Apply Standing Waves & Resonance — Apply C.4 Standing waves and resonance in problem-solving and explanations - IB Physics

SECTION 1

Historical Context & Motivation

Humans have been making music for thousands of years, but the physics behind musical instruments remained mysterious until scientists began studying how waves behave inside confined spaces. When you pluck a guitar string or blow into a flute, you are creating standing waves — patterns that seem to vibrate in place rather than travel from one end to the other. Understanding these patterns turned out to be crucial not only for music, but also for engineering, architecture, and even quantum mechanics.

~500 BCE • Pythagoras & the Monochord

Pythagoras discovered that vibrating strings of specific length ratios produce harmonious sounds, linking mathematics to music for the first time.

1636 • Mersenne's Laws

Marin Mersenne published quantitative relationships between string length, tension, mass per unit length, and frequency — the first equations for vibrating strings.

1787 • Chladni Patterns

Ernst Chladni sprinkled sand on vibrating plates and revealed stunning nodal patterns, providing the first visual evidence of two-dimensional standing waves.

1842 • Doppler & Wave Theory Advances

Christian Doppler's work on wave frequency shifts deepened the mathematical framework used to analyse resonance in acoustic systems.

1924 • De Broglie's Matter Waves

Louis de Broglie proposed that electrons behave as standing waves in atoms, extending the concept of resonance from macroscopic strings to quantum physics.

The central question this topic addresses is: How do we predict the frequencies and wavelengths at which standing waves form in strings and air columns, and how can we use those predictions to solve real-world problems? By the end of this lesson you will be able to answer that question confidently and apply the ideas in IB Physics exam-style problems.

SECTION 2

Core Principles & Definitions

Before diving into calculations, you need a solid grasp of the key ideas behind standing waves. A standing wave is not a single travelling wave — it results from the superposition of two waves of the same frequency and amplitude travelling in opposite directions. When these waves overlap, certain points along the medium experience zero displacement at all times (called nodes), while other points oscillate with maximum amplitude (called antinodes). The wave appears to "stand still" because the nodes never move.

Superposition

When two or more waves overlap, the resultant displacement at any point is the vector sum of the individual displacements. This principle is the foundation of standing wave formation.

Nodes & Antinodes

Nodes are points of permanent zero displacement; antinodes are points of maximum displacement. The distance between two adjacent nodes equals half a wavelength (λ/2).

Boundary Conditions

The type of boundary (fixed or free) determines whether a node or antinode forms at the end. A fixed end produces a node; an open end produces an antinode.

Harmonics

A harmonic is a specific standing wave pattern. The first harmonic (fundamental) has the lowest frequency. Higher harmonics are integer multiples of the fundamental frequency.

Resonance

Resonance occurs when an external driving frequency matches a natural frequency of the system. At resonance the amplitude of oscillation reaches a maximum, and energy transfer is most efficient.

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KEY TAKEAWAY

Think of a standing wave like two people shaking a jump rope at the same time from opposite ends. Where their shakes perfectly cancel, the rope stays still (node). Where their shakes add up, the rope swings wildly (antinode). Resonance is what happens when you shake at just the right rhythm — the rope's motion becomes huge with very little effort.

SECTION 3

Visualising Standing Waves on a String

The diagram below shows the first three harmonics on a string that is fixed at both ends. Notice how each successive harmonic adds one more node and one more antinode. The fundamental (1st harmonic) fits exactly half a wavelength between the two fixed ends. The 2nd harmonic fits one full wavelength, and the 3rd harmonic fits one and a half wavelengths.

The first three harmonics of a string fixed at both ends. Red dots mark nodes (N) and green circles mark antinodes (A). Dashed curves show the wave at its opposite phase. Each harmonic adds one more half-wavelength segment.

