Chapter 14 · Waves

14.3 Superposition, Reflection and Standing Waves

Overlapping waves simply add. Waves reflected from a rigid end return inverted, and waves trapped between two boundaries form standing waves with fixed nodes and antinodes — the normal modes $\nu_n=\dfrac{nv}{2L}$ of strings and open pipes, and the odd harmonics of closed pipes.

Pluck a guitar string and it does not sound just any frequency: it sounds one particular note, plus a set of fainter, higher notes at exactly two, three, four… times that frequency, which give the guitar its voice. Blow across the top of an empty bottle and you get a single hollow note that drops as you pour water in; blow harder into a flute and the note leaps up an octave. In a concert hall, a pair of noise-cancelling headphones silences the hum of an air-conditioner by adding more sound. All of these come from one simple rule for what happens when two waves are in the same place at the same time.

This section states that rule, the principle of superposition, and uses it to add waves of the same frequency (interference). It then follows a wave to the end of a string and finds how it is reflected — upside down from a fixed end, upright from a free one. When a wave and its reflection overlap they form a standing wave, with points that never move (nodes) and points that swing widely (antinodes). A string fixed at both ends, or a column of air in a pipe, can hold standing waves only at certain frequencies — its normal modes — and these are the notes and harmonics of every string and wind instrument.

The principle of superposition

Send a pulse from each end of a long stretched spring so that they run towards each other. Where they overlap, the spring takes a shape that is simply the two pulses added together; afterwards each pulse continues on its way exactly as if it had never met the other. Two waves do not bounce off each other or damage each other — they pass straight through.

📐 Principle 14.8 — The principle of superposition

When two or more waves overlap in a medium, the resultant displacement of each element at each instant is the algebraic sum of the displacements that the individual waves would produce on their own:

$$y(x,t)=y_1(x,t)+y_2(x,t)+\dots$$

Each wave travels on unaltered by the others. The principle holds for waves of small amplitude, for which the restoring forces are proportional to the displacements (linear media) — true of everyday sound and of gentle waves on strings.

If two pulses overlap crest on crest, the displacement at the overlap is larger than either; if a crest meets a trough of the same size, they cancel and the string is momentarily straight. The energy has not vanished: at that instant the straight part of the string is moving fast, so its energy is all kinetic. Overlapping waves that reinforce each other are said to interfere constructively; those that cancel, destructively.

📘 Definition 14.6 — Interference

Interference is the change in the resultant amplitude when two or more waves overlap. It is constructive where the waves arrive in phase (displacements add to a larger amplitude) and destructive where they arrive in antiphase (displacements cancel, wholly or partly).

Two waves of the same frequency. Let two waves of the same amplitude $a$, angular frequency $\omega$ and wave number $k$ travel in the $+x$ direction with a phase difference $\phi$: $y_1=a\sin(kx-\omega t)$ and $y_2=a\sin(kx-\omega t+\phi)$. Using $\sin A+\sin B=2\sin\dfrac{A+B}2\cos\dfrac{A-B}2$:

$$y=y_1+y_2=2a\cos\frac\phi2\,\sin\!\left(kx-\omega t+\frac\phi2\right).$$

📐 Result 14.9 — Superposition of two waves of the same frequency

Two waves of equal amplitude $a$ and equal frequency, travelling in the same direction with a phase difference $\phi$, add to a single wave of the same frequency and speed, with amplitude

$$A(\phi)=2a\left|\cos\frac\phi2\right| .$$

For $\phi=0,\,2\pi,\,4\pi,\dots$ the interference is fully constructive ($A=2a$); for $\phi=\pi,\,3\pi,\dots$ it is fully destructive ($A=0$).

✏️ Example 1 — Two pulses meet

Two pulses of the same shape and width travel towards each other along a stretched string. One has a peak displacement of $3.0\ \text{cm}$ upwards, the other $2.0\ \text{cm}$ upwards. (a) What is the largest displacement of the string when the pulses exactly overlap? (b) Repeat if the second pulse is inverted ($2.0\ \text{cm}$ downwards). (c) What are the pulses like after they have passed through each other?

💡 Show solution▶

(a) Same shape, so the peaks coincide: $3.0+2.0=5.0\ \text{cm}$ upwards (constructive).

(b) $3.0+(-2.0)=1.0\ \text{cm}$ upwards (destructive, but not complete — the pulses are of different sizes).

(c) Exactly as before: $3.0\ \text{cm}$ and $2.0\ \text{cm}$ pulses (one inverted in (b)), each still moving in its original direction. Superposition only adds displacements where the pulses overlap.

