Stationary waves on strings will be described in terms of harmonics. The…

The idea

A stationary wave on a string forms when two identical progressive waves travelling in opposite directions superpose. For a string fixed at both ends, the ends must be nodes, so only certain wavelengths fit: L = nλₙ/2, giving λₙ = 2L/n. The allowed frequencies are fₙ = nv/(2L), where v = √(T/μ), T is tension and μ is mass per unit length. AQA requires these modes to be described as harmonics: first harmonic (n = 1), second harmonic (n = 2), third harmonic (n = 3), and so on. The labels fundamental and overtone must not be used.

For example: A 0.80 m string with v = 240 m s⁻¹ has first-harmonic frequency 150 Hz, second harmonic 300 Hz and third harmonic 450 Hz.

What earns the marks

• A stationary wave on a string is produced by the superposition of two progressive waves of the same frequency and similar amplitude travelling in opposite directions along the string.
• For a string fixed at both ends, each end is a node, so the length L contains a whole number n of half-wavelengths: L = nλₙ/2, hence λₙ = 2L/n.
• The harmonic number n equals the number of half-wavelengths on the string and also the number of antinodes present.
• The frequency of the nth harmonic is fₙ = nv/(2L), where v is the wave speed on the string.
• The wave speed is given by v = √(T/μ), where T is the tension and μ is the mass per unit length, so fₙ = (n/2L)√(T/μ).
• The first harmonic is the lowest-frequency stationary-wave mode; the second harmonic has twice that frequency and the third harmonic three times it, for an ideal string.
• Answers must use harmonic terminology: first harmonic, second harmonic, third harmonic; the terms fundamental and overtone are not used in this specification.

Common mistakes and how to put them right

• Writing that the string length equals one wavelength for the first harmonic.

  Put it right: The length equals one half-wavelength, so λ₁ = 2L.

• Calling the lowest mode the fundamental or labelling higher modes as overtones.

  Put it right: Use first harmonic, second harmonic and third harmonic.

• Assuming frequency is directly proportional to tension.

  Put it right: fₙ ∝ √T, because v = √(T/μ).

• Treating the harmonic number as the number of nodes.

  Put it right: n is the number of half-wavelengths and the number of antinodes, while the number of nodes is n + 1.

Exam tips

• Sketch the string for n = 1, 2 and 3, labelling nodes N and antinodes A, then count half-wavelengths to choose the correct relationship.
• When a question gives tension and mass per unit length, calculate v = √(T/μ) first, then substitute into fₙ = nv/(2L).
• Use harmonic vocabulary throughout; avoid the words fundamental and overtone even if they appear in general reading.
• Check unit consistency: convert grams per metre to kilograms per metre and centimetres to metres before calculating.