Year 11 Physics Module 3: Waves 40 min Lesson 7 of 18

Standing Waves and Resonance

A standing wave is not a wave travelling along the medium. It is the stable pattern formed when two identical waves travel in opposite directions. That pattern reveals nodes, antinodes, harmonics, and the idea of resonance — the reason a guitar string can ring out a clear note while other frequencies die away almost instantly.

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Think First

A guitar string can appear to vibrate in a stable pattern with some points staying still while others move a lot. How can a wave pattern contain points that never move at all?

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Key Relationships — This Lesson

$L = \dfrac{n\lambda}{2}$
L = string length n = harmonic number \lambda = wavelength of the standing wave mode
Fundamental: $L = \lambda/2$   |   Second harmonic: $L = 2\lambda/2$   |   Resonance: driving frequency = natural frequency

Formula Reference — Standing Waves

$L = n\lambda/2$
Standing Waves on a String Fixed at Both Ends
n = 1,2,3,\dots
Use when: finding allowed wavelengths on a string with nodes at both ends.
Common trap: do not set $L = n\lambda$ for a string fixed at both ends. Each loop is half a wavelength, not a full wavelength.
resonance when driving frequency = natural frequency
Resonance Condition
Maximum amplitude occurs when the forcing frequency matches a natural mode.
Use when: explaining why some frequencies produce very large responses.
Common trap: resonance does not mean "large frequency." It means matching frequency.

Know

  • How standing waves form
  • What nodes and antinodes are
  • What resonance means
  • The string relationship $L = n\lambda/2$

Understand

  • Why standing waves are formed by opposite-travelling waves
  • Why some points remain at zero displacement
  • Why resonance causes large amplitudes
  • Why only certain harmonics fit on a fixed string

Can Do

  • Identify nodes and antinodes on a diagram
  • Work out wavelengths for harmonics on a string
  • Explain resonance in practical contexts
  • Apply $L = n\lambda/2$ correctly

Misconceptions to Fix

Wrong: Work and energy are completely different concepts.

Right: Work is the transfer of energy; they share the same unit (joules) and are fundamentally linked.

📚 Core Content

Key Terms
standing wavenot a wave travelling along the medium
Each loophalf a wavelength, not a full wavelength
Why standing wavesformed by opposite-travelling waves
Work and energycompletely different concepts
Workthe transfer of energy; they share the same unit (joules) and are fundamentally linked
rate at which workdone or energy is transferred; P = W/t
01How Standing Waves Form

How Standing Waves Form

Standing waves are formed by the superposition of two identical waves travelling in opposite directions.

On a string fixed at both ends, a wave pulse generated at one end travels toward the far end, reflects, and travels back. If the outgoing and reflected waves have the same frequency, amplitude, and speed, they interfere continuously as they pass through each other. At some positions the displacements always add to zero — these become nodes. At other positions they always add constructively — these become antinodes. The result is a stationary pattern that appears to stand still, even though the medium is oscillating.

The key condition is that the reflected wave must be identical to the incident wave. This is why fixed ends are important: a fixed end inverts the reflected pulse, ensuring that the superposition produces nodes exactly at the boundary. Without this inversion, the boundary would not become a node and the standing-wave pattern would not lock into place.

Real-World Anchor Surf lifesavers on Australian beaches use standing-wave knowledge to identify rip currents. When incoming and reflected swells overlap near a headland, they create regions of unusually calm water (nodes) next to regions of intense chop (antinodes). Recognising these patterns helps locate where the water is moving out to sea.
02Nodes and Antinodes

Nodes and Antinodes

Nodes stay at zero displacement. Antinodes oscillate with maximum amplitude.

Nodes are the fixed points in a standing wave where destructive interference occurs at all times. The two opposite-travelling waves arrive at a node exactly out of phase by half a wavelength, so their displacements cancel continuously. Antinodes are the points midway between nodes where constructive interference produces the largest oscillation. In diagrams, nodes are often shown at the ends of a string fixed at both ends.

