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, f_{o}. 

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 antinode 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 antinode
must be λ/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" (often called
the end correction) 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 antinode 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 1½ 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 
