Topic 9.710 min read
Stationary Waves in an Air Column
An air column resonates when an incident sound wave and its reflected wave form a stationary pattern at one of the column's natural frequencies. For air displacement, a closed end is a node and an open end is an antinode.
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What happens inside the tube?
A vibrating tuning fork near the open end sends a sound wave into the air column. The wave reflects from the closed end. The original and reflected waves travel in opposite directions and combine to form a stationary wave.
Remember the ends
- Closed end: displacement node because the air cannot move back and forth.
- Open end: displacement antinode because the air can move freely.
- A loud sound is heard when the air column resonates with the source.
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Formation in an air column
Standing waves can be generated in the air contained in tubes such as organ pipes. An organ pipe is a long tube that produces sound when its air column is set into vibration. Flutes, trombones, clarinets, trumpets and similar instruments also use vibrating air columns.
Incident and reflected sound waves
Consider a tube closed at one end and open at the other. When a vibrating tuning fork is held near the open end, an incident sound wave travels through the tube and reflects from the closed end.
The reflected wave has the same frequency and wavelength as the incident wave but travels in the opposite direction. Their superposition can produce a stationary wave inside the tube.
Boundary conditions for air displacement
| End of tube | Motion of air | Displacement point |
|---|---|---|
| Closed end | Air molecules cannot move longitudinally | Node |
| Open end | Air molecules can move freely | Antinode |
Resonance of the air column
The air column has natural frequencies. When the frequency of the incident sound equals one of these natural frequencies, resonance occurs. The stationary wave then has a large amplitude and a loud sound is heard.
Simplest mode in a closed-open tube
In the lowest mode, the stationary pattern has a displacement node at the closed end and a displacement antinode at the open end. The distance from a node to its nearest antinode is one quarter of a wavelength.
Using the wave relation v=fλ for this lowest mode:
| Symbol | Meaning | SI unit |
|---|---|---|
| L | Length of the resonating air column | m |
| λ₁ | Wavelength of the lowest mode | m |
| v | Speed of sound in air | m s⁻¹ |
| f₁ | Fundamental frequency | Hz |
Worked example
Finding the fundamental frequency
A tube is closed at one end and open at the other. Its resonating air column is 0.85 m long. Find its lowest resonant frequency when the speed of sound is 340 m s⁻¹.
- L=0.85m
- v=340m s-1
λ1=4L=4(0.85m)=3.40m
f1=vλ1=340m s-13.40m=100Hz
Final answer: The lowest resonant frequency is 100 Hz.
Why the sound becomes loud
At resonance, energy is supplied to the air column at exactly one of its natural frequencies. Successive vibrations reinforce the motion, so the amplitude increases and the sound becomes much louder.
Common mistakes
Check your understanding
Q. Why is the closed end a displacement node?
A. The rigid end prevents air molecules from moving longitudinally, so their displacement is zero there.
Q. Why is the open end a displacement antinode?
A. Air at the open end can move freely, so its displacement can become maximum.
Q. What is required for a loud resonant sound?
A. The driving frequency must equal a natural frequency of the air column.
Quick revision
- Organ pipes and similar tubes produce musical sounds by setting their air columns into vibration.
- A sound wave sent into a tube reflects from the closed end and travels back in the opposite direction.
- The incident and reflected waves have the same frequency and wavelength and can form a stationary wave.
- At a closed end, air cannot move longitudinally, so a displacement node is formed.
- At an open end, air molecules can move freely, so a displacement antinode is formed.
- Resonance occurs when the driving frequency equals a natural frequency of the air column, producing a loud sound.
- For the simplest mode of a tube closed at one end and open at the other, the tube contains one quarter-wavelength: L=λ₁/4.
- The corresponding wavelength and frequency are λ₁=4L and f₁=v4L.
Exam Focus
Exam Focus
Important points
- Formation by incident and reflected waves.
- Closed end: displacement node.
- Open end: displacement antinode.
- Resonance at a natural frequency.
- For the simplest closed-open mode, L=λ₁/4.
Important formulas
- L=λ14
- λ1=4L
- f1=v4L
Common exam traps
- Identify a diagram as displacement or pressure before labeling nodes.
- The node-to-antinode distance is one quarter-wavelength.
Long question structure
Organ pipe or tube → incident and reflected waves → boundary conditions → resonance → quarter-wavelength relation
Common Mistakes
Common Mistakes
- Mistake
Putting a displacement antinode at the rigid closed end.
- Mistake
Using the half-wavelength relation for the simplest closed-open tube.
- Mistake
Confusing the direction change on reflection with a change of frequency.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
Organ pipes and similar tubes produce musical sounds by setting their air columns into vibration.
A sound wave sent into a tube reflects from the closed end and travels back in the opposite direction.
The incident and reflected waves have the same frequency and wavelength and can form a stationary wave.
At a closed end, air cannot move longitudinally, so a displacement node is formed.
Stationary waves are produced in an air column when a sound wave and its reflection, having the same frequency and wavelength, travel in opposite directions and superpose.
- Definition
- Stationary waves are produced in an air column when a sound wave and its reflection, having the same frequency and wavelength, travel in opposite directions and superpose.
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Exam Practice
Important Question
Q1. How are stationary waves produced in an air column?
Important Question
Q2. State and explain the displacement boundary conditions at the open and closed ends of a tube.
Important Question
Q3. What is resonance in an air column?
Important Question
Q4. Derive the wavelength relation for the simplest mode of a tube closed at one end and open at the other.
Numerical
Q5. A 0.85 m closed-open air column resonates in its lowest mode. Find its frequency if v=340 m s⁻¹.