Between any two adjacent nodes, all particles move in phase — they reach maximum displacement together and pass through equilibrium together. However, particles on opposite sides of a node move in antiphase: when one loop moves upward, the adjacent loop moves downward. This phase jump of 180° at each node is a defining signature of a standing wave.

node antinode

Nodes remain still. Antinodes oscillate the most.

Common Misconceptions

A node is a point where no energy exists.
A node has zero displacement at all times, but energy still passes through it. In a standing wave, kinetic energy is maximum at the node when the string is flat, and potential energy is maximum at the antinode at maximum displacement. Energy shuttles back and forth between kinetic and potential forms.
Standing waves are two waves that stop moving.
The individual waves keep travelling. It is the pattern that appears stationary. Each point on the string still oscillates (except at nodes) — the wave components do not freeze.
Nodes and antinodes are spaced a full wavelength apart.
Adjacent nodes are half a wavelength apart, and adjacent antinodes are also half a wavelength apart. A full wavelength contains two loops, each loop being half a wavelength.
03Resonance

Resonance

Resonance occurs when the driving frequency matches a natural frequency of the system.

When a system is driven at one of its natural frequencies, energy transfer becomes especially efficient and amplitude rises dramatically. This happens because each push from the driver arrives in phase with the motion, adding constructively to the oscillation rather than fighting against it. Over many cycles the amplitude grows until energy input is balanced by energy losses such as sound radiation and internal damping.

Resonance is not about the size of the frequency — it is about the match. A tiny driving force at exactly the right frequency can eventually produce a large amplitude, whereas a very large force at the wrong frequency may produce almost no response. This is why engineers must design bridges and buildings to avoid resonant frequencies that match typical wind or earthquake frequencies. The 1940 collapse of the Tacoma Narrows Bridge is a classic — though somewhat debated — example of how wind excitation at a resonant frequency can drive catastrophic oscillations in a structure.

The sharpness of a resonance is described by its quality factor, or Q-factor. A high-Q system (like a tuning fork or a guitar string) has a very narrow resonance peak: only frequencies extremely close to the natural frequency produce a large response, and the sound rings on for a long time after the driving force stops. A low-Q system (like a damped car suspension) has a broad resonance peak and the oscillations die away quickly. Musicians prefer high-Q strings because they sustain notes; acoustic engineers often prefer low-Q room surfaces because they prevent unwanted resonant ringing.

Real-World Anchor A guitar string responds strongly at specific resonant frequencies, which is why it can produce clear notes rather than every possible frequency equally well. When a guitarist plucks the string, the broad initial disturbance contains many frequencies, but only the resonant modes survive and ring on.
Real-World Anchor Concert sound engineers in Sydney must check that bass frequencies do not excite resonant modes of the Opera House stage structures. If a subwoofer happens to match a natural frequency of a lighting rig or floor panel, that structure can vibrate visibly — an unwanted acoustic resonance.
04Harmonics on a String

Harmonics on a String

Only certain wavelengths fit on a string with nodes at both ends.

For a string fixed at both ends, the ends must be nodes. The simplest pattern that satisfies this is the fundamental, where the string vibrates in one loop with an antinode in the middle. The second harmonic fits two loops on the string, the third harmonic fits three loops, and so on. Because each loop is half a wavelength, the length of the string must equal an integer number of half-wavelengths. This pattern is captured by $L = n\lambda/2$.

The frequency of each harmonic is related to the wave speed on the string by $v = f\lambda$. Since $v$ depends on tension and mass per unit length, a guitarist can tune a string by changing its tension, which shifts all harmonic frequencies together while keeping their ratios fixed.

