Lesson 5: Musical
Instruments
OpenEnd Air
Columns
Many woodwind instruments
consist of an air column enclosed inside of a hollow
metal tube. Though the metal tube may be more than a
meter in length, it is often curved upon itself one or
more times in order to conserve space. If the end
of the tube is uncovered such that the air at the end of
the tube can freely vibrate when the sound wave reaches
it, then the end is referred to as an open end. If
both ends of the tube are uncovered or
open, the musical instrument is said to contain an
openend air
column. A variety of
instruments operate on the basis of openend air columns;
examples include the brass instruments such as the flute
and trombone and woodwinds such as the saxophone and
oboe. Even wind chimes and some organ pipes serve as
openend air columns.
As
has already been
mentioned, a musical instrument
has a set of natural frequencies at which it vibrates at
when a disturbance is introduced into it. These natural
frequencies are known as the harmonics of the instrument;
each harmonic is associated with a standing wave pattern.
In Lesson 4 of Unit
10, a standing wave
pattern was defined as a vibrational pattern
created within a medium when the vibrational frequency of
the source causes reflected waves from one end of the
medium to interfere with incident waves from the source in such a manner that
specific points along the medium appear to be standing
still. In the case of stringed instruments (discussed earlier), standing wave
patterns were drawn to depict the amount of movement of
the string at various locations along its length. Such
patterns show nodes  points of no displacement or
movement  at the two fixed ends of the string. In the
case of air columns, a closed end in a column of air is
analogous to the fixed end on a vibrating string. That
is, at the closed end of an air column, air is not free
to undergo movement and thus is forced into assuming the
nodal positions of the standing wave pattern. Conversely,
air is free to undergo its backandforth longitudinal
motion at the open end of an air column; and as such, the
standing wave patterns will depict antinodes at the open
ends of air columns.
So the basis for drawing the standing
wave patterns for air columns is that antinodes will be
present at any open end and nodes will be present at any
closed end. If this principle is applied to openend air
columns, then the pattern for the fundamental frequency
(the lowest frequency and longest wavelength pattern)
will have antinodes at the two open ends and a single
node in between. For this reason, the standing wave
pattern for the fundamental frequency (or first harmonic)
for an openend air column looks like the diagram
below.
The distance between antinodes on a
standingwave pattern is equivalent to onehalf of a
wavelength. A careful analysis of
the diagram above shows that adjacent antinodes are
positioned at the two ends of the air column. Thus, it
the length of the air column is equal to onehalf of the
wavelenth for the first harmonic.
The standing wave
pattern for the second harmonic of an openend air column
could be produced if another antinode and node was added
to the pattern. This would result in a total of three
antinodes and two nodes. This pattern is shown in the
diagram below. Observe in the pattern that there is one
full wave in the length of the air column. That is twice
the number of waves in the first harmonic. For this
reason, the frequency of the second harmonic is two times
the frequency of the first harmonic.
And finally, the
standing wave pattern for the third harmonic of an open
end air column could be produced if still another
antinode and node was added to the pattern. This would
result in a total of four antinodes and three nodes. This
pattern is shown in the diagram below. Observe in the
pattern that there are one and onehalf waves present in
the length of the air column. That is three times the
number of waves in the first harmonic. For this reason,
the frequency of the third harmonic is three times the
frequency of the first harmonic.
The process of
adding another antinode and node to each consecutive
harmonic in order to determine the pattern and the
resulting lengthwavelength relationship could be
continued. If doing so, it is important to keep antinodes
on the open ends of the air column and to maintain an
alternating pattern of nodes and antinodes. When
finished, the results should be consistent with the
information in the table below.
The
relationships between the standing wave pattern for a
given harmonic and the lengthwavelength relationships
for open end air columns are summarized in the table
below.
Harm.
#

