5-4 Vibrational Mode of an Air Column
How is sound produced in an open cylindrical tube?
When you blow across an open tube, the air transmits down the tube. At the end of the tube there is a pressure node where part of the air is reflected back while the rest of the air flows out of the tube's opening. With the proper amount of air the waves will interfere and create standing waves. The pitch produced by the tube tells us the frequency that the air inside the tube is vibrating.
How can you find the frequency of the harmonics in an open cylindrical tube?
The frequency of the harmonics of an open cylindrical tube is found with the formula:
fn = (v/ (wavelength X n)) = (n/(2L)) X v
where v is the velocity (20.1 X the square root of (Ta) can be substituted for v)
How can you find the effective length of an open cylindrical tube?
Leff = Lactual + 1.226R
where Leff is the effective length, Lactual is the actual length, and r is the radius
How is sound produced in a semiclosed cylindrical tube?
A semiclosed cylindrical tube has one end open (where a pressure node exists) and one end closed (where an antinode exists). Like an open cylindrical tube, when someone blows across the tube, air transmits down the tube where some air will reflect back and cause superposition of waves while the rest of the air will flow out of the tube. When the air waves superimpose, standing waves occur. However, in a semiclosed cylindrical tube, the air can vibrate at many harmonically related frequencies, but the evens are absent. Also, the fundamental frequencies produced are an octave lower than an open pipe of the same length.
How can you find the frequency of the harmonics in a semiclosed cylindrical tube?
The frequency of the harmonics in a semiclosed cylindrical tube is found with the formula:
fn = v / (wavelength X n) = n X (20.1/(4L)) X (the square root of (Ta))
fn = (5.025/L) X (the square root of (Ta)) n = 1, 3, 5, 7,... (odd)
where v is velocity and L is length
At room temperature (23 degrees C):
fn = n (86/L) n = 1, 3, 4, 7,... (odd)
where L equals length
How can you find the effective length of a semiclosed cylindrical tube?
Leff = Lactual + 0.613R
where Leff is the effective length, Lactual is the actual length, and r is the radius
Are there any other tube shapes where air vibrates frequencies that are multiples of some fundamental frequency?
1) Flared tubes have a radius which increased by a factor of 11.31 for every doubling distance from the end. (For example, if 2 cm from the end the radius is 1 cm, then 4 cm from the end the radius is 11.31 cm.)
2) Conical tubes have straight sides which meet at a point. They will vibrate with the same frequencies as an open cylindrical tube with the same length.
3) Truncated cones are conical tubes with a portion cut off. The fundamental frequency decreases from that of a conical tube.
Resonance Frequencies for Truncated Cone (where I0 = 0.4I)
| f1 = 0.75 X (v / (2L)) |
| f2 = 1.60 X (v / (2L)) = 2.13f1 |
| f3 = 2.56 X (v / (2L)) = 3.41f1 |
| f4 = 3.55 X (v / (2L)) = 4.73f1 |
| f5 = 4.54 X (v / (2L)) = 6.05f1 |
| f6 = 5.53 X (v / (2L)) = 7.37f1 |
Practice Problems:
1. Suppose you have two long tubes with radii of 2 cm and 5 cm. You would like to cut off pieces of these long tubes so that the enclosed air in the resulting pieces will vibrate with a fundamental frequency of 220 Hz at 22 degree C. What should their respective lengths be?
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