HarmonicRelationshipWavelengthNumber of loops
1st (fundamental)$L = \lambda/2$$\lambda = 2L$1
2nd$L = 2\lambda/2$$\lambda = L$2
3rd$L = 3\lambda/2$$\lambda = 2L/3$3
4th$L = 4\lambda/2$$\lambda = L/2$4
5th$L = 5\lambda/2$$\lambda = 2L/5$5
Vector Protocol — harmonic wavelength problems
Step 1 — Confirm the boundary condition: string fixed at both ends → nodes at both ends
Step 2 — Write $L = n\lambda/2$ before substituting any numbers
Step 3 — Solve for $\lambda$ and check that the answer gets smaller as $n$ increases
05Standing Waves in Pipes

Standing Waves in Pipes

Strings are not the only place standing waves appear — air columns inside pipes support them too.

In a pipe open at both ends, the air at each opening is free to move, so the openings behave like antinodes. The simplest standing-wave mode has an antinode at each end and a node in the middle, which means the length of the pipe equals half a wavelength — just like a string. Therefore, the harmonics in an open-open pipe satisfy the same relationship as a string: $L = n\lambda/2$, with all integer harmonics allowed.

In a pipe closed at one end, the closed end is a node (no air motion) while the open end is an antinode. The simplest mode then fits only a quarter of a wavelength inside the pipe, giving $L = \lambda/4$. Because the closed end must always be a node, only odd harmonics are possible: $L = n\lambda/4$ where $n = 1, 3, 5, \dots$. This is why a closed pipe produces a duller, less bright timbre than an open pipe — the even harmonics are missing entirely.

Pipe typeBoundary conditionsHarmonic relationshipAllowed harmonics
Open at both endsAntinode at each end$L = n\lambda/2$All integers $n = 1, 2, 3, \dots$
Closed at one endNode at closed end, antinode at open end$L = n\lambda/4$Odd integers $n = 1, 3, 5, \dots$
Real-World Anchor Australian didgeridoos are traditionally played as closed-open pipes. The player’s lips create the vibration and the tube length determines the fundamental pitch. Because only odd harmonics are present, the didgeridoo has a distinctive hollow, drone-like quality that differs from open-ended flutes.
Vector Protocol — pipe standing-wave problems
Step 1 — Identify the pipe ends: open = antinode, closed = node
Step 2 — Draw the simplest mode (fundamental) to count half-wavelengths
Step 3 — Check whether even harmonics are allowed (only for open-open pipes)
06Energy in a Standing Wave

Energy in a Standing Wave

A standing wave has no net energy transfer along the medium, but energy is continuously converted between kinetic and potential forms.

In a progressive wave, energy travels with the disturbance from one place to another. In a standing wave, the two travelling waves carry equal amounts of energy in opposite directions. The net energy flux along the string is zero — energy does not pile up at one end or drain away at the other. Instead, energy oscillates back and forth between the loops of the standing wave.

When the string passes through its equilibrium position (straight line), all the energy is kinetic — the particles are moving fastest. When the string reaches maximum displacement, all the energy is potential — the particles momentarily stop while the string is stretched into its curved shape. This exchange happens twice every period. Understanding this energy dance explains why a guitar string can keep ringing: once energy is injected, it sloshes between kinetic and potential with very little loss until damping drains it away.

Key Exam Move If a question asks whether energy is transferred along the medium in a standing wave, the answer is no net transfer. Energy is stored and exchanged within the pattern, not transported along it.
07From Wavelength to Frequency

From Wavelength to Frequency

Finding the wavelength is only half the story — musicians and physicists usually care about the frequency.

The wave speed $v$ on a string depends on the tension $T$ and the mass per unit length $\mu$ according to $v = \sqrt{T/\mu}$. Once $v$ is known, the frequency of each harmonic follows from $v = f\lambda$. Substituting $\lambda = 2L/n$ gives the set of allowed frequencies:

$$f_n = \frac{nv}{2L} = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$$

This equation explains three things a guitarist instinctively knows: tightening a string raises all harmonics (increases $T$), pressing the string against a fret shortens the vibrating length $L$ and raises the pitch, and thicker strings (larger $\mu$) produce lower notes at the same length and tension.

