C.4 Standing Waves and Resonance

IB DP PHYSICS Theme C: Wave Behaviour C.4 Standing Waves and Resonance

Revision Notes · Standard and Higher Level Fahad H. Ahmad

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Original notes prepared for the IB Diploma Programme Physics course (first assessment 2025)

What the syllabus requires

By the end of C.4 you should be able to work confidently with each of the following. Use this list as a final checklist before the exam.

Understanding You should be able to...

Formation of standing waves Explain a standing wave as the superposition of two identical waves travelling in opposite directions, usually a wave and its reflection.

Nodes and antinodes Locate them, state that adjacent nodes are λ/2 apart, and describe the phase of points between and across nodes.

Strings and pipes Sketch and interpret harmonic patterns; use λn = 2L/n (string, open pipe) and λn = 4L/n, n odd (pipe closed at one end).

Standing vs travelling waves Contrast energy transfer, amplitude and phase behaviour of the two types.

Natural frequency and resonance Describe free, damped and forced oscillations, and the large-amplitude response when the driving frequency matches the natural frequency.

Damping Distinguish light, critical and heavy damping and state the effect of damping on the resonance curve.

Exam note: C.4 content is common to SL and HL. Standing-wave patterns are a favourite for data-based questions, and the resonance-curve description is frequently asked as a 3-4 mark explain question.

1. How standing waves form

A standing (stationary) wave is produced by the superposition of two waves of the same frequency, wavelength and (ideally) amplitude travelling through the same medium in opposite directions. In practice the two waves are almost always an incident wave and its reflection from a boundary - the fixed end of a string, or the end of an air column.

At every point the displacements of the two waves add (principle of superposition). At some points the two waves always arrive in antiphase and cancel permanently: these are nodes. Midway between them the waves arrive in phase and reinforce, giving points of maximum amplitude: antinodes. The result is a pattern that oscillates but does not travel - the wave profile stays in place while the string or air within it vibrates.

Key phrase for the exam: “A standing wave forms when two waves of equal frequency and amplitude travelling in opposite directions superpose.” All four ideas - two waves, identical, opposite directions, superposition - are needed for full marks.

Standing waves compared with travelling waves

Property Travelling (progressive) wave Standing wave

Energy Transfers energy through the medium in the direction of travel

Stores energy in the oscillation; no net energy is transferred along the wave

Property Travelling (progressive) wave Standing wave

Amplitude Every point oscillates with the same amplitude (no absorption)

Amplitude varies with position: zero at nodes, maximum (2A) at antinodes

Phase Phase changes continuously with position: points one wavelength apart are in phase

All points between two adjacent nodes are in phase; points on opposite sides of a node are in antiphase

Wave profile Moves forward at the wave speed v Does not move; the pattern of loops is fixed in space

Frequency All points oscillate at the wave frequency All points oscillate at the same frequency (except nodes, which do not move)

2. Nodes, antinodes and phase

• A node is a point of permanently zero displacement (complete destructive superposition).
• An antinode is a point of maximum amplitude, equal to twice the amplitude of each travelling wave.
• Adjacent nodes are separated by λ/2; adjacent antinodes are also λ/2 apart.
• A node and the nearest antinode are separated by λ/4.

Picture one “loop” of a vibrating string between two nodes: every particle in that loop moves up together and down together - they are in phase (phase difference zero) although their amplitudes differ. Particles in the next loop do the opposite: when one loop is at its highest, the neighbouring loop is at its lowest. Points separated by one node are therefore in antiphase (phase difference 180°), and points separated by two nodes are back in phase.

Contrast to remember: In a travelling wave, phase difference depends on separation (Δx/λ of a cycle). In a standing wave there are only two possibilities: in phase (same loop, or an even number of nodes apart) or antiphase (an odd number of nodes apart).

The standing wave through one cycle

Freeze a vibrating string at successive instants. At t = 0 every point is at its extreme displacement; a quarter of a period later the whole string passes through the flat, undisplaced position - every particle is moving at its fastest, and the energy of the wave is entirely kinetic; at t = T/2 each loop has inverted. Twice in every cycle the string is momentarily completely straight. Each point oscillates inside a fixed envelope: its amplitude depends only on its position between the nodes, while its frequency is the same everywhere.

