Music & Acoustics
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Vibrating String Frequency

Fundamental pitch of a string from length, tension and mass.

Formula

f=12LTμf = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

Variables

fFrequency (Hz)
LString length (m)
TTension (N)
μMass per length (kg/m)

Example

L=0.65 m, T=80 N, μ=0.002: f ≈ 152 Hz

How it works

This formula gives the fundamental frequency, or lowest resonant pitch, of an ideal stretched string fixed at both ends. It comes from the fact that waves on the string travel at speed \(v=\sqrt{T/\mu}\), and the lowest standing wave that fits on the string has wavelength \(\lambda=2L\), so \(f=v/\lambda\). Increasing tension makes the string wave faster and raises the pitch, while increasing length or mass per unit length lowers the pitch.

When to use it

Use this when analyzing the pitch of guitar, violin, piano, or other stretched strings, especially when you know the string length, tension, and linear density. It is also useful in physics problems about standing waves, harmonics, and how changing a string's properties changes its tone.

Worked example, step by step

Given: L = 0.80 m T = 120 N μ = 0.006 kg/m Step 1: Substitute into the formula f = (1 / (2L)) \u221a(T/μ) Step 2: Calculate the tension-to-mass ratio T/μ = 120 / 0.006 = 20000 Step 3: Take the square root \u221a(20000) \u2248 141.4 Step 4: Calculate 1/(2L) 1 / (2 \u00d7 0.80) = 1 / 1.60 = 0.625 Step 5: Multiply f = 0.625 \u00d7 141.4 \u2248 88.4 Hz Final answer: f \u2248 88 Hz

Common mistakes to avoid

  • Using the total string mass instead of mass per unit length μ
  • Forgetting that length L is the vibrating length between fixed points, not the full instrument length
  • Confusing the fundamental frequency with higher harmonics, which are integer multiples of this value

Did You Know?

Tightening a guitar string raises pitch as the square root of tension — quadruple the tension only doubles the note.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026