Standing Waves on a String

Key idea: Use boundary conditions on a string to relate length to wavelength and frequency for standing-wave harmonics (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Explain standing-wave formation, nodes, antinodes and energy transfer.
  • Apply boundary conditions to standing waves on stretched strings.
  • Analyse displacement and pressure patterns in resonant air columns and determine sound wavelength.

1. Definitions (Must Know)

A. Standing wave (on a string)

A standing wave forms when two waves of the same frequency travel in opposite directions and superpose.

B. Boundary conditions (fixed ends)

For a string fixed at an end, the displacement at that end must be zero, so the end is a node.

C. Harmonic number, n

n = 1,2,3,… labels the allowed standing-wave modes:

  • n = 1: fundamental (first harmonic)
  • n = 2: second harmonic, etc.

2. Key Ideas (What Earns Marks)

  • A string fixed at both ends has nodes at both ends.
  • Allowed wavelengths (string length L): L = nλ/2 (n = 1,2,3,…) so: λₙ = 2L/n
  • Using v = fλ, the allowed frequencies are: fₙ = nv/2L
  • Higher n means shorter wavelength and higher frequency.
Link: general standing-wave ideas

3. Detailed Explanations

A. Why only certain wavelengths fit

With nodes at both ends, the string must contain an integer number of half-wavelengths: L = nλ/2

That is why only specific wavelengths (and therefore specific frequencies) produce large-amplitude standing waves (resonance).

B. Mode shapes (first three harmonics)

These are the displacement shapes at an instant (shape only).

Standing-wave harmonics on a string (shape only)

Example displacement shapes for the first three harmonics of a string fixed at both ends.

Scroll across the graph to read all labels.

Example displacement shapes for the first three harmonics of a string fixed at both ends.Example displacement shapes for the first three harmonics of a string fixed at both ends.
Standing-wave harmonics for a string fixed at both ends (shape only).
Open full-size graph
View figure data
Values for Standing-wave harmonics on a string (shape only)
Position along string (arbitrary units)n = 1n = 2n = 3
0000
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4. Common Mistakes

  • Using the double-slit formula x = λ D/a (wrong topic).
  • Mixing up L (string length) with λ (wavelength).
  • Forgetting that n starts at 1 (there is no n = 0 harmonic).
  • Forgetting that v = fλ uses SI units.

5. Exam Tips

  • Start by stating the boundary conditions: “both ends fixed, so nodes at both ends”.
  • Write L = nλ/2 first, then convert to λ or f.
  • If the question gives frequency and length, you can solve for wave speed using v = 2Lfₙ/n.

6. Worked Examples

Modelled example 1

Fundamental frequency

Core

Problem

A string fixed at both ends has length 0.80 m and wave speed 120 m s⁻¹. Find its fundamental frequency.
Study the worked solution
  1. Apply the boundary condition

    Method

    The fundamental fits one half-wavelength into the string.

    Reason

    Both fixed ends are displacement nodes.

    Working

    L = λ₁/2 ⇒ λ₁ = 2L
  2. Connect speed and frequency

    Method

    Use f₁ = v/(2L).

    Reason

    Substituting λ₁ = 2L into v = fλ gives the fundamental relation.

    Working

    f₁ = v/(2L)
  3. Calculate

    Method

    The fundamental frequency is 75 Hz.

    Reason

    The wave travels 120 m each second with wavelength 1.60 m.

    Working

    f₁ = 120/2(0.80) = 75 Hz

Guided practice 2

Third harmonic frequency

About 3 min

Problem

The same string has fundamental frequency 75 Hz. Find its third-harmonic frequency.

Try this before viewing the solution

Unit: Hz

Hints

Hint 1: use the harmonic factor
fₙ = nf₁ for a string fixed at both ends.
View solution step by step
  1. Use the allowed-frequency sequence

    Method

    The third harmonic has three times the fundamental frequency.

    Reason

    fₙ = nv/(2L) = nf₁ when v and L are unchanged.

    Working

    f₃ = 3f₁
  2. Calculate

    Method

    f₃ = 225 Hz.

    Reason

    Three half-wavelength loops fit into the same length.

