Topic 9.7.111 min read
Modes of Vibration in an Open Pipe
An open pipe supports stationary waves with displacement antinodes at both ends. Its length contains a whole number of half-wavelengths, so λₙ=2L/n and fₙ=nv/(2L)=nf₁ for n=1,2,3,…; therefore, every harmonic is present.
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How the air vibrates
An open pipe is open at both ends. Air can move freely at each end, so each end is a displacement antinode. Only patterns that fit these two end conditions can persist.
First three modes
- First harmonic: half a wavelength fits in the pipe.
- Second harmonic: one complete wavelength fits in the pipe.
- Third harmonic: three half-wavelengths fit in the pipe.
- An open pipe contains every harmonic.
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Open-pipe boundary condition
Let a vibrating tuning fork be held near one end of an open pipe of length L. Because both ends are open, air molecules can vibrate freely at both ends. Each open end is therefore a displacement antinode.
Fundamental frequency or first harmonic
The lowest-frequency pattern has one displacement node at the centre and an antinode at each end. The pipe length contains one half-wavelength.
Using v=fλ:
Second harmonic
The next allowed pattern contains two half-wavelengths, or one complete wavelength, along the pipe. It has antinodes at both ends and at the centre.
Third harmonic
The third allowed pattern contains three half-wavelengths along the pipe.
General nth harmonic
In the nth mode, n half-wavelengths fit into the pipe length. The mode number n can be any positive whole number.
Mode comparison
| Mode | Length relation | Wavelength | Frequency |
|---|---|---|---|
| First harmonic | L=λ12 | λ1=2L | f1=v2L |
| Second harmonic | L=λ₂ | λ₂=L | f₂=2f₁ |
| Third harmonic | L=3λ32 | λ3=2L3 | f3=3f1 |
| nth harmonic | L=nλn2 | λn=2Ln | fn=nf1 |
All harmonics are present
Because n may take every positive integer value, an open pipe supports f₁, 2f₁, 3f₁, 4f₁ and so on. Thus all harmonics are present.
Third harmonic of an open pipe
An open pipe is 0.80 m long. Find the wavelength and frequency of its third harmonic when the speed of sound is 320 m s⁻¹.
- L=0.80m
- v=320m s-1
- n=3
λ3=2L3=2(0.80m)3=0.533m
f3=3v2L=3(320m s-1)2(0.80m)=600Hz
Final answer: The third-harmonic wavelength is approximately 0.533 m and its frequency is 600 Hz.
Common mistakes
Check your understanding
Q. Why are both ends displacement antinodes?
A. Both ends are open, so air molecules can move freely at each end.
Q. How many half-wavelengths fit in the nth harmonic?
A. Exactly n half-wavelengths, so L=nλₙ/2.
Q. Which harmonics occur in an open pipe?
A. All harmonics: f₁, 2f₁, 3f₁, and so on.
Quick revision
- Both ends of an open pipe are displacement antinodes.
- In the fundamental mode, half a wavelength fits in the pipe: L=λ₁/2.
- The fundamental wavelength is λ₁=2L and the fundamental frequency is f₁=v2L.
- The second harmonic contains one complete wavelength: λ₂=L and f₂=2f₁.
- The third harmonic contains three half-wavelengths: λ₃=2L/3 and f₃=3f₁.
- For the nth harmonic, L=nλₙ/2 and λₙ=2L/n.
- The general frequency is fₙ=nv/(2L)=nf₁, where n=1,2,3,….
- All harmonics—first, second, third and higher—are present in an open pipe.
Exam Focus
Exam Focus
Important points
- Both open ends are displacement antinodes.
- L=λ₁/2 and f₁=v2L.
- λₙ=2L/n.
- fₙ=nv/(2L)=nf₁.
- All harmonics are present.
Important formulas
- L=nλn2
- λn=2Ln
- fn=nv2L=nf1
Common exam traps
- Open-pipe ends are antinodes in a displacement diagram.
- Even harmonics are allowed.
Long question structure
Open-end conditions → first harmonic → second harmonic → third harmonic → nth harmonic → all harmonics
Common Mistakes
Common Mistakes
- Mistake
Putting displacement nodes at open ends.
- Mistake
Using the closed-pipe quarter-wavelength formula.
- Mistake
Omitting even harmonics from an open pipe.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
Both ends of an open pipe are displacement antinodes.
In the fundamental mode, half a wavelength fits in the pipe: L=λ₁/2.
The fundamental wavelength is λ₁=2L and the fundamental frequency is f₁=v2L.
The second harmonic contains one complete wavelength: λ₂=L and f₂=2f₁.
An open pipe has open ends at both sides, so both ends are displacement antinodes. Its allowed stationary-wave frequencies are all whole-number multiples of the fundamental frequency.
- Definition
- An open pipe has open ends at both sides, so both ends are displacement antinodes. Its allowed stationary-wave frequencies are all whole-number multiples of the fundamental frequency.
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Exam Practice
Important Question
Q1. Derive the fundamental frequency of an open pipe.
Important Question
Q2. Derive the frequency of the second harmonic in an open pipe.
Important Question
Q3. Derive the frequency of the third harmonic in an open pipe.
Important Question
Q4. Obtain the general expressions for wavelength and frequency in an open pipe.
Numerical
Q5. An open pipe is 0.80 m long. Find its third-harmonic frequency if v=320 m s⁻¹.