Because the node positions are fixed, standing waves only “fit” on a string or in a pipe for particular wavelengths - those that satisfy the boundary conditions. This is why each system has a discrete set of natural frequencies, the harmonics.

3. Standing waves on strings

A string fixed at both ends must have a node at each end. The simplest pattern that fits is a single loop - one antinode in the middle. This is the fundamental (first harmonic): the string length is half a wavelength, so λ1 = 2L. Adding one more node at a time gives the higher harmonics.

Harmonic n Pattern (both ends fixed) Nodes Antinodes Wavelength λn Frequency fn

(fundamental) one loop: N-A-N 2L f1 = v / 2L two loops: N-A-N-A-N L 2f1 three loops 2L/3 3f1 n n loops n + 1 n 2L/n nf1 The allowed frequencies follow from v = fλ with the wave speed v on the string fixed by its physical properties:

fn = nv / 2L = nf1, n = 1, 2, 3, ...

What decides the fundamental frequency?

Since f1 = v / 2L, anything that changes the length or the wave speed changes the pitch. The speed of a transverse wave on a string is v = √(T/μ), where T is the tension and μ the mass per unit length (this speed formula supports the reasoning; the syllabus expects the qualitative conclusions):

• Shorter string (fretting a guitar): f1 ∝ 1/L, so pitch rises.
• Greater tension (tightening the tuning peg): v increases, so f1 rises. Quadrupling T doubles f1.
• Thicker, heavier string (larger μ): v decreases, so f1 falls - bass strings are thick or wire-wound.

A real plucked string vibrates in a mixture of harmonics at once. The fundamental determines the note; the blend of higher harmonics determines the characteristic quality (timbre) that distinguishes a guitar from a violin playing the same note.

Figure 1. Harmonics of a string fixed at both ends (node N at each end). The fundamental has one loop; each higher harmonic adds a node. With v = 143 m s-1 and L = 0.65 m, fn = n·v/(2L) gives f1 = 110 Hz, f2 = 220 Hz, f3 = 330 Hz.

Worked example 1 - guitar string harmonics

A guitar string of length 0.65 m fixed at both ends sounds its fundamental at 110 Hz. Find (a) the speed of waves on the string, (b) the frequency and wavelength of the third harmonic.

• Fundamental: λ1 = 2L = 1.30 m, so v = f1λ1 = 110 × 1.30 = 143 m s-1.
• f3 = 3f1 = 330 Hz; λ3 = 2L/3 = 0.43 m. Check: 330 × 0.433 ≈ 143 m s-1, as required - the speed is a property of the string, not of the harmonic.

Worked example 2 - identifying a harmonic

A string of length 0.75 m fixed at both ends vibrates in a standing wave with four nodes in total (including the ends). The frequency is 480 Hz. Find the wavelength, the wave speed, and the fundamental frequency.

Four nodes means three loops: the third harmonic, so λ3 = 2L/3 = 2 × 0.75 / 3 = 0.50 m.

v = fλ = 480 × 0.50 = 240 m s-1.

f1 = f3 / 3 = 160 Hz.

4. Standing waves in pipes (air columns)

Sound reflecting up and down an air column superposes with itself to form longitudinal standing waves. The boundary conditions are:

• Closed end: air cannot move along the pipe, so there is a displacement node.
• Open end: air moves freely, so there is a displacement antinode (in reality just beyond the rim - see the end correction in Section 5).

Pipe open at both ends

Antinode at each end. The simplest pattern is A-N-A: half a wavelength fits in the pipe, exactly as for a string (with nodes and antinodes swapped). So an open-open pipe supports all harmonics:

λn = 2L / n, fn = nv / 2L, n = 1, 2, 3, ...

Pipe closed at one end

Node at the closed end, antinode at the open end. The simplest pattern is N-A: only a quarter of a wavelength fits, so λ1 = 4L. The next pattern that satisfies both boundary conditions (N-A-N-A) fits three quarter-wavelengths, giving frequency 3f1. A closed pipe therefore produces odd harmonics only:

λn = 4L / n, fn = nv / 4L, n = 1, 3, 5, ...