It also reveals that the harmonics of a string fixed at both ends are integer multiples of the fundamental: $f_1, 2f_1, 3f_1, \dots$. This simple integer ratio is what makes string instruments sound harmonious — the brain perceives these precise multiples as musical tones rather than noise.

Real-World Anchor Piano tuners in Australia must account for the fact that steel strings have a slightly non-ideal stiffness that raises the upper harmonics a little above exact integer multiples. This "inharmonicity" means that a concert piano is not tuned to perfect mathematical harmonics, but to a stretched scale that sounds pleasing to the human ear.

✏️ Worked Examples

Worked Example 1 Type 7 — Harmonics

Problem Setup

Scenario: A string fixed at both ends has length 1.2 m. Find the wavelength of the first three harmonics.

  • $L = 1.2\ \text{m}$

Solution

1
1st harmonic: $\lambda = 2L = 2.4\ \text{m}$
The fundamental fits half a wavelength on the string.
2
2nd harmonic: $\lambda = L = 1.2\ \text{m}$
The second harmonic fits one full wavelength along the string.
3
3rd harmonic: $\lambda = 2L/3 = 0.80\ \text{m}$
Use $L = 3\lambda/2$ and rearrange.

What would change if...

If the string were shorter, all allowed wavelengths would decrease because each harmonic has to fit inside a smaller length. If the tension were increased, the wave speed would rise and the frequencies of all harmonics would increase, but their wavelengths would stay the same because the boundary conditions have not changed.

Vector Protocol — checking your harmonic answer
Step 1 — Did you use $L = n\lambda/2$ for a string fixed at both ends?
Step 2 — Does your wavelength decrease as $n$ increases?
Step 3 — Are your units in metres and is your answer smaller than $2L$?
Worked Example 2 Type 7 — Resonance

Problem Setup

Scenario: Explain why a driven string suddenly reaches a much larger amplitude when the driving frequency matches one of its allowed harmonic frequencies.

Solution

1
Driving frequency matches natural frequency
This is the resonance condition.
2
Energy transfer becomes highly efficient
Each push arrives in step with the motion rather than opposing it.
3
Amplitude grows dramatically
The system stores more oscillation energy at that matched frequency because constructive interference between the driven and reflected waves reinforces the motion at every cycle.

What would change if...

If the driving frequency did not match a natural mode, the amplitude would stay smaller because the forcing would not stay in step with the motion. After a few cycles the driver would sometimes add energy and sometimes subtract it, leading to no sustained buildup.

Visual Break

Decision Flowchart — Standing Wave Problems

Standing wave problem String fixed at both ends? Nodes must exist at each end Yes Use $L=n\lambda/2$ No Check boundaries Count loops Each loop = $\lambda/2$

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Formation

  • Standing waves form from opposite-travelling identical waves
  • Superposition creates a stationary interference pattern
  • Fixed ends invert the reflection, ensuring nodes at boundaries
  • No net energy transfer along the medium

Nodes and Antinodes

  • Nodes = permanent zero displacement (destructive interference)
  • Antinodes = maximum displacement (constructive interference)
  • Adjacent nodes are $\lambda/2$ apart
  • Particles in the same loop move in phase

Harmonics on a String

  • $L = n\lambda/2$ for a string fixed at both ends
  • $n$ = number of loops = harmonic number
  • Wavelength decreases as harmonic number increases
  • Frequency ratios are simple whole numbers

Resonance

  • Resonance when driving frequency = natural frequency
  • Energy transfer is most efficient at resonance
  • Amplitude grows dramatically at matched frequencies
  • Only harmonics that fit the boundary conditions resonate

🏃 Activities

Activity 1

Node or Antinode?

For a standing wave diagram, identify which labelled points are nodes and which are antinodes. Then explain how you know by referring to interference.

