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Stationary waves

Two waves of the same frequency travelling in opposite directions superpose into a pattern that stays put. Fix both ends of a string and only certain wavelengths fit, which gives the string a set of natural frequencies. The same argument applied to a column of air explains the notes a wind instrument can play.

Year 12AQA 3.3.1.3 CIE 8.1.1, 8.1.2, 8.1.3, 8.1.4 OCR A 4.4.3(a)(i), 4.4.3(b), 4.4.4(a), 4.4.4(b), 4.4.4(c), 4.4.4(d), 4.4.4(e)(i), 4.4.4(e)(ii), 4.4.4(f), 4.4.4(g)EDEXCEL 63, 67, 68

Builds on Progressive waves.

IN THIS TOPIC

Describe how two travelling waves superpose to form a stationary wave, and define node and antinode.
Use the node and antinode spacings, $\lambda/2$ and $\lambda/4$ , to get from a measured distance to a wavelength.
Find the allowed wavelengths of a string fixed at both ends, and use $v = \sqrt{T/\mu}$ and the first-harmonic frequency equation.
Work out the harmonics of an air column in a closed and in an open tube, and say why a closed tube sounds only the odd ones.
Describe the resonance-tube method for the speed of sound, and say why it uses two lengths rather than one.
State the amplitude and phase relationships that separate stationary waves from progressive ones.

COMMON MISCONCEPTION

A stationary wave is a wave that has stopped moving.

Two waves, one string

Send a single pulse along a rope tied to a wall and it reflects and comes back. Drive the rope continuously instead, and the reflected wave has to travel back through the incoming one. The rope now carries two waves at the same time. They have the same frequency and the same speed, their amplitudes are similar, and they move in opposite directions.

Wherever two waves overlap, the displacement of the rope is the sum of the displacements each wave would produce on its own. That sentence is the principle of superposition, and it is the only new idea this topic needs. Two waves of the same frequency and similar amplitude travelling opposite ways sum to a pattern that stays where it is, a stationary wave, also called a standing wave.

On a driven string a second idea joins the first. Only at the frequencies whose pattern fits the fixed ends do the repeated reflections reinforce cycle after cycle, building the pattern to a size worth seeing. That build-up is resonance, the driven string answering at one of its natural frequencies, and it is a separate idea from the superposition that shapes the pattern.

FIG. 1Two identical waves, one travelling each way (faint), and their point-by-point sum (cyan). The sum never travels. It pumps in place, pinned at the amber nodes, which never move, and swings hardest between them. Every three periods the travelling pair fades and the dashed envelope appears, the outline the oscillation never leaves.

Be careful with the name. The rope itself is still moving, in places more violently than either travelling wave would move it alone. What has stopped is the pattern. The crests no longer travel along the rope. Instead, each point of the rope oscillates about its rest position with a fixed amplitude of its own.

Nodes and antinodes

At certain points the two waves always arrive in antiphase, so they cancel and the rope there never moves at all. These points are called nodes. Midway between them the waves always arrive in phase and reinforce, so the rope swings with the largest amplitude found anywhere on the pattern. These points are antinodes.

That cancellation is total only when the two waves carry equal amplitudes, which is the condition for a pure stationary wave and the case every exam question means. A real support absorbs a little, so the returning wave comes back slightly weaker and the pattern is a partial one, its minima shallow rather than dead still. The spacing does not care. Nodes and antinodes sit where they always sat, and only the depth of the minima changes.

FIG. 2The envelope of the motion. Nodes (N) never move; antinodes (A) oscillate with the largest amplitude. Adjacent nodes are half a wavelength apart.

Two distances are worth learning. Adjacent nodes are $\lambda/2$ apart, because one complete wavelength contains two loops of the pattern, not one. The distance from a node to the nearest antinode is $\lambda/4$ .

A stationary wave also transfers no net energy along the rope. Each section keeps its own energy, which passes back and forth between kinetic and potential forms as the rope oscillates. Contrast this with a progressive wave, whose function is to carry energy from one place to another.

WORKED EXAMPLE

From node spacing to wave speed

Adjacent nodes on a vibrating string are 0.25 m apart, and the string is driven at 660 Hz. Find the wavelength and the wave speed on the string.

Adjacent nodes are half a wavelength apart, never a whole one, so λ = 2 × 0.25 = 0.50 m.

The speed belongs to the travelling waves the pattern is built from. v = fλ = 660 × 0.50 = 330 m s⁻¹.

That factor of two in the first line is the most commonly dropped number in the topic. Picture two loops to a wavelength and you will keep it.

Harmonics on a stretched string

A string fixed at both ends must have a node at each end, because a fixed point cannot move. That single condition restricts the wavelengths the string can support: a pattern only fits if it places a whole number of loops between the two ends.