    Working

    f₃ = 3(75) = 225 Hz

Common misconception 3

Find wave speed from a resonance

Find and correct the mistake

Learner claim

A 0.60 m string resonates in its second harmonic at 200 Hz. A learner uses v = 2Lf as though it were the fundamental. Diagnose the method and find v.

Try this before viewing the solution

Unit: m s^-1

View solution step by step
  1. Keep the harmonic number

    Method

    Use fₙ = nv/(2L) with n = 2.

    Reason

    The relation f = v/(2L) applies only to the fundamental.

    Working

    200 = 2v/[2(0.60)]
  2. Simplify the second mode

    Method

    For n = 2, f₂ = v/L.

    Reason

    The second harmonic contains one full wavelength along the fixed-end string.

    Working

    v = f₂L
  3. Calculate

    Method

    The wave speed is 120 m s⁻¹.

    Reason

    Multiply the resonant frequency by the one-wavelength string length.

    Working

    v = (200)(0.60) = 120 m s⁻¹

Examiner practice 4

Identify the harmonic number

3 marks

Examination question

A 0.80 m fixed-end string with wave speed 120 m s⁻¹ resonates at 300 Hz. Find and identify the harmonic number. [3 marks]

Try this before viewing the solution

View solution step by step
  1. Use the modal relation

    1 mark

    Method

    fₙ = nv/(2L).

    Reason

    Both string ends are fixed nodes.

    Working

    fₙ = nv/(2L)
  2. Rearrange and calculate

    1 mark

    Method

    n = 4.

    Reason

    The mode number is obtained from the measured frequency relative to the fundamental spacing.

    Working

    n = 2Lfₙ/v = 2(0.80)(300)/120 = 4
  3. Identify the mode

    1 mark

    Method

    This is the fourth harmonic.

    Reason

    Allowed fixed-end modes use positive integer n beginning at one.

    Working

    n = 4

Challenge 5

Wavelength of a given harmonic

Minimal support

Independent transfer

A fixed-end string has length 0.60 m. Find the wavelength of its fifth harmonic.

Try this before viewing the solution

Unit: m

Hints

Hint 1: count half-wavelengths
Use L = nλ/2 with n = 5.
View solution step by step
  1. Apply the fifth-mode geometry

    Method

    Five half-wavelengths fit into 0.60 m.

    Reason

    The fixed ends are nodes and the fifth harmonic has five loops.

    Working

    L = 5λ/2
  2. Calculate wavelength

    Method

    λ = 0.24 m.

    Reason

    Rearrange the boundary relation for wavelength.

    Working

    λ = 2(0.60)/5 = 0.24 m

7. Mind Stretchers

Mind stretcher 1: Why does increasing n increase frequency?Extension

Explain (without heavy maths) why higher harmonics have higher frequency.

Show Answer

Higher harmonics fit more half-wavelengths into the same length, so the wavelength is shorter. With the same wave speed, v = fλ, a smaller λ means a larger f.

Mind stretcher 2: What happens if the wave speed increases?Extension

For the same string length L, suppose the wave speed v increases (e.g. by increasing tension).

How do the resonant frequencies change? Explain using fₙ = nv/(2L).

Show Answer

From fₙ = nv/(2L), each resonant frequency is directly proportional to v.

So if v increases by some factor, all the harmonics’ frequencies increase by the same factor.

Mind stretcher 3: Optional (Enrichment)Extension

A. One end free (node–antinode)

If one end is fixed (node) and the other end is free (antinode), the allowed lengths are: L = ((2n-1)λ)/4

This is more commonly tested using air columns: see Standing Waves in Air Columns.

Mind stretcher 4: Simulation Bridge: Standing Wave ExplorerExtension

Concept Explorer: Standing Wave Explorer

Toggle string and air-column boundary conditions, change harmonic mode, and test wavelength-frequency relations.

BetaA LevelWavesBest for: A Level waves and superposition
  • Boundary Conditions
  • Harmonics
  • Resonance Spacing
  • f–λ–v Links

Open the full interactive simulation on its own page

Use the standalone simulation page for the live controls, SVG scene, run modes, and scoring flow.

The lesson stays lightweight and links out to the dedicated simulation page.

Investigate strings in the Standing Wave Explorer.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027