The general pattern for a string fixed at both ends is that the nth harmonic fits exactly n half-wavelengths into the length L of the string. This gives us the relationship L = nλ/2, which you can rearrange to find the wavelength of any harmonic. Because every harmonic is an integer multiple of the fundamental frequency, the series of allowed frequencies is called the harmonic series.

SECTION 4

Mathematical Framework

Now let's formalise the relationships you saw in the diagrams. The equations below are the tools you need for IB Physics problems involving standing waves on strings and in air columns.

WAVE SPEED EQUATION

v = f × λ

v = wave speed (m s⁻¹), f = frequency (Hz), λ = wavelength (m). This equation applies to all waves and is the starting point for most problems.

STRING FIXED AT BOTH ENDS — HARMONICS

λₙ = 2L / n and fₙ = n × v / (2L)

n = harmonic number (1, 2, 3, …), L = length of string (m). All harmonics are allowed because both boundaries are identical (fixed ends → nodes).

OPEN PIPE (OPEN AT BOTH ENDS)

λₙ = 2L / n and fₙ = n × v / (2L)

n = 1, 2, 3, … Both ends are antinodes. The formula is identical to the fixed-string formula, and all harmonics are present.

CLOSED PIPE (ONE END CLOSED, ONE END OPEN)

λₙ = 4L / n and fₙ = n × v / (4L)

n = 1, 3, 5, … (odd harmonics only). The closed end is a node and the open end is an antinode. Only odd harmonics are allowed because an even number of quarter-wavelengths cannot satisfy these asymmetric boundary conditions.

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IB Exam Tip

Many students lose marks by using n = 1, 2, 3, … for a closed pipe. Remember: a pipe closed at one end only supports odd harmonics (n = 1, 3, 5, …). The harmonic number always refers to the frequency ratio: the 3rd harmonic has frequency 3f₁, the 5th harmonic has frequency 5f₁, and so on. For a closed pipe, the allowed harmonics are the 1st (n = 1), 3rd (n = 3), and 5th (n = 5) — so if a question asks for the "third harmonic of a closed pipe," that is n = 3, giving f₃ = 3v/(4L). The even harmonics (n = 2, 4, 6, …) simply do not exist for a closed pipe.

SECTION 5

Standing Waves in Open and Closed Pipes

Standing waves in air columns follow the same superposition principle as strings, but the boundary conditions differ. At a closed end of a pipe, air cannot move freely, so a displacement node forms. At an open end, the air is free to oscillate, so a displacement antinode forms. These boundary conditions determine which harmonics are possible.

Comparison of standing wave patterns in open and closed pipes. The thick vertical bar on the closed pipe represents the sealed end where a node always forms. Note that the closed pipe skips even harmonics entirely.

A practical consequence of these patterns is that a closed pipe produces a fundamental frequency that is half that of an open pipe of the same length. This is because the closed pipe only fits one quarter-wavelength (λ/4) at the fundamental, whereas the open pipe fits a half-wavelength (λ/2). The missing even harmonics in a closed pipe also give it a distinctly different tone — this is why a clarinet (effectively a closed pipe) sounds different from a flute (an open pipe), even when they play the same note.

SECTION 6

Worked Example — Resonance in a Closed Pipe

A pipe closed at one end has a length of 0.85 m. The speed of sound in air is 340 m s⁻¹. Calculate the frequencies of the first three resonant modes of this pipe.

Closed Pipe — First Three Resonant Frequencies

1
Step 1 — Identify Given Values

L = 0.85 m, v = 340 m s⁻¹. The pipe is closed at one end and open at the other.
2
Step 2 — Choose the Correct Formula

For a closed pipe: fₙ = n × v / (4L), where n = 1, 3, 5, … (odd harmonics only). The first three resonant modes correspond to n = 1, n = 3, and n = 5.
3
Step 3 — Calculate f₁ (Fundamental)

f₁ = 1 × 340 / (4 × 0.85) = 340 / 3.40 = 100 Hz

f₁ = 100 Hz
4
Step 4 — Calculate f₃ (Third Harmonic)

f₃ = 3 × 340 / (4 × 0.85) = 1020 / 3.40 = 300 Hz. Notice this is exactly 3 × f₁.