FIG. 3The first three harmonics of a string of length L. Each pattern adds one loop, and each loop is half a wavelength.

Each loop is half a wavelength, so a pattern of n loops on a string of length L has

\lambda_n = \frac{2L}{n}

NOT ON THE AQA DATA SHEET: LEARN IT

The simplest pattern, n = 1, is the first harmonic. Many books call it the fundamental, and the two names mean the same pattern. It has the longest wavelength the string allows, 2L, and therefore the lowest frequency. The string fixes the wave speed, so $v = f\lambda$ forces the harmonic frequencies into whole-number multiples of the first, $f_n = nf_1$ .

Two things fix that speed, and neither of them is the oscillator. A transverse wave travels along a stretched string at

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

where T is the tension in newtons and $\mu$ is the mass per unit length in kg m⁻¹. Tighten the string and the wave runs faster; hang the same tension on a heavier string and it runs slower. Be clear about which speed this is. It is the speed of the travelling waves the pattern is built from, moving along the string, and not the speed of any point of the string, which is bobbing across it and passes through zero at every node. Edexcel prints this form and sets questions straight on it; the other boards print only the frequency it leads to.

That frequency is one substitution away. The first harmonic has $\lambda_1 = 2L$ , so $f_1 = v/2L$ , and putting the speed in gives

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

ON THE AQA DATA SHEET

Raising the tension or using a lighter string raises the frequency. A guitarist works both handles, tightening the tuning pegs to change T and reaching for the thicker, heavier strings to play low notes.

WORKED EXAMPLE

A wave speed, then a note

A wire of length 0.65 m is stretched to a tension of 84 N. Its mass per unit length is 3.4 × 10⁻³ kg m⁻¹. Find the speed of transverse waves on it, and its first-harmonic frequency.

Speed first, because the frequency is built from it. v = √(T/μ) = √(84/(3.4 × 10⁻³)) = √(2.47 × 10⁴) = 157 m s⁻¹.

The first harmonic fits one loop, so λ = 2L = 1.30 m, and f₁ = v/λ = 157/1.30 = 121 Hz, a note sitting between the open A and D strings of a guitar.

Doing it in two steps is worth the extra line. The speed belongs to the wire alone, so it stays the same however you fret or damp the wire, and every harmonic in the ladder is that one speed divided by a different wavelength.

GUIDED PRACTICE

The harmonic ladder on one string

A string 0.60 m long carries waves at 48 m s⁻¹. Find the first-harmonic frequency and the second harmonic, starting from the first harmonic's wavelength.

Show the working

The fundamental fits one loop, so λ = 2L = 1.2 m, and f₁ = v/λ = 48/1.2 = 40 Hz.

The second harmonic fits two loops, halving the wavelength and doubling the frequency: 80 Hz. The ladder climbs in integer multiples of the fundamental.

INDEPENDENT PRACTICE

Standing waves in a microwave oven

The melted spots on a chocolate bar in a microwave oven (turntable removed) sit 6.1 cm apart. Find the microwaves' wavelength and frequency, taking c = 3.0 × 10⁸ m s⁻¹.

Show the working

The hot spots are antinodes, and adjacent antinodes sit half a wavelength apart, so λ = 2 × 0.061 = 0.12 m.

f = c/λ = 3.0 × 10⁸/0.122 = 2.5 GHz, the real operating frequency of a domestic oven. A chocolate bar and a ruler have just measured the wavelength that, with the known frequency, gives the speed of light.

Air columns in pipes

A string is not the only thing that can be pinned at its ends. Blow across the mouth of a bottle, or into an organ pipe, and the note you hear is a stationary wave in the column of air inside. The mechanism is the one you already have. A sound wave runs down the tube, reflects at the far end and superposes with the wave still arriving. What is new is the boundary conditions, because a pipe has two kinds of end and a string had only one.

A closed end is a solid wall, and the air touching it cannot shuttle back and forth along the tube, so a closed end is always a node. An open end is free, and the air there moves further than anywhere else in the pipe, so an open end is always an antinode. Those are displacement nodes and antinodes, which is how pipe diagrams are always drawn, though the pressure picture is their exact inverse, the pressure swinging hardest at the very end where the displacement is pinned.

Take a tube closed at one end and open at the other. It needs a node at one end and an antinode at the other, and the shortest pattern that manages that is a single quarter of a wavelength, since node to nearest antinode is $\lambda/4$ . So the first harmonic of a closed tube of length L has

\lambda_1 = 4L

The next pattern that fits must still start at a node and finish at an antinode, so it adds half a wavelength: $L = 3\lambda/4$ , then $5\lambda/4$ , and onwards. Every length the tube allows is an odd number of quarter wavelengths, so with n odd,

\lambda_n = \frac{4L}{n}

and the frequencies climb $f_1$ , $3f_1$ , $5f_1$ , with the even harmonics missing altogether. The reason is the mismatch between the ends. An even number of quarter wavelengths is a whole number of half wavelengths, and half a wavelength always runs node to node or antinode to antinode, which would demand the same kind of end at both ends of the tube. A closed tube has one of each and can never fit one. The gap is audible, and it is what gives a stopped organ pipe and a clarinet their hollow tone.

