$ and $f_2$($f_1≠f_2$) are added together to form a superimposed wave then what will be the frequency of the resultant wave if -

1. the velocity of the two wave is same

2. the velocity is not same.

Answer 1

Since you have not tagged your question "special relativity", the answer is simple: wave travel velocity is totally unrelated to frequency, because frequency is what you detect at one single resting point. I.e. frequency is the same, no matter with what velocity the wave travels.

So the question boils down to what the character of a mixture of two frequencies is. The result will be amplitude modulation, i.e. the mixture will be an average frequency that is modulated at a frequency that is (half) the difference between the two component frequencies.

Let $$\omega_1 = \bar \omega + \Delta \omega \qquad \omega_2 = \bar \omega - \Delta \omega$$ or, equivalently, $$\bar \omega=\frac{1}{2}(\omega_1+\omega_2) \qquad \Delta \omega=\frac{1}{2}(\omega_1-\omega_2)$$ Then you can see the above fact most clearly in complex representation: $$\psi(t) = A_1\cdot \exp(i\omega_1 t) + A_2\cdot \exp(i\omega_2 t)=$$ $$=A_1\cdot \exp(i\bar \omega t+i\Delta \omega t)+A_2\cdot \exp(i\bar \omega t-i\Delta \omega t)=$$ $$=A_1\cdot \exp(i\bar \omega t)\cdot \exp(i\Delta \omega t)+A_2\cdot \exp(i\bar \omega t)\cdot \exp(-i\Delta \omega t)=$$ $$=\exp(i\bar \omega t)\cdot (A_1\cdot \exp(i\Delta \omega t)+A_2\cdot \exp(-i\Delta \omega t))$$ that is, $$\psi(t)=\exp(i\bar \omega t)\cdot \left[(A_1+A_2)\cdot \cos(\Delta \omega t)+(A_1-A_2)\cdot \sin(\Delta \omega t)\right]$$ The exponential factor in front of the square brackets is the average frequency oscillation, while the sines and cosines in the square brackets is the modulation with (half) the difference frequency, which has effectively a phase that depends on the amplitudes $A_1$ and $A_2$.

If the two frequencies are close together, you would see the modulation very well in a plot. The greater the frequency distance between both, the less the modulation will be visible in a plot, and you would probably just say that the faster wave sails on top of the slower one.