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Standing waves link two chapters, waves and sound, and examiners use them in both theory and practical questions. The formulas are short, but students mix up which harmonics are allowed in which pipe, and where the nodes sit. Then the resonance tube practical adds an "end correction" that is rarely explained. This post builds each case from its boundary conditions and works through a full resonance tube calculation.
What does the syllabus include?
The JEE (Main) 2026 syllabus lists "standing waves in strings and organ pipes, fundamental mode and harmonics, beats", and its experiment list includes the speed of sound in air using a resonance tube. The CBSE 2026-27 curriculum has the same theory, plus sonometer experiments and "speed of sound in air at room temperature using a resonance tube by two resonance positions" for Class 11. The JEE (Advanced) 2026 brochure lists vibration of strings and air columns, and the resonance column experiment. Check the current documents for your exam.
How do you find the allowed harmonics?
Every case follows from one rule: a fixed or closed end is a displacement node, a free or open end is a displacement antinode, and adjacent nodes are half a wavelength apart. The NCERT chapter Waves uses exactly these conditions.
String fixed at both ends: nodes at both ends, so L = nλ/2 and f = nv/2L for n = 1, 2, 3... All harmonics.
Pipe open at both ends: antinodes at both ends, so again f = nv/2L. All harmonics.
Pipe closed at one end: a node at the closed end and an antinode at the open end, so L = λ/4, 3λ/4, 5λ/4... and f = (2n − 1)v/4L. Only odd harmonics.
For a string, the wave speed is v = √(T/μ), where T is the tension and μ the mass per unit length.
Why do students get nodes in pipes wrong?
Diagrams in pipes show wavy lines, but sound in air is a longitudinal wave: air moves back and forth along the pipe. Zeng and colleagues (2014) found in pilot studies that many textbook illustrations of standing waves in pipes hindered understanding, because students misunderstood how air molecules move inside the pipe. In their study of 30 students, most said they understood standing waves better with an illustration showing the air molecules' motion than with the traditional textbook picture.
There is a second trap. NCERT notes that at the closed end, displacement is zero but pressure change is largest. At the open end it is the other way round. So a "node" in a pipe question depends on whether the question is about displacement or pressure. Read the question twice.
Worked examples
Example 1: a string. A wire of length 0.50 m and mass per unit length 0.010 kg/m is under 100 N tension. Then v = √(100 / 0.010) = √10,000 = 100 m/s, so the fundamental is f₁ = 100 / (2 × 0.50) = 100 Hz. The next harmonics are 200 Hz and 300 Hz. To double the fundamental, you need four times the tension, because f goes as √T.
Example 2: open and closed pipes. A pipe is 85 cm long, and v = 340 m/s.
Closed at one end: f₁ = 340 / (4 × 0.85) = 100 Hz. Allowed frequencies are 100, 300 and 500 Hz.
Open at both ends: f₁ = 340 / (2 × 0.85) = 200 Hz. Allowed frequencies are 200, 400 and 600 Hz.
Closing one end halves the fundamental and removes the even harmonics. A common question gives two successive resonant frequencies and asks which type of pipe it is. If their ratio is 3:5 or 5:7, it is closed. If it is 2:3 or 3:4, it is open.
What is the end correction, and how do you use it?
At an open end, the antinode sits slightly outside the pipe, so the air column behaves as if it were a little longer. In a classic paper, Levine and Schwinger (1948) solved the problem for an unflanged circular pipe and found that at low frequencies the pipe acts as if lengthened by 0.6133 times its radius. Textbooks round this to about 0.6r.
In the resonance tube, you do not need the radius. You find two successive resonance lengths and the end correction e cancels:
First resonance: L₁ + e = λ/4
Second resonance: L₂ + e = 3λ/4
Subtract: λ = 2(L₂ − L₁). Then e = (L₂ − 3L₁) / 2.
Example 3. A 512 Hz tuning fork gives resonance at L₁ = 15.8 cm and L₂ = 49.4 cm.
λ = 2 × (49.4 − 15.8) = 2 × 33.6 = 67.2 cm, so v = 512 × 0.672 ≈ 344 m/s.
e = (49.4 − 3 × 15.8) / 2 = (49.4 − 47.4) / 2 = 1.0 cm.
Check: L₁ + e = 16.8 cm, which is λ/4 = 67.2 / 4 = 16.8 cm. If e came out negative, a reading would be wrong.
Which mistakes cost marks?
Even harmonics in a closed pipe. They do not exist.
Mixing up overtones and harmonics. In a closed pipe, the first overtone is the third harmonic and the second overtone is the fifth.
Using λ = 4L for the second resonance. It is 3λ/4 = L₂ + e.
Forgetting μ is mass per length. Convert grams per metre to kg/m.
Standing waves link closely with simple harmonic motion and with the Doppler effect and its sign rules. If you find it hard to picture nodes and antinodes forming from two travelling waves, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
A practice routine
Day 1: Four string problems, changing tension, length and μ in turn.
Day 2: Four pipe problems, including two where you identify the pipe type from two resonant frequencies.
Day 3: Three resonance tube calculations from given readings, finding v and e each time.
Day 4: A mixed set with beats between a string and a pipe.
Tonight, sketch the first three modes of a closed pipe, marking displacement nodes with N and antinodes with A, then label where pressure varies most.