f₃ = 300 Hz
5
Step 5 — Calculate f₅ (Fifth Harmonic)

f₅ = 5 × 340 / (4 × 0.85) = 1700 / 3.40 = 500 Hz. Again, this is 5 × f₁.

f₅ = 500 Hz
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Step 6 — Verify & Reflect

The three resonant frequencies are 100 Hz, 300 Hz, and 500 Hz. They are all odd multiples of the fundamental, which confirms we used the correct formula. There are no resonant modes at 200 Hz or 400 Hz for this closed pipe.

SECTION 7

Comparing Standing Wave Systems

Different systems produce standing waves under different constraints. The table below summarises the key features of the three main systems you encounter in IB Physics C.4.

Feature	String (fixed both ends)	Open Pipe	Closed Pipe
Boundary at each end	Node – Node	Antinode – Antinode	Node – Antinode
Fundamental wavelength λ₁	2L	2L	4L
Fundamental frequency f₁	v / (2L)	v / (2L)	v / (4L)
Harmonics present	All (n = 1, 2, 3, …)	All (n = 1, 2, 3, …)	Odd only (n = 1, 3, 5, …)
Frequency formula	fₙ = nv / (2L)	fₙ = nv / (2L)	fₙ = nv / (4L)
Real-world example	Guitar string, piano wire	Flute, organ pipe (open)	Clarinet, bottle

Comparison of three standing wave systems

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KEY TAKEAWAY

The boundary conditions are the whole story. If both ends are the same type (both nodes or both antinodes), you get all harmonics and use 2L. If the ends are different (one node, one antinode), you get only odd harmonics and use 4L. Once you know the boundaries, the rest is just plugging into v = fλ.

SECTION 8

Connections to Advanced Topics

Standing waves and resonance are not just exam topics — they connect to ideas across physics and engineering. In quantum mechanics, the allowed energy levels of an electron in an atom can be understood as standing wave patterns of the electron's probability wave. The mathematics is remarkably similar: only certain wavelengths "fit" inside the potential well, just as only certain wavelengths fit on a vibrating string.

IB C.4 Concept	Advanced / University Extension
Nodes and antinodes on a string	Quantum probability density nodes in hydrogen atom orbitals
Resonance at natural frequencies	Resonance in RLC electrical circuits; structural resonance in bridges
Harmonic series (f₁, 2f₁, 3f₁, …)	Fourier analysis — decomposing any periodic signal into its harmonics
Boundary conditions determine allowed modes	Solutions to the Schrödinger equation require boundary conditions to yield quantised energy levels

From IB C.4 to university physics

If you continue to study physics at university, you will find that the skills you build in C.4 — identifying boundary conditions, selecting the correct harmonic formula, and interpreting wave patterns — transfer directly to quantum mechanics, electromagnetism, and signal processing. The concept of resonance alone appears in almost every branch of physics.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A guitar string vibrating at its second harmonic has three nodes (including the two fixed ends). Explain why a node must form at each fixed end and describe what happens at the point exactly halfway along the string.

Show answer

A node must form at each fixed end because the string is clamped and cannot move at those points — the displacement is always zero. At the second harmonic, the string vibrates in two half-wavelength segments, so there is an additional node at the exact midpoint. This midpoint node separates two antinodes, each located one-quarter of the string length from the ends. The two halves of the string oscillate in opposite directions (antiphase), and the midpoint remains stationary at all times.

Question 2: BASIC CALCULATION

A string of length 1.20 m is fixed at both ends. The wave speed on the string is 240 m s⁻¹. Calculate the frequency of the third harmonic.

