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Standing Waves in Open Tubes
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Standing Waves in Open Tubes
An open tube has an open end at both ends. Sound waves reflect inside it and can combine to form a standing wave. In displacement patterns, A marks a displacement antinode (greatest movement) and N marks a displacement node (no movement). For an open tube, both ends are antinodes. The first harmonic has one internal node at the midpoint. The second harmonic has internal nodes at one-quarter and three-quarters of the tube length from the left end. Use fₙ = n × v ÷ 2L, where n is the harmonic number, v is the speed of sound and L is the tube length. Use v = 340 m/s unless a question gives another value.
Explore the patterns
Use the open-tube standing-wave information in the instructions to answer Questions 1–5.
1.At an open end of a tube, what kind of displacement point must occur?
- A node
- An antinode
- Either a node or an antinode
2.In the first harmonic, how many nodes are inside the tube?
3.In the second harmonic, name the positions of the internal nodes as fractions of the tube length measured from the left end.
4.True or false: In an open tube, the first harmonic has a wavelength equal to the tube length.
TrueFalse
5.For an open tube, the standing-wave relationship is L = nλₙ/2. Complete the relationship: the wavelength of harmonic n is λₙ = .
Calculate and reason
Show your working and include units in calculation answers.
6.A resonance tube open at both ends is 0.40 m long. Calculate its first-harmonic frequency using fₙ = n × v ÷ 2L and v = 340 m/s.
7.For the same 0.40 m tube, calculate the frequency of the third harmonic.
8.A flute behaves approximately like an open tube. A note has a frequency of 680 Hz and is the second harmonic. If the speed of sound is 340 m/s, calculate the flute's effective length.
9.A musician plays a higher harmonic without changing the tube. Explain what happens to the frequency and wavelength, and why.
10.A second open tube is twice as long as the 0.40 m tube. Compare its first-harmonic frequency with 425 Hz. Give the new frequency and explain the relationship.
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