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Let's assume two waves with same phases and same wave speed $c$ in the same plane overlap, thus creating a constructive interference.

The elongation and speed of both waves is added to create the new interference wave:

$$s_{1} + s_{2} = s_{3}$$ and $$v_{1} + v_{2} = v_{3}$$

Due to energy conservation, the sum of kinetic energies and potential energies of waves 1 and 2 should be equal to the kinetic and potential energy of wave 3. Simplifying and using the same mass each time:

$$\frac{mv_{1}^{2}}{2} + mgh_{1} + \frac{mv_{2}^{2}}{2} + mgh_{2} = \frac{m(v_{1}+v_{2})^2}{2} + mg(h_{1} + h_{2})$$ $$\frac{mv_{1}^{2}}{2} + \frac{mv_{2}^{2}}{2} = \frac{m(v_{1}^2+v_{2}^2 + 2v_{1}v_{2})}{2}$$ $$0 = mv_{1}v_{2}$$ Obviously I'm forgetting something here, but what?

Narusan
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  • @ThePhoton The linked question is talking about EM-Waves, where the form of energy is different (and thus the energy conversation based on different formulas) so I'd argue against this being an exact duplicate – Narusan Sep 13 '17 at 21:29
  • What are $v_1$ , $v_2$, $s_2$, $s_2$, $h_2$, $h_2$? You said that wave speed is $c$. If $v$ is particle speed, what makes you think that the particle speeds add? – Bill N Sep 13 '17 at 21:43
  • @Narusan, nonetheless the answer is the same: If there is constructive interference in one place, there will be destructive interference somewhere else. – The Photon Sep 13 '17 at 23:18
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    Also, if it is important, you should probably say what kind of wave you're asking about. Taking potential energy as $mgh$ makes me think you're talking about waves on the surface of a liquid, but maybe you could also be talking about longitudinal compression waves in a material, or vibrations on a guitar string, or something else entirely. – The Photon Sep 13 '17 at 23:20

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