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Activity 2

Harmonic Check

A string has length 0.90 m. Find the wavelength of the second and third harmonics. Show your working using $L = n\lambda/2$.

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Activity 3

Resonance Reasoning

Explain in one or two sentences why pushing a swing at random times is less effective than pushing it in rhythm. Use the word "resonance" in your answer.

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Activity 4

Energy Exchange in a Standing Wave

Describe where the kinetic energy and potential energy are maximum in a standing wave on a string. Explain why the energy shuttles between these two forms rather than travelling along the string.

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Interactive: Standing Wave Harmonics Builder
Interactive: Standing Wave Spotter
Revisit Your Thinking

Earlier you were asked how a wave pattern can contain points that never move.

The full answer: in a standing wave, opposite-travelling identical waves superpose. At some positions, destructive interference happens continuously, creating nodes of permanent zero displacement. At other positions, constructive interference creates antinodes of maximum oscillation.

Now revisit your prediction. How does superposition explain the still points?

Annotate your prediction in your book with what you now understand differently.

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Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

✅ Check Your Understanding

Multiple Choice

6 MARKS

1. A standing wave is formed by:

A
One wave stopping completely
B
Two identical waves travelling in opposite directions
C
Two waves of different frequencies only
D
A wave crossing a boundary

2. A node in a standing wave is a point of:

A
Maximum displacement
B
Maximum speed of the whole wave
C
Permanent constructive interference
D
Permanent zero displacement

3. An antinode is a point of:

A
Maximum displacement
B
Zero displacement
C
No oscillation at all
D
Zero frequency

4. For a string fixed at both ends, the allowed wavelengths satisfy:

A
$L = n\lambda$
B
$L = \lambda/n$
C
$L = n\lambda/2$
D
$L = 2n/\lambda$

5. Resonance occurs when:

A
The amplitude is zero
B
The driving frequency matches a natural frequency
C
The wavelength becomes infinite
D
Two unrelated sources interfere

6. A string of length 1.0 m vibrates in the fundamental. The wavelength is:

A
0.50 m
B
1.0 m
C
1.5 m
D
2.0 m

Short Answer

10 MARKS

7. Explain the difference between a node and an antinode in a standing wave. 3 MARKS

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8. A string fixed at both ends has length 1.5 m. Find the wavelength of the third harmonic. 3 MARKS

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9. Explain why resonance causes a dramatic increase in amplitude in a mechanical system. 4 MARKS

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Comprehensive Answers

Multiple Choice

1. B — standing waves form from opposite-travelling identical waves.

2. D — nodes remain at zero displacement.

3. A — antinodes have maximum displacement.

4. C — for a string fixed at both ends, $L = n\lambda/2$.

5. B — resonance is matching the driving and natural frequencies.

6. D — the fundamental has $\lambda = 2L = 2.0\ \text{m}$.

Short Answer — Model Answers

Q7 (3 marks): A node is a point of permanent zero displacement in a standing wave. It remains still because destructive interference happens there at all times. An antinode is a point of maximum displacement, where constructive interference produces the largest oscillation.

Q8 (3 marks): Use $L = n\lambda/2$. So $1.5 = 3\lambda/2$, giving $\lambda = 1.0\ \text{m}$.

Q9 (4 marks): Resonance causes a dramatic increase in amplitude because the driving frequency matches a natural frequency of the system. This means each input of energy arrives in step with the oscillation and adds efficiently to the motion rather than disrupting it. As more energy is added in phase, the amplitude grows much larger than it would at a non-matching frequency. In a mechanical system such as a swing or a guitar string, resonance allows a small periodic force to build up a large oscillation over many cycles, limited only by energy losses such as friction, air resistance, and sound radiation.

🏎️
Speed Race

Race Through Standing Waves!

Answer questions on standing waves, nodes, antinodes and resonance. Pool: lessons 1–7.

Mark lesson as complete

Tick when you have finished the activities and checked the answers.