# Incident and reflected waves

Let's consider a transversal wave on a string, which is reflected on a wall. I understand that the velocity of the incident and reflected wave are equal. However, I don't understand why the frequencies of both waves are the same. Can anyone please explain this fact?

• Is this about electromagnetic waves or some other kind of waves? – Vadim Apr 11 at 5:56
• My doubt was with respect to waves travelling on a string. – Jonathan Apr 11 at 5:59
• Elastic string or a string in the string theory? Could you post a more detailed question? – Vadim Apr 11 at 6:11

The spatial boundary conditions on the fields must hold for all times, something not possible unless the incident, reflected and transmitted waves have the same temporal part that “cancels out” for all times. You have a spatial boundary (or obstacle) so this changes the spatial part of the wave, but why should it change the temporal part?

• boundary conditions speak of total wave displacement and partial of y wrt x remaining continuous I don't understand how it directly implies frequency remaining constant. – Jonathan Apr 11 at 6:02
• From the boundary conditions, we'd have two equations dealing with 1. the algebraic sum of incident wave and reflected wave equated with the transmitted wave, and 2. their partial derivatives wrt time. So if we write their taylor expansions and then try to show that only when their frequencies are equal then the equations will be consistent I feel it would be a decent proof, what do you think? – Jonathan Apr 11 at 10:06

we know that frequency when a wave travels in string is $$f$$ $$=$$ $$\cfrac{v}{\lambda}$$

after reflection from fixed or unfixed end end velocity ($$v$$) and wavelength($$\lambda$$) remains same so by above equation frequency ($$f$$) also remains same. I hope it helps

In physics we like to apply the harmonic oscillator model, because this is the only model we actually understand. Hence, this is what I'd like to do:

• The wall is the "origin" of the reflected wave. Hence, we should consider the wall as an harmonic oscillator. The reflected wave is just depicting the chronological sequence of the harmonic oscillator.
• Also, the incident wave acts as an external periodic force, acting upon the harmonic oscillator.

Once we understand this analogy the fact that the incident and the reflected wave must possess the same frequency follows directly: An harmonic oscillator always oscillates with the frequency, with which it is excited. So, even if we consider a hypothetical setup, where the two waves have different velocities, the frequency of the two wave would remain the same. This also explains why the frequency of light (=transversal wave) does not change, if it enters a medium.

• This analogy of forced harmonic oscillator is interesting. I've never thought about it this way. – Jonathan Apr 11 at 10:08

... the velocity of the incident and reflected wave are equal... why the frequencies of both waves are the same.

Take it as a common process. The transported energy in the wave is a function of its amplitude and forward motion (velocity). As long as there is no loss of energy, the relation of both parameters must remain constant after reflection for reasons of energy conservation.

• Assuming a classical string the energy is $A^2 \omega^2$ so energy conservation is not enough to live that the frequency does not change. The amplitude A could change. – my2cts Apr 11 at 14:06
• @my2cts That you are right, but the empirical fact shows, that a rigid wall let stay constant the amplitude. For changing amplitude we have to discuss the elasticity and other parameters of the string, the wall and the air. – HolgerFiedler Apr 11 at 14:20
• @HolgerFriedler You are right about the rigid case. Here the reflected wave must have equal but opposite amplitude, which implies equal frequency but 180n degrees phase change. It does not explain why any elastically reflected wave has the same frequency as the incoming wave. – my2cts Apr 11 at 16:23
• @my2cts What is an elastically reflected wave? For example a string, attached to a rubber wall? Or to a balloon? – HolgerFiedler Apr 11 at 18:52
• That would be an excellent question, @HolgerFiedler . – my2cts Apr 11 at 20:27