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Resonance in Air Columns
Resonance in Tubes Closed at One End  
Stationary (or standing) waves can occur in many different situations.  
One important example, especially for musicians, is the stationary waves set up in columns of air, as in, for example trumpets, saxophones, organs etc  
   
When you blow across the top of an open water bottle, the sound you hear (if you get it right!) is due to a stationary wave being established inside the bottle.  
Turbulence around the edge of the bottle produces a more or less random set of frequencies of oscillation.  
These oscillations cause sound waves to travel along the bottle.  
The waves are reflected at the ends (both the closed and the open* end) of the bottle.  
For certain of the frequencies of oscillations, interference between these reflected waves amplifies the sound.  
When this occurs, we say the air column is resonating.  
These frequencies correspond to situations in which the effective distance travelled by a disturbance is a whole number of wavelengths.  
 
The frequencies at which resonance occurs depend on  
1. the length of the air column  
2. the speed of sound in the column  
   
The first of these is easy to verify for yourself:  
first  blow across the open end of an empty bottle and then do the same thing when the bottle is, say, half full.  
 
* Waves are reflected from an open end because sound travels a little more slowly inside a tube than in free air.  
   
Imagine a small loudspeaker placed near the end of a tube which is closed at one end.  
The frequency of the sound produced by the speaker is varied, starting from a very low frequency.  
At a certain frequency a much louder sound is heard: this is the first resonant frequency, the fundamental frequency, fo.  
The diagram below represents the oscillations of "layers" of air in the column when it is resonating at its lowest frequency.  
   
 
   
The length of the arrow represents (approximately) the amplitude of the oscillation at that point.  
Situations of resonance in air columns are often represented as shown in the next diagram.  
   
 
   
Diagrams like the one above, to represent resonance in air columns are inspired by looking at resonating strings.  
They can be confusing because the string wave is transverse and sound is a longitudinal wave.  
They should be thought of as being something like a sketch graph of amplitude of oscillation of "layers" of air, against position, measured along the axis of the tube, as shown below.  
   
 
   
At the closed end, waves are reflected with a phase change of 180.  
There is no displacement at this point: a displacement node exists at the closed end.  
   
At the open end, the air is free to move.
Here, waves are reflected with no phase change so a displacement anti-node exists at the open end.  
 
Therefore, if waves travel twice the length of the tube in half a time period, they will arrive back at the open end in phase and resonance will occur.  
This means that the fundamental frequency of resonance occurs when  
 
N.B. When considering stationary waves on strings we concluded that the distance between two adjacent nodes is equal to half the wavelength of the waves producing the stationary wave.  
This means that the distance between a node and the adjacent anti-node must be l/4, so we arrive at the same conclusion.  
   
The above statement is approximate because there is some air beyond the end of the tube which should be considered as part of the air column.  
The length of this "extra bit of air" depends on the diameter of the tube (we will ignore this for now).  
 
and these two expressions give us  
 
for the fundamental (lowest) frequency of resonance.  
   
As with the string under tension, the air column will also resonate at higher harmonics, however, the relation between these higher frequencies of resonance is a little different.  
This is because the string is fixed at both ends meaning that there must be a node at both ends.  
In this tube, open at one end and closed at the other, the next frequency at when resonance occurs must still have a node at the closed end and an anti-node at the open end.  
The situation for the second harmonic is therefore as shown in the next diagram  
   
 
   
Here, the waves travel twice the length of the tube in 1 time periods  
Therefore, in this case  
 
which gives us  
Therefore, in general, for a tube closed at one end, we have  
 
and  
 
where, n = 0, 1, 2 etc  
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