Show answer

For a string fixed at both ends, fₙ = n × v / (2L). Substituting: f₃ = 3 × 240 / (2 × 1.20) = 720 / 2.40 = 300 Hz. The third harmonic frequency is 300 Hz. As a check, the fundamental is f₁ = 240 / 2.40 = 100 Hz, and 3 × 100 = 300 Hz, which confirms the result.

Question 3: INTERMEDIATE

An organ pipe open at both ends produces a fundamental frequency of 262 Hz (middle C). The speed of sound is 343 m s⁻¹. (a) Calculate the length of the pipe. (b) If one end of the same pipe is now closed, what is the new fundamental frequency?

Show answer

(a) For an open pipe, f₁ = v / (2L), so L = v / (2f₁) = 343 / (2 × 262) = 343 / 524 ≈ 0.654 m. (b) When one end is closed, the pipe becomes a closed pipe with f₁ = v / (4L) = 343 / (4 × 0.654) = 343 / 2.616 ≈ 131 Hz. This is exactly half of the original fundamental frequency. The closed pipe's fundamental is one octave lower than the open pipe's fundamental, which makes physical sense because the closed pipe only fits a quarter-wavelength at the fundamental, not a half-wavelength.

Question 4: APPLIED

A student blows across the top of a test tube that is 18.0 cm long and closed at the bottom. The air temperature is 20 °C (speed of sound = 343 m s⁻¹). She hears a clear tone. (a) Model the test tube as a closed pipe and calculate the fundamental frequency. (b) She then pours water into the test tube until the air column is only 12.0 cm long. Calculate the new fundamental frequency and state whether the pitch goes up or down.

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(a) Using f₁ = v / (4L) = 343 / (4 × 0.180) = 343 / 0.720 ≈ 476 Hz. (b) With water added, L = 0.120 m: f₁ = 343 / (4 × 0.120) = 343 / 0.480 ≈ 715 Hz. The pitch goes up because shortening the air column increases the fundamental frequency. This is the same principle behind blowing across bottles filled with different amounts of water — more water means a shorter air column, a shorter wavelength, and therefore a higher pitch.

Question 5: CRITICAL THINKING

Two pipes of identical length L are placed side by side. Pipe A is open at both ends and pipe B is closed at one end. A student claims that the 2nd harmonic of pipe A has the same frequency as the 3rd harmonic of pipe B. Evaluate whether this claim is correct, showing your reasoning mathematically.

Show answer

For pipe A (open): f₂ = 2v / (2L) = v / L. For pipe B (closed), the 3rd harmonic means n = 3, giving f₃ = 3v / (4L). Setting them equal: v / L = 3v / (4L) gives 4 = 3, which is false. The claim is incorrect. The 2nd harmonic of the open pipe (frequency v/L) does not equal the 3rd harmonic of the closed pipe (frequency 3v/4L) when the pipes have the same length. Note that for a closed pipe, 'the 3rd harmonic' correctly means n = 3 (frequency 3f₁ = 3v/4L); the even harmonics simply do not exist for a closed pipe, but the harmonic number still refers to the frequency ratio. Students should take care not to confuse harmonic number (which defines the frequency ratio) with the ordinal count of allowed resonant modes.

SUMMARY

Lesson Summary

Summary

Standing waves form when two identical waves travelling in opposite directions undergo superposition, creating fixed nodes (zero displacement) and antinodes (maximum displacement). The boundary conditions of the system determine which harmonics are allowed: systems with matching boundaries (string fixed at both ends, open pipe) support all harmonics using fₙ = nv/(2L), while systems with mixed boundaries (closed pipe) support only odd harmonics using fₙ = nv/(4L).

Resonance occurs when a driving frequency matches one of these natural frequencies, causing a large amplitude response. To solve IB C.4 problems, always start by identifying the boundary conditions, then select the correct formula, and remember that the wave equation v = fλ links frequency, wavelength, and wave speed. The distance between adjacent nodes is always λ/2, which provides a powerful shortcut for reading wavelengths directly from diagrams.