Most common C.4 error: For a closed pipe the harmonic number n counts 1, 3, 5, ... - there is no second harmonic. The “first overtone” of a closed pipe is the third harmonic at 3f1, not 2f1. Never write fn = nv/4L with n = 2.

Figure 2. Fundamental modes of an open pipe (antinode A at both ends, all harmonics, fn = n·v/2L) versus a pipe closed at one end (node at the closed end, antinode at the open end, odd harmonics only, fn = n·v/4L, n odd). For v = 340 m s-1 and L =

0.50 m the open pipe sounds f1 = 340 Hz but the closed pipe only 170 Hz - an octave lower.

Comparing the three systems

Feature String, both ends fixed Pipe, both ends open Pipe, one end closed

Ends Node - Node Antinode - Antinode Node (closed) - Antinode (open)

Fundamental λ1 = 2L, f1 = v/2L λ1 = 2L, f1 = v/2L λ1 = 4L, f1 = v/4L

Harmonics present all: f1, 2f1, 3f1, ... all: f1, 2f1, 3f1, ... odd only: f1, 3f1, 5f1, ...

Wavelengths 2L/n, n = 1, 2, 3... 2L/n, n = 1, 2, 3... 4L/n, n = 1, 3, 5...

Which v? speed of the wave on the string speed of sound in the air speed of sound in the air

Note the practical consequence: for the same length and the same wave speed, a closed pipe sounds an octave lower than an open pipe (f1 is halved), and its missing even harmonics give it a distinctive hollow tone - the clarinet (effectively closed at the reed end) compared with the flute (open).

Worked example 3 - open pipe

An organ pipe open at both ends is 0.85 m long. The speed of sound is 340 m s-1. Find the fundamental frequency and the frequencies of the next two harmonics.

λ1 = 2L = 1.70 m, so f1 = v / λ1 = 340 / 1.70 = 200 Hz.

All harmonics are present: f2 = 400 Hz, f3 = 600 Hz.

Worked example 4 - closed pipe

A pipe 0.17 m long is closed at one end. Taking the speed of sound as 340 m s-1, find the three lowest frequencies at which it resonates.

Fundamental: λ1 = 4L = 0.68 m, so f1 = 340 / 0.68 = 500 Hz.

Only odd harmonics exist: f3 = 3 × 500 = 1500 Hz and f5 = 5 × 500 = 2500 Hz. (There is no resonance at 1000 Hz or 2000 Hz.)

Worked example 5 - which harmonic is this?

A pipe of length 0.60 m closed at one end resonates strongly at 425 Hz (speed of sound 340 m s-1). Which harmonic is sounding?

λ = v / f = 340 / 425 = 0.80 m.

For a closed pipe λn = 4L/n, so n = 4L / λ = (4 × 0.60) / 0.80 = 3: the third harmonic - consistent, because n must be odd.

5. Measuring the speed of sound: the resonance tube

A classic experiment (and IA favourite): a tuning fork of known frequency f is held over a vertical tube whose air-column length can be changed by raising or lowering the water level inside it. The water surface acts as the closed end. As the column is lengthened, the sound becomes loud whenever the column resonates - that is, whenever a closed-pipe standing wave of the fork's frequency fits the column.

• First (shortest) resonance: L1 ≈ λ/4
• Second resonance: L2 ≈ 3λ/4
• Therefore L2 - L1 = λ/2 exactly, and v = fλ = 2f(L2 - L1).

The antinode actually forms a small distance e (the end correction, roughly 0.6 × the tube radius) above the open end, so L1 + e = λ/4. Using the difference of two resonance lengths eliminates e - which is why the two-resonance method is better than using L1 alone.

Worked example 6 - speed of sound by resonance tube

A 512 Hz tuning fork is held over a resonance tube. The sound is loudest at column lengths of 16.0 cm and 49.2 cm. Find the speed of sound and the end correction.

λ/2 = L2 - L1 = 49.2 - 16.0 = 33.2 cm, so λ = 0.664 m.

v = fλ = 512 × 0.664 = 340 m s-1.

