OCR · A-Level · Physics
Course version: H556 · Version from Sept 2015
(i) stationary wave patterns for a stretched string and air columns in…
Full specification wording
Stationary waves: (i) stationary wave patterns for a stretched string and air columns in closed and open tubes
Stationary-wave patterns depend on the boundary conditions.
Checked against the OCR H556 specification (opens in a new tab) on .
- Point 6 of 8
- 3 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Stationary-wave patterns depend on the boundary conditions. A stretched string fixed at both ends has nodes at both ends, so its wavelengths are given by L = nλ/2, where n is a whole number of half-wavelengths. An open tube has antinodes at both open ends, giving the same relationship L = nλ/2. A closed tube has a node at the closed end and an antinode at the open end, so the shortest pattern is a quarter wavelength: L = (2n − 1)λ/4. These patterns determine the resonant frequencies, and the end correction must be added to the measured length for real tubes.
What earns the marks
- A stretched string fixed at both ends has nodes at both ends and L = nλ/2.
- An open tube has antinodes at both open ends and L = nλ/2.
- A closed tube has a node at the closed end and an antinode at the open end, giving L = (2n − 1)λ/4.
- The fundamental frequency of a closed tube is half that of an open tube of the same length.
- An end correction must be added to the measured tube length to account for the antinode lying just outside the open end.
Common mistakes and how to put them right
Using L = nλ/2 for a closed tube.
Put it right: A closed tube has a node at one end and an antinode at the other, so L = (2n − 1)λ/4.
Placing an antinode at the closed end of a tube.
Put it right: The closed end must be a node because the air cannot move there.
Forgetting the end correction.
Put it right: The antinode forms slightly outside the open end, so the effective length is longer than the measured length.
Assuming an open tube and a closed tube of equal length have the same fundamental frequency.
Put it right: The closed tube has half the fundamental frequency.
Exam tips
- Sketch the pattern first, marking nodes and antinodes, then write the length in terms of λ.
- Check each boundary: fixed or closed gives a node, open or free gives an antinode.
- Include the end correction when comparing measured resonance lengths with theory.
- Use the harmonic number to check whether the pattern is the fundamental or an overtone.
Try the lesson
The start of the Stationary waves: (i) stationary wave patterns for a stretched string and air columns in closed and open tubes lesson, made from checked teaching notes for this specification point.
Listen
Stationary-wave patterns depend on the boundary conditions. A stretched string fixed at both ends has nodes at both ends, so its wavelengths are given by L = nλ/2, where n is a whole number of half-wavelengths. An open tube has antinodes at both open ends, giving the same relationship L = nλ/2. A closed tube has a node at the closed end and an antinode at the open end, so the shortest pattern is a quarter wavelength: L = (2n − 1)λ/4. These patterns determine the resonant frequencies, and the end correction must be added to the measured length for real tubes. A strong answer earns marks for these points. A stretched string fixed at both ends has nodes at both ends and L = nλ/2. An open tube has antinodes at both open ends and L = nλ/2. A closed tube has a node at the closed end and an antinode at the open end, giving L = (2n − 1)λ/4. The fundamental frequency of a closed tube is half that of an open tube of the same length. Exam tip: Sketch the pattern first, marking nodes and antinodes, then write the length in terms of λ.
See it
Key words
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Length–wavelength relationships for strings and tubes
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Fundamental frequency and harmonics
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End correction in real tubes
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See every definition
- Length–wavelength relationships for strings and tubes
- A stretched string fixed at both ends and an open tube both give L = nλ/2, while a closed tube gives L = (2n − 1)λ/4.
- Fundamental frequency and harmonics
- The patterns determine the resonant frequencies, and the fundamental frequency of a closed tube is half that of an open tube of the same length.
- End correction in real tubes
- An end correction must be added to the measured tube length to account for the antinode lying just outside the open end.
Label the diagram
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1Box 1
2Box 2
3Box 3
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