Standing Waves on Strings & in Pipes
- Determine the harmonic frequencies of a string fixed at both ends
- Distinguish open-open from open-closed pipes and their harmonic series
- Relate the number of nodes and antinodes to the harmonic number
Boundary conditions choose the frequencies
A standing wave forms only at frequencies whose wavelength fits the boundary conditions. A fixed end must be a node — it cannot move. An open end must be an antinode — it is free to move maximally. Which frequencies are allowed follows entirely from fitting the pattern between those constraints, which is why three different systems give three different harmonic series from the same physics.
Why a closed pipe skips the even harmonics
A pipe closed at one end needs a node at the closed end and an antinode at the open end. The shortest pattern satisfying both is a quarter wavelength, so λ₁ = 4L and f₁ = v/4L — half the frequency of an open pipe of the same length. Adding further nodes always adds a half wavelength, so the allowed wavelengths are 4L, 4L/3, 4L/5 and so on: only the odd multiples. This is why a stopped organ pipe sounds an octave lower and has a distinctly different timbre from an open one of the same length.
Counting nodes and antinodes
For a string in its nth harmonic there are n antinodes and n + 1 nodes, counting the two fixed ends. For an open-open pipe in the nth harmonic there are n + 1 antinodes and n nodes. For an open-closed pipe the pattern is easiest read off the diagram. Being able to count them lets you identify the harmonic from a picture, which is a standard multiple-choice format.
What changes the wave speed
The harmonic frequencies depend on v, so anything that changes the wave speed changes them all. On a string, v depends on tension and linear density — tightening a guitar string raises every harmonic, which is how tuning works. In a pipe, v is the speed of sound in the gas, which rises with temperature and depends on the gas: this is why a wind instrument goes sharp as it warms up, and why helium raises the pitch of a voice.
A pipe 0.60 m long is closed at one end. Using v = 343 m/s, find the fundamental frequency and the next two frequencies at which it resonates.
- 1.Closed at one end, so f_n = nv/4L with n odd.
- 2.Fundamental: f₁ = 343/(4 × 0.60) = 343/2.40 = 143 Hz.
- 3.Next allowed is n = 3: f₃ = 3 × 143 = 429 Hz.
- 4.Then n = 5: f₅ = 5 × 143 = 715 Hz.
- 5.Note 286 Hz (n = 2) is NOT a resonance for this pipe.
The fundamental of a string fixed at both ends is a half wavelength, not a whole one — so λ₁ = 2L, not L. Using λ = L gives every harmonic wrong by a factor of two, and it is the most common error in the topic.
A pipe closed at one end resonates at 200 Hz as its fundamental. Which frequency is also a resonance?
The fundamental wavelength of a string fixed at both ends of length L is:
A guitar string is tightened. Every harmonic frequency:
Sketch the standing-wave pattern before computing. Marking the nodes and antinodes fixes the wavelength geometrically, which is more reliable than recalling which formula has 2L and which has 4L.
Answer the 3 checkpoints as you read.
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