#
of
Waves in
Column

# of
Nodes

# of
Antinodes

Length
Wavelength
Relationship

1

1/2

1

2

Wavelength = (2/1)*L

2

1 or 2/2

2

3

Wavelength = (2/2)*L

3

3/2

3

4

Wavelength = (2/3)*L

4

2 or 4/2

4

5

Wavelength = (2/4)*L

5

5/2

5

6

Wavelength = (2/5)*L

Now the aim of
the above discussion is to internalize the mathematical
relationships for openend air columns in order to
perform calculations predicting the length of air column
required to produce a given natural frequency. And
conversely, calculations can be performed to predict the
natural frequencies produced by a known length of air
column. Each of these calculations requires a knowledge
of the speed of a wave in a air (which is approximately
340 m/s at room temperatures). The graphic below depicts
the relationships between the key variables in such
calculations. These relationships will be used to assist
in the solution to problems involving standing waves in
musical instruments.
To demonstrate
the use of the above problemsolving scheme, consider the
following problem and its detailed solution.
Practice
Problem
The speed of sound waves in air is found
to be 340 m/s. Determine the fundamental
frequency (1st harmonic) of an openend air
column which has a length of 67.5 cm.

The solution to the problem
begins by first identifying known information, listing
the desired quantity, and constructing a diagram of the
situation.
Given:
v = 340 m/s
L = 67.5 cm = 0.675 m

Find:
f_{1} = ??

Diagram:

The problem statement asks us to
determine the frequency (f) value. From the
graphic
above, the only means of
finding the frequency is to use the wave equation
(speed=frequency*wavelength) and knowledge of the speed
and wavelength. The speed is given, but wavelength is not
known. If the wavelength could be found then the
frequency could be easily calculated. In this problem
(and any problem), knowledge of the length and the
harmonic number allows one to determine the wavelength of
the wave. For the first harmonic, the wavelength is twice
the length. This relationship is derived from the diagram
of the standing wave pattern (see
table above). The relationship,
which works only for the first harmonic of an openend
air column, is used to calculate the wavelength for this
standing wave.
Wavelength = 2 *
Length
Wavelength = 2 * 0.675
m
Wavelength = 1.35
m
Now that wavelength is known, it
can be combined with the given value of the speed to
calculate the frequency of the first harmonic for this
openend air column. This calculation is shown
below.
speed = frequency *
wavelength
frequency =
speed/wavelength
frequency = (340 m/s)/(1.35
m)
frequency = 253
Hz
Most problems can be solved in a
similar manner. It is always essential to take the extra
time needed to set the problem up; take the time to write
down the given information and the requested information,
and to draw a meaningful diagram.
Seldom in physics are two
problems identical. The tendency to treat every problem
the same way is perhaps one of the quickest paths to
failure. It is much better to combine good
problemsolving skills (part of which involves the
discipline to set the problem up) with a solid grasp of
the relationships among variables than to memorize
approaches to different types of problems. To further
your understanding of these relationships, examine the
following problem and its solution.
To demonstrate
the use of the above problemsolving scheme, consider the
following problem and its detailed solution.
Practice
Problem
Determine the length of an openend air
column required to produce a fundamental
frequency (1st harmonic) of 480 Hz. The speed of
waves in air is known to be 340 m/s.

The solution to the problem
begins by first identifying known information, listing
the desired quantity, and constructing a diagram of the
situation.
Given:
v = 340 m/s
f_{1} = 480 Hz

Find:
L = ??

Diagram:

The problem statement asks us to
determine the length of the air column. When inspecting
the
problem solving scheme described
above, one will notice
that the only means of finding the length of the air
column is from knowledge of the wavelength. But the
wavelength is not known. However, the frequency and speed
are given, so one can use the wave equation (speed =
frequency*wavelength) and knowledge of the speed and
frequency to determine the wavelength. This calculation
is shown below.
speed = frequency *
wavelength
wavelength =
speed/frequency
wavelength = (340 m/s)/(480
Hz)
wavelength = 0.708
m
Now that the wavelength is
found, the length of the air column can be calculated.
For the first harmonic, the length is onehalf the
wavelength . This relationship is derived from the
diagram of the standing wave pattern (see
table above); it may also be
evident to you by looking at the standing wave diagram
drawn above. This relationship between wavelength and
length, which works only for the first harmonic of an
openend air column, is used to calculate the wavelength
for this standing wave.
Length = (1/2) *
Wavelength
Length = (1/2) *
Wavelength
Length = 0.354
m
If you have successfully managed
the above two problems, take a try at the following
practice problems. As you proceed, be sure to be mindful
of the numerical relationships involved in such problems.
And if necessary, refer to the graphic
above.