A tube open at both ends needs an antinode at each end instead. Antinode to antinode is half a wavelength, so its first harmonic has $\lambda_1 = 2L$ and every whole number of half wavelengths fits after that, $\lambda_n = 2L/n$ with n taking every whole value. The open tube therefore sounds the complete ladder, $f_1$ , $2f_1$ , $3f_1$ , the same series as the stretched string but built between two antinodes rather than two nodes. Compare the two pipes at equal length and the closed one starts from the longer wavelength, 4L against 2L, so it sounds an octave lower and on half as many harmonics.

FIG. 4Two tubes of the same length L, each with the first two patterns its ends allow. A closed end pins the air still, a displacement node, and every open end is an antinode. The closed tube therefore holds an odd number of quarter wavelengths, one and then three, while the open tube holds any whole number of half wavelengths, one and then two. The amber brackets are those quarter and half wavelengths. The closed tube's second pattern is the third harmonic, not the second: the second would need a whole half wavelength across the tube, and half a wavelength always runs node to node or antinode to antinode, never one of each.

WORKED EXAMPLE

Two pipes of the same length

A tube 0.30 m long is closed at one end. Sound travels at 340 m s⁻¹. Find its first-harmonic frequency and the next frequency it can sound, then find what its first harmonic would be with both ends open.

Closed at one end means a node there and an antinode at the open top, so a quarter wavelength fits the tube: λ₁ = 4L = 1.20 m.

f₁ = v/λ = 340/1.20 = 283 Hz. The next resonance is the third harmonic, since no even one can fit, so the tube's next strong response is at 850 Hz; driven between the two it still responds, only far more quietly.

Open both ends and an antinode sits at each, so half a wavelength fits: λ₁ = 2L = 0.60 m and f₁ = 340/0.60 = 567 Hz, an octave above the closed pipe of identical length. That one is free to sound 1130 Hz next rather than skipping it.

Measuring the speed of sound in a resonance tube

The closed pipe gives a way to measure the speed of sound with nothing more than a tuning fork and a metre rule. Stand a long tube upright with its lower end in a tall cylinder of water, so the water surface closes the bottom while the top stays open. Raising or lowering the tube changes the length of the trapped air column, which makes it a closed pipe you can tune.

Strike a tuning fork of known frequency f and hold it flat, a centimetre or so above the open top, taking care that it never touches the glass. Start with the air column short and lengthen it slowly. At one length the sound swells suddenly, because the column's first harmonic has come into step with the fork and the column is resonating. Record that length $L_1$ , measured from the top of the tube down to the water surface. Keep going and a second, quieter resonance arrives at $L_2$ , where the same column is running in its third harmonic.

Those two lengths are a quarter and three quarters of the same wavelength, so their difference is half a wavelength, $\lambda = 2(L_2 - L_1)$ , and the speed follows.

v = 2f(L_2 - L_1)

Taking the difference rather than trusting $L_1$ alone is the reason for the second reading. The antinode actually forms slightly above the open end rather than exactly at it, so every length you measure is short by the same small amount, and subtracting one reading from the other cancels it out. CIE tells you to treat that offset as negligible, and this is the measurement that lets you.

Two habits sharpen the result. Approach each resonance from both directions, once lengthening the column and once shortening it, and average the pair, because the loudness peak is broad and easy to overshoot. Then swap the fork for others of different frequency and repeat: since $L_1$ is close to $v/4f$ , a graph of $L_1$ against $1/f$ comes out straight with gradient v/4, and its small negative intercept is the offset you have just been cancelling.

INDEPENDENT PRACTICE

The speed of sound from two lengths

A 512 Hz fork resonates with a closed air column first at 15.8 cm and again at 49.0 cm. Find the wavelength and the speed of sound, then work out what the first length alone would have given.

Show the working

The two resonances sit at λ/4 and 3λ/4, so the gap between them is half a wavelength: λ = 2 × (0.490 − 0.158) = 0.664 m.

v = fλ = 512 × 0.664 = 340 m s⁻¹, the textbook value, out of a fork and a ruler.

From $L_1$ alone, λ = 4 × 0.158 = 0.632 m and v = 324 m s⁻¹, five per cent low. The eight millimetres by which the antinode overshoots the open end is exactly what the subtraction removed.

