Does it only require two waves of the same frequency / wavelength travelling in opposite directions? My teacher says that one wave needs to be reflected and then superimpose with this reflection for a stationary wave to form. Also, aren't stationary waves only formed at resonant frequencies?


All a "perfect" reflection does for you is to guarantee a wave which is travelling in the opposite direction to the incident wave and also having the same frequency/wavelength and amplitude (which you missed) as the incident wave.
Even if the reflector is not perfect there will be variations in amplitude at different positions but there will be no positions of zero amplitude.

The idea of resonant frequencies crops up with waves which are bounded and the amplitudes of the standing wave of particular wavelengths (frequencies) which are produced are large.
At the boundaries certain conditions have to be satisfied eg node at the end of a clamped vibrating string, displacement node (pressure antinode) at the closed end of a tube and displacement antinode (pressure node) near the end of the open end of a tube, . . . . etc.
So you have to ensure that the standing wave "fits into" these boundary conditions which in turn means that the wavelength (frequency) of the wave can only have certain values.
The losses of energy at the reflecting surfaces and elsewhere can be made up by the energy supplied by a driver eg a bow, musician blowing air, etc and this will maintain the standing wave.

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  • $\begingroup$ Thank you for your answer. But you say that resonant frequencies are only related to waves which are bounded, but they are also used if we have a medium with two open ends ( which I assume is unbounded ). Also, a reflection makes the wave undergo a 180 degree phase shift at the point of contact, which is generally not the case with 2 waves travelling in opposite directions with the same freq/wavelength/amplitude. Lastly, in general, resonance is associated with increasing amplitudes, but in this case, the amplitudes are constant, are they not? $\endgroup$ – John May 7 '17 at 10:19
  • $\begingroup$ An open end is a constraint and hence I consider that as a boundary condition. The reversal of phase corresponds to a node. For a given driver amplitude the response of the system (amplitude) is a maximum at resonance. $\endgroup$ – Farcher May 7 '17 at 10:25
  • $\begingroup$ I see... so in general, we will always have a stationary wave if two waves are travelling in opposite directions with the same parameters, but if they are in a bounded medium, then resonance comes into play? I'm not clear about resonance.. if we push a swing at the resonant frequency, then the amplitude approaches infinity, not merely a high value. $\endgroup$ – John May 7 '17 at 10:30
  • $\begingroup$ There is usually a frictional force acting. $\endgroup$ – Farcher May 9 '17 at 16:59
  • $\begingroup$ Hold on. When you say "reflected wave", how does it necessarily indicate that it's the same wave, but just travelling in the opposite direction? For instance, if you have a rigid boundary causing reflection, the equation of the reflected wave would be asin(kx-wt+ pi) and not asin(kx+wt) if the incident wave is asin(kx-wt) . $\endgroup$ – kimi Feb 2 at 6:17

two progressive wave of same amplitude and freqency moving in opposite direction superimposed at a point

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