✏️ Example 2 — Interference of two waves

Two waves $y_1=4.0\sin(kx-\omega t)$ and $y_2=4.0\sin\left(kx-\omega t+\dfrac\pi3\right)$, in millimetres, travel along the same string. (a) Find the amplitude of the resultant wave and its phase relative to $y_1$. (b) What phase difference would make the resultant amplitude equal to $4.0\ \text{mm}$, the amplitude of each wave? (c) What phase difference makes the string stay still?

💡 Show solution▶

(a) $A=2\times4.0\cos\dfrac\pi6=8.0\times0.866=6.9\ \text{mm}$; the resultant leads $y_1$ by $\dfrac\phi2=\dfrac\pi6$.

(b) $8.0\cos\dfrac\phi2=4.0$: $\cos\dfrac\phi2=\dfrac12$, $\phi=\dfrac{2\pi}3$ ($120^\circ$).

Check At $kx-\omega t=0$: $y_1+y_2=0+4.0\sin\dfrac\pi3=3.46\ \text{mm}$ and $6.93\sin\dfrac\pi6=3.46\ \text{mm}$ ✓.

Wave 1 (blue) and wave 2 (red) are drawn thin; their sum — what the string actually does — is the thick curve. Start with two pulses and watch them pass through each other. Then add two waves travelling the same way and slide the phase difference $\phi$ from $0$ to $180^\circ$. Finally send the waves in opposite directions: the sum stops travelling, and nodes (●) appear that never move.

set-up:phase difference $\phi$:60°

set-up—

largest |y| of the sum now—

predicted amplitude—

interference—

Reflection of waves

A wave that reaches the end of its medium is reflected, at least in part. What comes back depends on what the end can do. Tie a rope firmly to a wall and send a hump along it: the hump returns upside down. As the hump arrives, the rope pulls up on the wall; by Newton's third law (Section 4.2) the wall pulls down on the rope, and that downward pull launches an inverted pulse back. Now tie the end to a light ring that slides freely on a smooth vertical pole: the ring is flung up by the arriving hump, overshoots, and pulls the rope after it — the hump returns upright.

For a sinusoidal wave, put the boundary at $x=0$, with the string along $x\gt0$, and let the incident wave $y_\text{i}=a\sin(kx+\omega t)$ travel towards it (towards $-x$). At a rigid end the total displacement at $x=0$ must be zero at every instant, so the reflected wave must give $-a\sin\omega t$ there. The wave travelling towards $+x$ that does this is $y_\text{r}=a\sin(kx-\omega t)$ (at $x=0$ it is $a\sin(-\omega t)=-a\sin\omega t$ ✓). At a free end the slope of the string must be zero instead, and the reflected wave arrives in step with the incident one: $y_\text{r}=-a\sin(kx-\omega t)$, which equals $+a\sin\omega t$ at $x=0$.

📐 Result 14.10 — Reflection at rigid and open boundaries

A wave reflected at a boundary returns with the same frequency, wavelength and speed, and (for a perfect reflector) the same amplitude.

At a rigid boundary (a fixed end) the reflected wave is in antiphase with the incident wave at the boundary: a phase change of $\pi$. A pulse comes back inverted, and the boundary is a point of zero displacement.
At an open boundary (a free end) there is no phase change: a pulse comes back upright, and the boundary moves with twice the amplitude of the incident wave.

Between these extremes, a wave reaching a junction with a different medium is partly reflected and partly transmitted. Passing from a light string to a heavy one (a slower medium), the reflected pulse is inverted, as at a wall; passing from a heavy string to a light one it is upright. The transmitted pulse is always upright. For sound in a pipe, a closed end is a rigid boundary for the air's displacement (the air cannot move into the wall), and an open end behaves very nearly as a free boundary.

✏️ Example 3 — A pulse and its echo

A rope $6.0\ \text{m}$ long, tied at its far end to a post, carries waves at $12\ \text{m/s}$. A student sends a pulse $5.0\ \text{cm}$ high from her end. (a) When does the pulse return to her, and which way up is it? (b) What changes if the far end is tied to a ring that slides freely on the post? (c) She sends a second, identical pulse $0.50\ \text{s}$ after the first (fixed end). Where do the returning first pulse and the outgoing second pulse meet, and what is the largest displacement there?

💡 Show solution▶

(a) Out and back is $12\ \text{m}$: $t=\dfrac{12}{12}=1.0\ \text{s}$. A rigid end inverts it: it returns $5.0\ \text{cm}$ downwards (ignoring losses).

