Standing waves — nodes, antinodes, harmonics for string, open pipe, closed pipe

Question

How do standing waves form, and what determines the pattern of nodes and antinodes? What harmonics are possible for a string fixed at both ends, an open pipe, and a closed pipe?

(CBSE 11, JEE Main, NEET — harmonic patterns and frequency calculations for strings and pipes are tested every year)

Solution — Step by Step

When two waves of the same frequency and amplitude travel in opposite directions, they superpose to form a standing wave. The resultant wave does not propagate — instead, certain points remain permanently at rest (nodes) and others oscillate with maximum amplitude (antinodes).

Node: Amplitude = 0 (destructive interference). Separation between adjacent nodes = $\lambda/2$
Antinode: Maximum amplitude (constructive interference). Located midway between nodes

Both ends must be nodes (fixed points cannot vibrate).

Condition: $L = n \cdot \frac{\lambda}{2}$ , where $n = 1, 2, 3, ...$

Frequencies:

f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}} = nf_1

$n = 1$ : fundamental (1st harmonic) — 1 antinode, 2 nodes
$n = 2$ : 2nd harmonic (1st overtone) — 2 antinodes, 3 nodes
$n = 3$ : 3rd harmonic (2nd overtone) — 3 antinodes, 4 nodes

All harmonics (integer multiples of fundamental) are present.

Both ends must be antinodes (air is free to vibrate at open ends).

Condition: $L = n \cdot \frac{\lambda}{2}$ , where $n = 1, 2, 3, ...$

Frequencies:

f_n = \frac{nv}{2L} = nf_1

Same frequency formula as the string. All harmonics are present.

$n = 1$ : 2 antinodes (at both ends), 1 node (at centre)
$n = 2$ : 3 antinodes, 2 nodes

Closed end = node, open end = antinode.

Condition: $L = (2n-1) \cdot \frac{\lambda}{4}$ , where $n = 1, 2, 3, ...$

Frequencies:

f_n = \frac{(2n-1)v}{4L}

$n = 1$ : fundamental ( $v/4L$ )
$n = 2$ : 3rd harmonic ( $3v/4L$ )
$n = 3$ : 5th harmonic ( $5v/4L$ )

Only odd harmonics are present (1st, 3rd, 5th, …). This is what gives closed pipes a distinctive, hollow sound.

The fundamental frequency of a closed pipe is half that of an open pipe of the same length.

flowchart TD
    A["Standing Wave System"] --> B{"What are the boundary conditions?"}
    B -->|"Both ends fixed (string)"| C["Node at both ends<br/>L = nλ/2<br/>All harmonics"]
    B -->|"Both ends open (open pipe)"| D["Antinode at both ends<br/>L = nλ/2<br/>All harmonics"]
    B -->|"One closed, one open (closed pipe)"| E["Node at closed end<br/>Antinode at open end<br/>L = (2n-1)λ/4<br/>Odd harmonics only"]

Why This Works

Standing waves are constrained by boundary conditions. A fixed end forces a node (zero displacement); a free/open end forces an antinode (maximum displacement). Between these constraints, only specific wavelengths “fit” — these are the resonant modes. Each allowed wavelength corresponds to a specific harmonic frequency.

The closed pipe gets only odd harmonics because fitting a node at one end and an antinode at the other requires an odd number of quarter-wavelengths. This geometric constraint eliminates all even harmonics.

Common Mistake

Students confuse “harmonic number” with “overtone number.” The fundamental = 1st harmonic = 0th overtone. The 2nd harmonic = 1st overtone. For a closed pipe, the 1st overtone is the 3rd harmonic (not the 2nd harmonic). When a JEE question asks for the “frequency of the 2nd overtone of a closed pipe,” the answer is $5v/4L$ (5th harmonic), not $3v/4L$ . The harmonic-overtone offset trips up many students.

Quick comparison: Open pipe and string — same formula ( $f_n = nv/2L$ , all harmonics). Closed pipe — different formula ( $f_n = (2n-1)v/4L$ , odd harmonics only). If you remember that closed pipes lack even harmonics, you will never confuse the two.

Question

Solution — Step by Step

Why This Works

Common Mistake

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