End correction: L1 + e = λ/4 = 16.6 cm, so e = 16.6 - 16.0 = 0.6 cm.

6. Standing waves in the laboratory

Vibrating string (Melde's experiment)

A string runs from a signal-generator-driven vibration generator over a pulley to a hanging load, which sets the tension T. As the driving frequency is slowly increased, clear standing-wave patterns appear one after another whenever the frequency matches a harmonic of the stretched string - this is itself an example of resonance. Measuring the loop length gives the wavelength (one loop = λ/2), and changing the hanging mass tests how the wave speed depends on tension: doubling f1 requires four times the tension, confirming v = √(T/μ).

Worked example 7 - vibration generator and pulley

A string of mass per unit length 4.0 g m-1 is stretched between a vibration generator and a pulley 1.5 m away, with tension 90 N provided by a hanging load. The frequency is raised until the string vibrates in three loops. Find the wave speed on the string and the generator frequency.

v = √(T/μ) = √(90 / 0.0040) = √(22500) = 150 m s-1.

Three loops on length 1.5 m: λ = 2L/3 = 1.0 m, so f = v/λ = 150 Hz (the third harmonic of this string).

Microwaves: measuring the speed of light in a kitchen

With the turntable removed, the microwave field inside an oven forms a standing wave. A bar of chocolate left in place melts first at the antinodes, which are λ/2 apart. Melted spots typically about 6.1 cm apart give λ ≈ 0.122 m; with the oven frequency f = 2.45 GHz printed on the back panel, c = fλ ≈ 2.45 × 109 × 0.122 ≈ 3.0 × 108 m s-1. The same physics - fixed hot and cold spots - is why ovens use rotating turntables at all.

Other classic demonstrations

• Kundt's tube: fine powder in a horizontal tube collects in small heaps at the displacement nodes of a sound standing wave, making the node spacing (λ/2) directly measurable.
• Chladni plates: sand on a vibrating plate gathers along nodal lines, showing two-dimensional standing-wave patterns.
• Resonance tube (Section 5): the standard IB method for the speed of sound.

Experiment logic to quote: Every standing-wave measurement works the same way: locate two adjacent nodes or antinodes, take their separation as λ/2, then combine with a known frequency using v = fλ.

7. Resonance and forced oscillations

Free, damped and forced oscillations

• A free oscillation occurs when a system is displaced and released: it oscillates at its natural frequency f0, set only by the system's own properties (mass and stiffness of a spring system; length of a pendulum; length, tension and μ of a string).
• A damped oscillation is a free oscillation losing energy to resistive forces, so its amplitude decreases with time.
• A forced oscillation occurs when a periodic external force (the driver) is applied: the system settles into oscillation at the driver's frequency, not its own.

Resonance

Resonance occurs when the driving frequency equals (or is very close to) the natural frequency of the system. Energy is then transferred from the driver to the oscillator most efficiently - each push arrives in step with the motion - so the amplitude builds up to a maximum, limited only by damping.

The amplitude-frequency (resonance) curve

Plot the steady amplitude of the driven oscillator against the driver frequency:

• At very low driver frequency, the oscillator simply follows the driver: amplitude roughly equals the driver amplitude.
• As the driver frequency approaches f0, the amplitude rises steeply to a sharp peak at (very near) f0.
• Beyond f0 the amplitude falls away, tending to a very small value at high frequency - the system cannot keep up with the driver.

Phase of the driven oscillator

The phase relationship between driver and oscillator changes across the curve: well below f0 the oscillator moves in phase with the driver; at resonance it lags the driver by a quarter of a cycle (90°), which puts the driving force in phase with the velocity - the condition for maximum power transfer; well above f0 it approaches antiphase. Barton's pendulums show all of this at once: of several paper-cone pendulums hung from the same cord as a heavy driver pendulum, only the one whose length (and hence natural frequency) matches the driver builds up a large amplitude.

Figure 3. Resonance response: steady amplitude versus driving frequency for a system with natural frequency f0 = 50 Hz. Lighter damping gives a taller, narrower peak; heavier damping a lower, broader one. The peak sits just below f0 (at f ≈ 49.7

Hz for the light curve).

