Consider a pipe of length L that is closed at one end and open at the other. The rigid closed end prevents longitudinal motion of air and is a displacement node. Air at the open end can move freely and forms a displacement antinode.
Fundamental frequency or first harmonic
The lowest-frequency pattern extends from a displacement node to the nearest displacement antinode. This distance is one quarter of a wavelength.
Key equation
L=λ14⇒λ1=4L
Key equation
f1=vλ1=v4L
Key equation
f1=v4L
Second vibration mode: third harmonic
The next permitted pattern contains one half-wavelength plus one quarter-wavelength, giving a total of three quarter-wavelengths. This is the second vibration mode, but its frequency is three times the fundamental, so it is the third harmonic and first overtone.
Key equation
L=λ22+λ24=3λ24
Key equation
λ2=4L3
Key equation
f2=vλ2=3v4L
Key equation
f2=3f1
Third vibration mode: fifth harmonic
The next permitted pattern contains two half-wavelengths plus one quarter-wavelength, giving five quarter-wavelengths. It is the third vibration mode, the fifth harmonic and the second overtone.
Key equation
L=2(λ32)+λ34=5λ34
Key equation
λ3=4L5
Key equation
f3=vλ3=5v4L
Key equation
f3=5f1
General nth mode
The allowed patterns contain 1, 3, 5, … quarter-wavelengths. For mode number n, the number of quarter-wavelengths is 2n−1.
Key equation
L=(2n-1)λn4
Key equation
λn=4L2n-1
Key equation
fn=vλn=(2n-1)v4L
Key equation
fn=(2n-1)v4L=(2n-1)f1,\qquad n=1,2,3,…
Mode and harmonic comparison
Vibration mode
Quarter-wavelengths
Wavelength
Frequency
Harmonic
First
1
λ1=4L
f1=v4L
First
Second
3
λ2=4L3
f2=3f1
Third
Third
5
λ3=4L5
f3=5f1
Fifth
nth
2n−1
λn=4L2n-1
fn=(2n-1)f1
(2n−1)th
Why even harmonics are absent
An even number of quarter-wavelengths cannot place a displacement node at one end and an antinode at the other. Therefore frequencies 2f₁, 4f₁, 6f₁, … do not satisfy the boundary conditions of an ideal closed pipe.
First three allowed frequencies
A closed pipe is 0.50 m long. Find its first three allowed frequencies when the speed of sound is 340 m s⁻¹.
1
Given
L=0.50m
v=340m s-1
2
Working
f1=v4L=3404(0.50)=170Hz
3
Step 2
f2=3f1=3(170)=510Hz
4
Step 3
f3=5f1=5(170)=850Hz
Final answer: The first three allowed frequencies are 170 Hz, 510 Hz and 850 Hz.
Common mistakes
Check your understanding
Q. Why does a closed pipe contain only odd harmonics?
A. Only odd numbers of quarter-wavelengths can connect a displacement node at the closed end to an antinode at the open end.
Q. What is the first overtone of a closed pipe?
A. It is the second vibration mode with frequency 3f₁, so it is the third harmonic.
Q. What is the nth allowed frequency?
A.fₙ=(2n−1)v4L=(2n−1)f₁.
Quick revision
The closed end is a displacement node and the open end is a displacement antinode.
In the fundamental mode, one quarter-wavelength fits in the pipe: L=λ₁/4.
The fundamental wavelength is λ₁=4L and the fundamental frequency is f₁=v4L.
The second vibration mode contains three quarter-wavelengths: λ₂=4L/3 and f₂=3f₁; it is the third harmonic.
The third vibration mode contains five quarter-wavelengths: λ₃=4L/5 and f₃=5f₁; it is the fifth harmonic.
For mode number n, the pipe contains (2n−1) quarter-wavelengths.
The general relations are λₙ=4L/(2n−1) and fₙ=(2n−1)v4L=(2n−1)f₁.
Only odd harmonics—first, third, fifth and higher odd multiples—are present in a closed pipe.
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Exam preparation
Exam Focus
Important points
Closed end: displacement node; open end: displacement antinode.
L=λ₁/4 and f₁=v4L.
λₙ=4L/(2n−1).
fₙ=(2n−1)v4L=(2n−1)f₁.
Only odd harmonics are present.
Common mistakes
Calling the second vibration mode the second harmonic.
Including even harmonics.
Using the same displacement condition at both ends.
Quick revision
Revision Zone
Review the key facts, formulas and checks before you move on.
Modes of Vibration in a Closed Pipe
Sticky Notes · Quick Revision
1.
Key Point 1
The closed end is a displacement node and the open end is a displacement antinode.
2.
Key Point 2
In the fundamental mode, one quarter-wavelength fits in the pipe: L=λ₁/4.
3.
Key Point 3
The general relations are λₙ=4L/(2n−1) and fₙ=(2n−1)v4L=(2n−1)f₁.
4.
Key Point 4
Only odd harmonics—first, third, fifth and higher odd multiples—are present in a closed pipe.
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8 concept questions
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Explanation after each answer
Exam Practice
Important Question
Q1. Derive the fundamental frequency of a closed pipe.
Important Question
Q2. Derive the frequency of the second vibration mode in a closed pipe.
Important Question
Q3. Derive the frequency of the third vibration mode in a closed pipe.
Important Question
Q4. Obtain the general expressions for wavelength and frequency in a closed pipe.
Numerical
Q5. A closed pipe is 0.50 m long. Find its first three frequencies if v=340 m s⁻¹.