Phase along a stationary wave

On a progressive wave, every point oscillates with the same amplitude, and there is a steady phase difference between one point and the next. A stationary wave differs on both counts. The amplitude depends on position, running from zero at a node to a maximum at an antinode, and only two phase relationships exist anywhere on the pattern.

FIG. 5A snapshot between the extremes. P and Q sit between the same pair of nodes and move together; R, in the next loop along, moves the opposite way.

All the points between one pair of adjacent nodes move in phase. They reach their maximum displacements at the same instant, though each has its own amplitude. Points on opposite sides of a node move in antiphase, a phase difference of $\pi$ rad or 180°. Cross two nodes and the motion is back in phase.

ASSESSMENT FOCUS

Describing formation, say that two waves of the same frequency and equal amplitude, travelling in opposite directions, superpose. Answers about the wave “bouncing back and interfering with itself” usually drop a mark.
Adjacent nodes are $\lambda/2$ apart, never $\lambda$ . If in doubt, sketch two loops and read the wavelength off your own diagram.
Write first harmonic on AQA papers. “Fundamental” is the same thing, but their mark schemes are written in their own vocabulary.
In the first-harmonic equation, $\mu$ is mass per unit length. Grams convert to kilograms before you divide by the length, and the unit kg m⁻¹ is there to remind you.
Edexcel prints $v = \sqrt{T/\mu}$ on its own and asks for the wave speed on a string as a question in its own right. The other boards print only $f_1 = (1/2L)\sqrt{T/\mu}$ , which is the same physics with $\lambda_1 = 2L$ already substituted in, so recover the speed from $v = f_1 \times 2L$ if a paper wants it.
A phase-difference question about a stationary wave has only two possible answers, 0 or $\pi$ rad.
In a pipe, mark the ends before you draw anything. Closed end a node, open end an antinode, and the pattern then draws itself. Sketching a pretty curve first and labelling the ends afterwards is how the odd-harmonic rule gets lost.
A closed tube sounds only the odd harmonics, $f_1$ , $3f_1$ , $5f_1$ . Doubling a closed pipe's first harmonic names a frequency it cannot produce, so the next note up is three times the first, not twice.
In the resonance tube, quote $\lambda = 2(L_2 - L_1)$ rather than $\lambda = 4L_1$ . The difference of two lengths is the version of the answer that needs no correction at the open end, and saying why is often the explain mark.
The required practical asks for the proportionalities, so know them cold. At fixed length, $f_1 \propto \sqrt{T}$ . At fixed tension, $f_1 \propto 1/L$ and $f_1 \propto 1/\sqrt{\mu}$ . A graph of $f_1$ against $\sqrt{T}$ should go through the origin, and saying so is often the last mark.

CHECK YOURSELF

A string 1.20 m long is fixed at both ends and vibrates in its third harmonic. (a) What is the wavelength? (b) At the same tension, how does its frequency compare with the first harmonic?

Show a hint

How many half-wavelengths fit into the 1.20 m?

Show the answer

(a) The third harmonic has three loops. Each loop is half a wavelength, so $L = 3\lambda/2$ , which gives $\lambda = 2L/3$ = 2 × 1.20 / 3 = 0.80 m.

(b) The wave speed depends only on the tension and the mass per unit length, so it has not changed. In the first harmonic the wavelength is 2L = 2.40 m. The wavelength has fallen by a factor of three, so by $v = f\lambda$ the frequency has risen by the same factor: the third harmonic is three times the frequency of the first.

Notice what decided the answer. The fixed ends select which wavelengths can resonate, and the fixed wave speed turns each of those wavelengths into a frequency. The oscillator sets the driving frequency; the string's ends and speed decide which driving frequencies get a resonant answer.

Nothing travels along the string.

Adjacent nodes are half a wavelength apart.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the stationary waves questions page.

15 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Required practical 1: stationary waves on a string puts this topic in the lab, and the written papers ask about it.

CHECK YOUR PROGRESS

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Describe how two travelling waves superpose to form a stationary wave, and define node and antinode.
Use the node and antinode spacings, $\lambda/2$ and $\lambda/4$ , to get from a measured distance to a wavelength.
Find the allowed wavelengths of a string fixed at both ends, and use $v = \sqrt{T/\mu}$ and the first-harmonic frequency equation.
Work out the harmonics of an air column in a closed and in an open tube, and say why a closed tube sounds only the odd ones.
Describe the resonance-tube method for the speed of sound, and say why it uses two lengths rather than one.
State the amplitude and phase relationships that separate stationary waves from progressive ones.

Open the full revision checklist to track your progress across the whole unit.

Stationary waves

Two waves, one string

Nodes and antinodes

Harmonics on a stretched string

Air columns in pipes

Measuring the speed of sound in a resonance tube

Phase along a stationary wave

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Question workbook

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