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Guitar string: tension to wave speed to the fifth harmonic

SPH3U Grade 11 Physics · Waves and Sound

For the waves unit, a student brings a guitar into the lab and the class turns one string into the experiment. It is a wound string, and its packaging lists a linear mass density of 4.00 g/m; a ruler laid along the neck puts the vibrating length — stretched between the nut and the bridge — at 65.0 cm. Brought up to pitch on the tuning machine, the string carries a tension measured at 67.6 N. Plucked open, it sounds its fundamental; touched lightly at the right spot and plucked again, it rings instead with a high, flute-like harmonic. Find the speed of a wave travelling along the string, the fundamental frequency it sounds when plucked, the frequency of its fifth harmonic, and the wavelength of that harmonic on the string.

T = 67.6 Nfifth harmonic: 500.0 Hzμ = 4.00 g/mL = 65.0 cmλ₅ = 0.260 m

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • μ = 4 g/m — Linear mass density of the string
  • L = 65 cm — Vibrating length, nut to bridge
  • F = 67.6 N — Tension in the string
Determine
  1. (a)the speed of a wave on the string
  2. (b)the fundamental frequency when plucked
  3. (c)the frequency of the fifth harmonic
  4. (d)the wavelength of that harmonic on the string
Step 1 of 4(a) · solve for Wave speed

Wave speed is set by the string itself — tension pulling back, mass resisting — and by nothing else. The length has not entered yet, and neither has any frequency.

FFvμ
Rearranged for v
v=Fμv = \sqrt{\tfrac{F}{\mu}}
Your values, in your units
v=(67.6 N)(4 g/m)v = \sqrt{\tfrac{\left( 67.6\ \text{N} \right)}{\left( 4\ \text{g/m} \right)}}
Converted to base units
v=(67.6 N)(0.004 kg/m)v = \sqrt{\tfrac{\left( 67.6\ \text{N} \right)}{\left( 0.004\ \text{kg/m} \right)}}
Answer
v=130 m/sv = 130\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Wave Speed on a String solver →

Step 2 of 4(b) · solve for Fundamental frequency

Now the length matters. Fixed at both ends, the longest standing wave that fits is one half-wavelength, so λ₁ = 2L and f₁ = v/2L. A scale length is measured off the fretboard in centimetres and the speed from step 1 is in metres per second, so the 65.0 cm has to become 0.650 m before they can be divided.

vfL
Rearranged for f
f=v2Lf = \tfrac{v}{2L}
130 m/scarried from step 1
Your values, in your units
f=(130 m/s)2 (65 cm)f = \tfrac{\left( 130\ \text{m/s} \right)}{2 \, \left( 65\ \text{cm} \right)}
Converted to base units
f=(130 m/s)2 (0.65 m)f = \tfrac{\left( 130\ \text{m/s} \right)}{2 \, \left( 0.65\ \text{m} \right)}
Answer
f=100 Hzf = 100\ \text{Hz}

Carried onward at full precision, not this rounded figure.

Open the Fundamental Frequency of a String solver →

Step 3 of 4(c) · solve for Harmonic frequency

Every harmonic is a whole-number multiple of the fundamental, so the fifth is simply five times step 2.

f1fnn
Rearranged for fₙ
fn=nf1f_n = n f_1
100 Hzcarried from step 2
Your values, in your units
fn=(5) (100 Hz)f_n = \left( 5 \right) \, \left( 100\ \text{Hz} \right)
Answer
fn=500 Hzf_n = 500\ \text{Hz}

Carried onward at full precision, not this rounded figure.

Open the Harmonic Frequencies solver →

Step 4 of 4(d) · solve for Wavelength

The universal wave equation, run backwards, with the speed from step 1 and the frequency from step 3. The answer must equal 2L/5 = 1.30 m/5 — check it.

λvf
Rearranged for λ
λ=vf\lambda = \tfrac{v}{f}
130 m/scarried from step 1
500 Hzcarried from step 3
Your values, in your units
λ=(130 m/s)(500 Hz)\lambda = \tfrac{\left( 130\ \text{m/s} \right)}{\left( 500\ \text{Hz} \right)}
Answer
λ=260 mm\lambda = 260\ \text{mm}

Carried onward at full precision, not this rounded figure.

Open the Wave Speed (v = fλ) solver →

Answer

Therefore waves travel the string at 130.0 m/s, the open string sounds its 100.0 Hz fundamental, the fifth harmonic rings at 500.0 Hz, and that harmonic fits the string with a wavelength of 0.260 m — exactly 2L/5.

Why this order

The chain runs in this order because each stage adds exactly one new fact about the string. Step 1 uses only what the string is made of and how hard it is pulled; a luthier can change the speed with a tuning peg and nothing else. Step 2 adds the boundary conditions — clamped at both ends — which is what quantises the frequencies in the first place. Step 3 adds the integer. The commonest error is trying to shortcut from tension straight to pitch, which hides the square root: to raise a string's fundamental by an octave you must quadruple the tension, and guitars would be unplayable if that were the only way to change notes. It is not — fretting halves L instead, which is linear and costs nothing.

Step 4 exists as a cross-check, and it is the kind of check worth building the habit of. Two independent routes give the wavelength of the fifth harmonic: divide the wave speed by the harmonic's frequency, or note that five half-wavelengths must fit into 65.0 cm so λ₅ = 2L/5. Both land on 0.260 m, and if they had not, one of the earlier steps was wrong. This ladder of integer harmonics is also what makes a guitar sound like a guitar rather than a sine generator: all of them ring at once, in proportions set by where the string is plucked, and touching the string lightly one fifth of the way along silences everything except the multiples of five — the flute-like "harmonic" every guitarist learns in their first year.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.