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?
[Physics] Conditions for forming a stationary wave
frequencyresonancesuperpositionwavelengthwaves
Related Solutions
First off, much of this answer is based on the amazingly helpful University of New South Wales website on acoustic impedance. This is a topic I'm very interested in and this is a great opportunity to reduce the complex topic of impedance down for my own learning purposes as well as to provide an answer.
You ask,
does that mean that high frequency sound waves have a lower acoustic impedance than low frequencies (hence why high frequencies are reflected and more than low frequencies?)
Actually, specific acoustic impedance, $z$, is a property of the medium, not of the waves themselves.
It has both a real part, resistance $r(\omega)$ and an imaginary part, reactance $x(\omega)$:
$$z(\omega) = r(\omega) + i \space x(\omega)$$
Where $\omega$ is the angular frequency, $2 \pi f$.
Reactance has a positive and a negative part: compliance (capacative) reactance $x_C(\omega)$, and inertive (inductive) reactance $x_L(\omega)$:
$$x(\omega) = x_L(\omega) - x_C(\omega)$$
Without going into exhaustive detail, I want to point out the relevant fact that inertiance, associated with the density of the medium, is proportional to frequency:
$$x_L(\omega) \propto \omega$$
While compliance, associated with the elasticity of the medium, is inversely proportional to frequency:
$$x_C(\omega) \propto \frac{1}{\omega}$$
In other words,
- denser media are more likely to impede high-frequency waves
- more elastic media are more likely to impede high-frequency waves
In the example of your concrete wall - a dense, inelastic media relative to air - we might expect more of an impedance missmatch (between the two media) for higher-frequency energy than for the lower frequency energy. This would result in the higher frequency energy being reflected more.
The answer to the first part of my question is on wiki, so I will not answer that here.
After some research I have come up with the following answer to the second part of my questions. There are 4 types of waves we need to consider:
- Travelling waves (also known as progressive waves). (TW)
- Compound waves (CW)
- Stationary waves (SW)
- Infinite (harmonic) plane waves. (IPW)
The IPW are basically the 'normal modes' of the infinite string. They are the only type that can be superimposed to form any TW. IPW can also be superimposed to form the other two types of wave (SW and CW) (in a given region where a finite string lies). SW's can be superimposed to form CW only but not TW. All of this information can be summarised on the following diagram.
Where an arrow pointing from wave A to wave B indicates the wave B can be made from the superposition of wave A.
Sources:
- http://encyclopedia2.thefreedictionary.com/compound+wave
- https://books.google.co.uk/books?id=Geqi-oR8lR4C&pg=PA95&dq=superposition+of+standing+waves&hl=en&sa=X&ei=WfEPVdCDIq7B7AbpvYGADA&ved=0CCcQ6AEwAQ#v=onepage&q=superposition%20of%20standing%20waves&f=false
3.http://www.colorado.edu/physics/EducationIssues/baily/courses/BailySP12_GroupVelocity.pdf
- Vibrations and Waves by A.P. French pg 205
Best Answer
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.