A-Level — physicsMedium4 min read

Stationary Waves and Resonance

Nodes and antinodes; harmonics on strings and in pipes; resonance conditions

Tutor AI Team2026-02-25T05:23:43.913Z

# Stationary Waves and Resonance — A-Level Physics

Stationary (standing) waves are formed when two progressive waves of the same frequency and amplitude travel in opposite directions and superpose. Unlike progressive waves, they don't transfer energy — they store it. They're fundamental to understanding musical instruments, microwave ovens, and laser cavities.

1. Formation of Stationary Waves

Two waves travelling in opposite directions with the same frequency, wavelength, and amplitude superpose
Common scenario: a wave reflects off a boundary and interferes with the incoming wave

Characteristics

Property	Progressive Wave	Stationary Wave
Energy transfer	Yes (along wave direction)	No (energy stored)
Amplitude	Same everywhere	Varies from 0 (node) to max (antinode)
Phase	Changes continuously along wave	All points between adjacent nodes in phase
Wavelength	Distance between consecutive points in phase	Distance between alternate nodes = $\lambda/2$

2. Nodes and Antinodes

Node: Point of zero displacement (destructive interference). Spacing between nodes = $\lambda/2$ .

Antinode: Point of maximum displacement (constructive interference). Located midway between nodes.

3. Harmonics on a String

A string fixed at both ends vibrates in harmonics (modes):

Fundamental (1st harmonic): $L = \lambda/2$ → $f_1 = v/(2L)$

2nd harmonic: $L = \lambda$ → $f_2 = 2v/(2L) = 2f_1$

nth harmonic: $L = n\lambda/2$ → $f_n = nv/(2L) = nf_1$

Wave Speed on a String

$v = \sqrt{\frac{T}{\mu}}$

Where $T$ = tension (N) and $\mu$ = mass per unit length (kg/m).

So: $f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

4. Stationary Waves in Pipes

Open Pipe (open at both ends)

Antinodes at both open ends
All harmonics present: $f_n = nv/(2L)$
Same formula as a string

Closed Pipe (closed at one end)

Node at closed end, antinode at open end
Only odd harmonics present: $f_n = nv/(4L)$ where $n = 1, 3, 5, ...$
Fundamental: $L = \lambda/4$ → $f_1 = v/(4L)$

5. Required Practical: Stationary Waves on a String

Method

Set up a vibration generator connected to a string over a pulley with hanging masses
Adjust the frequency until stationary wave patterns form
Measure the length of the vibrating section
Count nodes/antinodes to determine the harmonic
Calculate $\lambda$ and wave speed

Measuring Speed of Sound (Resonance Tube)

Hold a tuning fork of known frequency over an open tube
Adjust the length of the air column (by lowering water level)
Find lengths where resonance occurs (loudest sound)
First resonance at $L_1 \approx \lambda/4$ , second at $L_2 \approx 3\lambda/4$
$\lambda = 2(L_2 - L_1)$ and $v = f\lambda$

Worked Example: String Harmonics

Problem

A guitar string of length 65 cm vibrates at its fundamental frequency of 330 Hz. Find the wave speed.

$f_1 = v/(2L)$ → $v = 2Lf_1 = 2 \times 0.65 \times 330 = 429$ m/s

Solution

Worked Example: Closed Pipe

Problem

A closed pipe is 0.85 m long. The speed of sound is 340 m/s. Find the first three resonant frequencies.

$f_1 = v/(4L) = 340/(4 \times 0.85) = 100$ Hz $f_3 = 3f_1 = 300$ Hz (3rd harmonic) $f_5 = 5f_1 = 500$ Hz (5th harmonic)

No even harmonics!

Solution

Worked Example: Resonance Tube

Problem

A tuning fork of 512 Hz resonates with an air column at lengths 16.2 cm and 50.1 cm. Find the speed of sound.

$\lambda = 2(L_2 - L_1) = 2(0.501 - 0.162) = 2 \times 0.339 = 0.678$ m $v = f\lambda = 512 \times 0.678 = 347$ m/s

Solution

7. Practice Questions

1. A string vibrates in its 3rd harmonic. The string length is 90 cm. Calculate the wavelength. (2 marks)
1. A closed pipe has a fundamental frequency of 256 Hz. What are the frequencies of the next two resonant modes? (2 marks)
1. A string of length 1.2 m has a mass of 3 g and is under 50 N tension. Calculate the fundamental frequency. (3 marks)
1. Explain the difference between a node and an antinode, and state how they are formed. (3 marks)
Answers
1. 3rd harmonic: $L = 3\lambda/2$ → $\lambda = 2L/3 = 2(0.9)/3 = 0.6$ m.

Want to check your answers and get step-by-step solutions?

Summary

Stationary waves: two identical waves in opposite directions → nodes + antinodes
No energy transfer; energy is stored
String/open pipe: $f_n = nv/(2L)$ — all harmonics
Closed pipe: $f_n = nv/(4L)$ — odd harmonics only
$v = \sqrt{T/\mu}$ for strings
Resonance tube: $\lambda = 2(L_2 - L_1)$

Tags:

#a-level#physics#waves#stationary-waves#standing-waves#harmonics#resonance

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# Stationary Waves and Resonance — A-Level Physics

1. Formation of Stationary Waves

Characteristics

2. Nodes and Antinodes

3. Harmonics on a String

Wave Speed on a String

4. Stationary Waves in Pipes

Open Pipe (open at both ends)

Closed Pipe (closed at one end)

5. Required Practical: Stationary Waves on a String

Method

Measuring Speed of Sound (Resonance Tube)

Worked Example: String Harmonics

Worked Example: Closed Pipe

Worked Example: Resonance Tube

7. Practice Questions

Answers

Summary

Related Topics

Energy Transfer and Productivity

Inheritance and Genetic Crosses

Populations and Succession

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