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.
Key equation
L=λ14
Key equation
λ1=4L
Using the wave relation v=fλ for this lowest mode:
Key equation
f1=vλ1=v4L
Key equation
f1=v4L
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.
1
Given
L=0.85m
v=340m s-1
2
Working
λ1=4L=4(0.85m)=3.40m
3
Step 2
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.
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Exam preparation
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.
Common mistakes
Putting a displacement antinode at the rigid closed end.
Using the half-wavelength relation for the simplest closed-open tube.
Confusing the direction change on reflection with a change of frequency.
Quick revision
Revision Zone
Review the key facts, formulas and checks before you move on.
Stationary Waves in an Air Column
Sticky Notes · Quick Revision
1.
Key Point 1
A sound wave sent into a tube reflects from the closed end and travels back in the opposite direction.
2.
Key Point 2
At a closed end, air cannot move longitudinally, so a displacement node is formed.
3.
Key Point 3
At an open end, air molecules can move freely, so a displacement antinode is formed.
4.
Key Point 4
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.
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8 concept questions
Easy → Medium → Hard
Explanation after each answer
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.
