Arturo O. answered • 05/27/18
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f = v/λ
L = 0.38 m
v = 20.1√T m/s = 20.1√(20 + 273) m/s ≅ 344 m/s
Fundamental:
If the pipe is open at both ends,
λ0/2 = L = 0.38 m ⇒
λ0 = 2L = 2(0.38) m = 0.76 m
f0 = v/λ0 = 344/0.76 Hz ≅ 453 Hz
Next harmonic:
λ1 = L = 0.38 m
f1 = v/λ1 = 344/0.38 Hz ≅ 905 Hz
Next higher harmonic:
(3/2)λ2 = L ⇒
λ2 = (2/3)L = (2/3)(0.38 m) = 0.253 m
f2 = v/λ2 = 344/0.253 Hz = 1360 Hz
Arturo O.
It is a matter of knowing how many nodes are present, and then counting how many wavelengths fit into the length of the tube. For a tube open at both ends, the lowest harmonic will have displacement antinodes at both ends and one node in between. (It becomes more clear if you draw a diagram.) This fits 2 quarter wavelengths in the tube, so half a wavelength equals the tube length. The speed of sound in air as a function of absolute temperature is computed from a simple formula. It will not change unless the medium is changed (or the air temperature). Then you can use basic relations between wave speed, wavelength, and frequency to get the frequency. For the next harmonic, you will have 2 nodes between the ends, and then 3 nodes for the harmonic above that one, etc. There is a formula for these harmonics, but I encourage the students to work it out rather than memorize the formula.
Similar reasoning may be applied to the case of a tube open at one end but closed at the other. In this case, there will always be a node at the closed end and an antinode at the open end. At the lowest harmonic, one quarter of a wavelength fits inside the tube. Then just add one node at a time to get the higher harmonic, count wavelengths inside the tube, and use the fundamental relations.
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05/28/18
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05/27/18