Effect of damping on the curve

Increasing the damping makes the peak lower and broader, and shifts the peak to a slightly lower frequency than f0. With very light damping the peak is tall and narrow - the response is dramatic but only over a narrow band of driving frequencies.

Type of damping Behaviour Example

Light (under-damping) Oscillates with gradually (approximately exponentially) decaying amplitude; period almost unchanged

Pendulum in air; guitar string ringing on

Critical damping Returns to equilibrium in the shortest possible time with no oscillation

Car suspension; analogue meter needles; some door closers

Heavy (over-damping) Returns to equilibrium slowly, without oscillating Oscillation in thick oil; heavily damped door closer

Resonance in the real world

Situation What resonates Useful or harmful?

Musical instruments Strings and air columns driven at their natural frequencies; the body of the instrument amplifies them

Useful - it is how the sound is produced

Microwave oven The 2.45 GHz field drives rotation of polar water molecules, transferring energy to the food

Useful

MRI scanner Nuclear magnetic resonance: nuclei absorb radio-frequency energy at their precession frequency

Useful - medical imaging

Wine glass and sound A loud tone at the glass's natural frequency can build a large enough amplitude to shatter it

Demonstration of harmful resonance

Bridges and buildings Wind, marching feet or earthquakes can drive structures near a natural frequency (e.g. the Millennium Bridge's lateral wobble, 2000)

Harmful - engineers add damping and change f0

Vehicle parts Mirrors, panels and aerials buzz at certain engine speeds Harmful - designed out with damping

Explain-question checklist: A full “explain resonance” answer states: (1) the system has a natural frequency; (2) it is driven by a periodic force; (3) when driver frequency = natural frequency, energy transfer to the system is maximum; (4) so amplitude becomes large / maximum.

8. Energy in damped oscillations

An undamped oscillator swaps energy back and forth between kinetic and potential forms with the total constant. Damping (friction, air resistance) converts that energy irreversibly to internal (thermal) energy, so the total mechanical energy falls with time.

• With light damping, the amplitude decays in an approximately exponential way: it falls by the same fraction in each successive cycle.
• Energy is proportional to amplitude squared (E ∝ A2), so the energy decays by a constant fraction per cycle too - and faster than the amplitude. If A halves, E falls to one quarter.
• The oscillation frequency is very slightly lowered by damping, but for light damping this shift is negligible.
• In a driven system at steady state, the driver supplies energy at exactly the rate damping dissipates it - which is why the amplitude is steady, and why weaker damping allows a larger resonant amplitude.

Worked example 8 - energy decay

A lightly damped pendulum has amplitude 40 mm. After 10 complete cycles the amplitude is 20

• (a) What fraction of the initial energy remains? (b) Estimate the amplitude after 20 cycles.
• E ∝ A2, so E/E0 = (20/40)2 = 0.25 - one quarter remains (75% has been dissipated as thermal energy).
• Exponential-style decay loses the same fraction in equal numbers of cycles: the amplitude halves every 10 cycles, so after 20 cycles A ≈ 10 mm.

9. Common pitfalls

• Using fn = nv/4L with even n for a closed pipe - closed pipes have odd harmonics only. Count n = 1, 3, 5, ...
• Writing the node spacing as λ instead of λ/2, or the node-to-antinode distance as λ/2 instead of λ/4.
• Confusing the number of loops (= harmonic number n for a string) with the number of nodes (n + 1) or antinodes (n).
• Claiming a standing wave transfers energy along the medium - it stores energy; there is no net transfer.
• Saying all points of a standing wave have the same amplitude. Amplitude depends on position (zero at nodes, 2A at antinodes); it is the frequency that is common to all points.
• Using the speed of sound for waves on a string, or the string-wave speed for the sound the string produces in air. The frequency is the same in both; the wavelengths differ.
• Forgetting that a displacement node in a sound wave is a pressure antinode (the closed end of a pipe is where pressure varies most).
• Saying the resonance peak of a damped system is exactly at f0 - damping lowers and broadens the peak and shifts it slightly below f0.

