## Animations of standing waves

As the basic example the development of a 1/4 wave standing wave is shown.
Click on the link to start the animation. This will be opened in a separate tab, to make it easier to move between the animation and the explanations here (and also later between different animations).
• Quarter wave

A sound-pressure disturbance, shown in red, enters the vocal tract at the glottis and propagates towards the lips. This input wave has been chosen to have a sinusoidal form, with the length of the complete wave equal to 4 x vocal tract length. Thus, when the intial value of 1 reaches the lips the value entering at the glottis is zero and the red curve forms 1/4 of a complete sinusoid.
On reaching the lips the wave is reflected WITH A CHANGE OF POLARITY (positive becomes negative).
This is a crucial feature of what happens at the open end of the tube.
The reflected wave is shown in green.
One way of thinking of this is that the pressure at the mouth opening has to be zero, since here the vocal tract is connected to a virtually infinite volume at atmospheric pressure (zero relative pressure).
Thus the sum of all waves just reaching the lips, and of those that have just been reflected has to be zero.
At this stage of the animation there is only one wave reaching the lips from the glottis (in red) and one reflected wave (in green). However, from now on the animation will show the SUM of all waves passing through the vocal tract in black. Thus the black curve is zero at the lip end of the tube and will remain so.

The animation made available by J. Wolfe (UNSW, Sydney) provides an alternative, perhaps more intuitive grasp of the concept of reflection with change of polarity at the open end of the tube:
• Reflections in closed/open tube
In our animation, when the reflected wave (green) reaches the glottis end of the tube, it is itself reflected, but at this closed end of the tube this occurs WITHOUT A CHANGE OF POLARITY (like bouncing off a wall). This reflection, that now travels from the glottis towards the lips, is shown in magenta.
When this reaches the lips it is reflected - with a polarity change - and travels back towards the glottis. (Throughout the animation reflections travelling towards the glottis are shown in green, those travelling towards the lips in magenta. The first pass through the vocal tract is always in red.)
Note that when this second green reflection is in turn reflected at the glottis then this is the first time since the start when a reflected wave is travelling from left to right with positive sign. This is also the time at which the input value at the glottis has returned to its initial value of 1, and starts its cycle again.
It has taken four passes through the vocal tract for this initial state to be reached again, moreover in such a way that the reflected waves match the input wave and are thus able to reinforce each other, rather than cancelling each other out.
This is the fundamental reason why resonances in this system are based on multiples of 1/4 wavelength.
(This idea of the pattern repeating itself after four passes through the vocal tract can also be followed very neatly in J. Wolfe's animation.)

Let us now turn our attention to the black curve (the sum of all reflections).
Over the first few milliseconds of the animation the black curve gets bigger and bigger until it reaches an amplitude of about 4. Notice also that as the reflections go to and fro through the vocal tract they decline in amplitude. This decay will happen in any realistic physical system due to loss of energy to the surroundings, and has the advantage here of making it easier to see multiple reflections.
The peak value of 4 in the black curve thus represents a state where energy input at the glottis is balanced by energy loss in the decaying reflections.
Look at the animation from a timepoint where the black curve is close to its maximum and also fairly smooth through the superposition of multiple reflections. e.g. from about 4ms.
How long does it take the black curve to go through one complete cycle, e.g. from its peak positive value, via the peak negative value back to the peak positive value?
You should come up with a value close to 2ms.
Thus the "standing" wave is an oscillation with a frequency of 1000/2 = 500Hz.
It is referred to as "standing" because the locations of zero pressure and maximum/minimum pressure in the tube do not change.
Notice that this period duration of 2ms is also the time required for four passes of the reflected wave through the vocal tract (follow the blue circle at the leading edge of the reflected wave). This must be so because the sound pressure disturbance (e.g. at the blue circle) moves at the speed of sound. With c=34000cm/s 2ms is the time required to travel 4x17 = 64cm. Thus we see here the link between frequency and wavelength captured in the formula f = c/lambda.

The following animation shows the next standing wave in a tube closed at one end and open at one end.

• Three-quarter wave
Once again try and pick out the time required for one complete cycle of the fully developed standing-wave (black curve). Clearly it is faster than the first one, but it may be difficult to estimate precisely. It is probably easiest to try and estimate how many cycles are completed in 2ms. The result should be three. In other words the frequency of the second standing-wave is three times that of the first one, i.e. 3 x 500 = 1500Hz.

Based on these first two examples we cannot exclude the possibility that a standing-wave will develop regardless of the nature of the input. The next example illustrates that this is not the case:

• Half wave
In this animation a half wave is used as input.
The crucial point here is that the black curve (the sum of all reflections) shows no tendency to increase in amplitude above the amplitude of the original input: It stays in the range of +/- 1.

This point can be emphasized by considering the following two animations:
The first one shows the original quarter-wave animation, but with a slower rate of decay in the amplitude of the reflections (the green and magenta lines are closer together than in the original animation).

• Quarter wave, slow decay
Note that the standing wave now reaches an amplitude of close to 10, i.e. considerably higher than in the original animation.
Now consider the half-wave animation with the same change made in the rate of decay:
• Half wave, slow decay
Here there is hardly any change in the amplitude of the black curve compared to the original animation.
This confirms that when the input signal does not match the boundary conditions of the tube system then the reflections tend to cancel each other out, rather than summing constructively to give a stable high-amplitude oscillation.