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Physics · 7. Waves and sound

Travelling waves, sound and standing waves

Connect frequency, wavelength and speed; explain reflection, resonance and harmonics with wave models.

45 minutesContent version: 2.2

Driving question

Why does a string resonate strongly only at particular frequencies?

Curriculum coverage

Vietnam Physics 11–12; Cambridge IGCSE 0625 and AS & A Level 9702: properties of waves, sound and stationary waves.

Model scope and limitations

The string model assumes a uniform string, small oscillations, fixed ends and negligible damping. For sound in air, speed depends on the medium and temperature, not on source frequency.

Learning objectives

  • Use v = fλ and distinguish source-controlled from medium-controlled quantities.
  • Identify transverse and longitudinal waves, nodes and antinodes.
  • Derive the natural frequencies of a string fixed at both ends.

Prerequisite knowledge

  • Period T = 1/f and unit conversion.
  • Tension, mass and square roots.

Core knowledge

Wave equation

A periodic wave obeys v = fλ. Across a boundary, source frequency stays constant while v and λ may change.

Standing wave

Two equal-frequency, equal-amplitude waves travelling oppositely superpose to form zero-amplitude nodes and maximum-amplitude antinodes.

String fixed at both ends

The condition L = nλ/2 gives fₙ = nv/(2L), n = 1, 2, 3, …; adjacent nodes are λ/2 apart.

Worked example

For a 0.80 m string with linear density μ = 5.00×10⁻³ kg/m and tension 80.0 N, v = √(F/μ) ≈ 126 m/s. The fundamental is f₁ = v/(2L) ≈ 79.1 Hz and the third harmonic is about 237 Hz.

Misconceptions to avoid

  • Greater amplitude does not make a wave travel faster in the linear model of the same medium.
  • Nodes and antinodes are fixed features of the pattern; medium particles still oscillate away from nodes.
  • A wave transfers energy without transporting the medium bodily over long distances.

Virtual experiment procedure: Standing waves on a string

  1. Choose string length and tension, then predict the fundamental before running.
  2. Sweep frequency slowly and record values producing 1, 2 and 3 clear antinodes.
  3. Test fₙ/f₁ ≈ n and measure adjacent-node spacing to infer λ.

Safety and cautions

  • In a real experiment, secure the string and masses and stay out of the path of a string that could snap.

Evidence to collect

  • A table of antinode count n, frequency fₙ, inferred wavelength and ratio fₙ/f₁.
  • A conclusion linking tension, speed and resonant frequency, with a model limitation.
Open the virtual experiment
📋 Embed this simulation in online lectures (For Teachers & LMS)

Copy the iframe code below to embed this interactive simulation into Canvas, Moodle, Google Sites or educational blogs:

<iframe src="https://visuallab.win/physics-lab/?experiment=g12-standingwave" width="100%" height="560" frameborder="0" allow="fullscreen" loading="lazy"></iframe>

Diagnostic check

1. Which quantity is set by the oscillator when a wave crosses into another medium?

  1. Frequency
  2. Speed
  3. Wavelength
Show answer and explanation

Correct answer: Frequency

Frequency remains continuous across the boundary; speed and wavelength adjust to the medium.

2. The distance between adjacent nodes is?

  1. λ/2
  2. λ
  3. λ/4
Show answer and explanation

Correct answer: λ/2

Adjacent nodes are separated by half a wavelength.

3. What is the unit of frequency?

  1. Hz = s⁻¹
  2. m/s
  3. m
Show answer and explanation

Correct answer: Hz = s⁻¹

Frequency is the number of cycles per second.

Application practice

1. A wave has f = 200 Hz and λ = 1.70 m. What is its speed?

  1. 340 m/s
  2. 118 m/s
  3. 201.7 m/s
Show answer and explanation

Correct answer: 340 m/s

v = fλ = 200×1.70 = 340 m/s.

2. A 0.60 m string fixed at both ends has wave speed 120 m/s. What is f₁?

  1. 100 Hz
  2. 200 Hz
  3. 72 Hz
Show answer and explanation

Correct answer: 100 Hz

f₁ = v/(2L) = 120/1.20 = 100 Hz.

3. If string tension is quadrupled while μ and L stay fixed, how does f₁ change?

  1. It doubles
  2. It quadruples
  3. It is unchanged
Show answer and explanation

Correct answer: It doubles

Since v = √(F/μ), quadrupling F doubles v and every natural frequency.

Academic references

  1. Cambridge IGCSE Physics 0625 — official syllabus and resources
  2. Cambridge AS & A Level Physics 9702 — official syllabus
  3. OpenStax Physics — peer-reviewed high-school physics
  4. BIPM — SI Brochure, 9th edition

This lesson is maintained against the cited sources below. Simulations are learning models and do not replace supervised physical-laboratory safety procedures.

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WebLab STEM
Technical maintainer
Đức Tiến — Control & Automation Engineer