10. Quick reference

Result Statement

Formation Two identical waves travelling in opposite directions superpose (wave + its reflection)

Node spacing adjacent nodes (or antinodes): λ/2; node to nearest antinode: λ/4

Phase rule same loop: in phase; opposite sides of a node: antiphase (180°)

String (fixed-fixed) / open-open pipe

λn = 2L/n, fn = nv/2L = nf1, n = 1, 2, 3, ...

Pipe closed at one end λn = 4L/n, fn = nv/4L = nf1, n = 1, 3, 5, ... only

Resonance tube v = 2f(L2 - L1); end correction e = λ/4 - L1

Resonance maximum amplitude when driver frequency = natural frequency f0

Result Statement

More damping resonance peak lower, broader, at slightly lower frequency; light damping: A decays exponentially, E ∝ A2

11. Test yourself

Attempt these without notes; full answers below.

• State the conditions needed for two waves to form a standing wave, and explain why a reflected wave usually satisfies them.
• A string 0.90 m long, fixed at both ends, has a fundamental frequency of 150 Hz. Find the speed of waves on the string and the frequency of the fourth harmonic.
• In a standing sound wave in a pipe, adjacent nodes are 0.25 m apart. Taking the speed of sound as 340 m s-1, find the frequency.
• An organ pipe open at both ends sounds a fundamental of 264 Hz (speed of sound 340 m s-1). Find its length.
• What length of pipe closed at one end would give the same fundamental of 264 Hz? Comment on the comparison.
• Two points on a vibrating string lie on opposite sides of a node; two other points lie within the same loop. State the phase difference for each pair.
• A tuning fork of frequency 480 Hz gives resonances of a closed tube at column lengths 17.2 cm and 52.6
• Find the speed of sound and the end correction.
• Explain why a pipe closed at one end cannot produce the second harmonic.
• The tension in a string is quadrupled without changing its length or mass per unit length. What happens to the fundamental frequency? Justify your answer.
• Sketch (or describe) resonance curves for the same oscillator with light and with heavier damping, and identify three differences between them.

Answers

• Same frequency and wavelength, similar amplitude, travelling in opposite directions through the same region. A reflected wave automatically has the same frequency and (nearly) the same amplitude as the incident wave and travels back through it - so incident + reflected waves superpose to form the standing wave.
• λ1 = 2L = 1.80 m, so v = 150 × 1.80 = 270 m s-1. Harmonics of a string are all integers: f4 = 4 × 150 = 600 Hz.
• Node spacing = λ/2, so λ = 0.50 m and f = v/λ = 340 / 0.50 = 680 Hz.
• λ1 = v/f = 340 / 264 = 1.29 m; L = λ1/2 = 0.64 m.
• Closed pipe: L = λ1/4 = 1.29 / 4 = 0.32 m - half the length of the open pipe for the same note. (This is why closed organ pipes save space, at the cost of missing even harmonics.)
• Opposite sides of a node: antiphase, phase difference 180° (half a cycle). Within the same loop: in phase, phase difference zero - even though their amplitudes differ.
• L2 - L1 = λ/2 = 52.6 - 17.2 = 35.4 cm, so λ = 0.708 m and v = 480 × 0.708 = 340 m s-1. End correction: e = λ/4 - L1 = 17.7 - 17.2 = 0.5 cm.
• The boundary conditions demand a node at the closed end and an antinode at the open end. A pattern with frequency 2f1 (wavelength 2L) would need either nodes or antinodes at both ends, which is impossible here - only patterns fitting an odd number of quarter-wavelengths (odd harmonics) satisfy node-at-one-end, antinode-at-the-other.
• v = √(T/μ), so quadrupling T doubles v. Since f1 = v/2L and L is unchanged, the fundamental frequency doubles (the note rises by an octave).
• Both curves peak near the natural frequency f0. With heavier damping the peak is (i) lower - smaller maximum amplitude, (ii) broader - the system responds over a wider range of driving frequencies, and (iii) at a slightly lower frequency than f0. At very low driving frequencies the two curves converge to roughly the driver amplitude.