Physics 2Waves, Sound and Physical Optics

Standing Waves on Strings

A string fixed at both ends resonates at fₙ = nv/(2L).

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Waves on a guitar string reflect from both fixed ends (Reflection at Boundaries), so waves run both ways at once. By Superposition and Interference they add, and for certain frequencies the result stands still in place: a standing wave.

Key idea

A string of length $L$ fixed at both ends resonates when a whole number of half-wavelengths fit: $$\lambda_n = \frac{2L}{n}, \qquad f_n = \frac{nv}{2L} = nf_1, \quad n = 1, 2, 3, \dots$$ Nodes never move; antinodes swing the most. Neighboring nodes are $\lambda/2$ apart.
The third harmonic: three loops, four nodes (counting the ends) and three antinodes.

Why it works

Both ends are fixed, so both must be nodes. Nodes come every half wavelength, so the string must hold a whole number of half wavelengths, $L = n\frac{\lambda}{2}$. The wave speed is set by the string, $v = \sqrt{F_T/\mu}$, so $f = v/\lambda = nv/(2L)$. Tighten the string and every note goes up; press a fret to shorten it and the pitch rises too. The lowest mode, $n = 1$, is the fundamental; the higher ones are harmonics, and the mix of harmonics a string plays at once gives each instrument its own sound.

Common mistake

Thinking a node is where the string moves the most. A node is a point that does not move at all; an antinode is where the string swings widest. Check: the fixed ends are nodes, and they never move.

Worked example: A guitar string

A string $0.65\ \text{m}$ long carries waves at $286\ \text{m/s}$. Find its fundamental and the next two harmonics.
  1. $f_1 = \dfrac{v}{2L} = \dfrac{286}{2(0.65)} = 220\ \text{Hz}$.
  2. $f_2 = 2f_1 = 440\ \text{Hz}$ and $f_3 = 3f_1 = 660\ \text{Hz}$.

Answer: $220$, $440$ and $660\ \text{Hz}$

Worked example: Counting loops

A string $1.2\ \text{m}$ long vibrates in 4 loops at $160\ \text{Hz}$. Find the wavelength and the wave speed.
  1. Four loops is $n = 4$: $\lambda = \dfrac{2L}{n} = \dfrac{2(1.2)}{4} = 0.60\ \text{m}$.
  2. $v = f\lambda = (160)(0.60) = 96\ \text{m/s}$.
  3. The nodes are $\lambda/2 = 0.30\ \text{m}$ apart.

Answer: $\lambda = 0.60\ \text{m}$, $v = 96\ \text{m/s}$

Standing waves are resonances: the string responds strongly only at its own frequencies, the way a swing builds up only when pushed in time. See resonance build in the Pendulums and Resonance lab.
Try it in the Science LabPendulums and Resonance

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Sources and further reading

  • OpenStax University Physics Volume 1, Section 16.6: Standing Waves and Resonance
  • OpenStax College Physics 2e, Section 16.10: Superposition and Interference
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