(b) Free end: the pulse returns upright, still after $1.0\ \text{s}$ — the speed does not depend on how the end is held.

(c) The first pulse reaches the post at $t=0.50\ \text{s}$, just as the second leaves her hand. Measured from her end, for $0.50\le t\le1.0\ \text{s}$ the returning first pulse is at $12-12t$ and the second pulse at $12(t-0.50)$ (metres). They meet when $12-12t=12t-6$: $t=0.75\ \text{s}$, at $3.0\ \text{m}$ — the middle of the rope. One is inverted, the other upright and the same size: the displacement there is $5.0-5.0=0$ at the instant they overlap exactly, while the rope at the overlap moves fast.

Check At $t=0.75\ \text{s}$ the first pulse has travelled $9.0\ \text{m}$ ($6.0$ out, $3.0$ back) and the second $3.0\ \text{m}$ ✓.

Standing waves and the normal modes of a string

Keep shaking one end of a string whose other end is fixed: the waves you send and the waves reflected back overlap all along the string. Take the incident and reflected waves found above, $y_\text{i}=a\sin(kx+\omega t)$ and $y_\text{r}=a\sin(kx-\omega t)$, and add them with $\sin A+\sin B=2\sin\dfrac{A+B}2\cos\dfrac{A-B}2$:

$$y=a\sin(kx-\omega t)+a\sin(kx+\omega t)=2a\sin kx\,\cos\omega t .$$

This is no longer of the form $f(x\mp vt)$: nothing travels. Every element oscillates in SHM with the same angular frequency $\omega$, but with an amplitude $2a|\sin kx|$ that depends on where it is. All the elements between two neighbouring zeros of $\sin kx$ move in phase; across a zero they move in antiphase.

📐 Result 14.11 — Standing wave

Two waves of equal amplitude $a$ and frequency travelling in opposite directions, $a\sin(kx-\omega t)$ and $a\sin(kx+\omega t)$, superpose to

$$y(x,t)=2a\sin kx\,\cos\omega t .$$

The element at $x$ executes SHM with amplitude $2a|\sin kx|$. The amplitude is zero at

$$x=0,\ \frac\lambda2,\ \lambda,\ \frac{3\lambda}2,\dots$$

and is greatest, $2a$, at

$$x=\frac\lambda4,\ \frac{3\lambda}4,\ \frac{5\lambda}4,\dots$$

📘 Definition 14.7 — Standing wave, node and antinode

A standing (stationary) wave is a pattern of oscillation in which the waveform does not travel: each element oscillates in SHM with an amplitude fixed by its position. A node is a point that is always at rest (zero amplitude); an antinode is a point of maximum amplitude. Neighbouring nodes (or antinodes) are $\dfrac\lambda2$ apart, and a node and the nearest antinode are $\dfrac\lambda4$ apart, where $\lambda$ is the wavelength of the two travelling waves.

No energy flows past a node — the element there never moves, so no work is done across it — and on average a standing wave carries no energy along the string. The energy in each loop sloshes between kinetic (all elements passing through $y=0$ together) and elastic potential (string at its most bent, all elements momentarily at rest).

A string fixed at both ends. A string of length $L$ is clamped at $x=0$ and $x=L$. Both ends must be nodes. The node at $x=0$ is already built into $\sin kx$; the node at $x=L$ needs $\sin kL=0$, i.e. $kL=n\pi$ with $n=1,2,3,\dots$ So only wavelengths that fit a whole number of half-wavelengths into the string are allowed: $L=n\dfrac{\lambda}2$.

📐 Result 14.12 — Normal modes of a string fixed at both ends

A string of length $L$, fixed at both ends, carrying waves at speed $v=\sqrt{T/\mu}$, can vibrate in standing waves only with

$$\lambda_n=\frac{2L}{n},\qquad \nu_n=\frac{nv}{2L}=\frac{n}{2L}\sqrt{\frac T\mu},\qquad n=1,2,3,\dots$$

The $n$th mode has $n$ loops, $n+1$ nodes (including the ends) and $n$ antinodes. All the frequencies are whole-number multiples of the lowest, $\nu_1=\dfrac{v}{2L}$.

📘 Definition 14.8 — Normal modes, fundamental, harmonics and overtones

The normal modes of a vibrating system are its natural patterns of standing-wave oscillation, in each of which every element moves in SHM at the same frequency. The mode of lowest frequency is the fundamental (or first harmonic). A mode whose frequency is $n$ times the fundamental is the $n$th harmonic. The modes above the fundamental, in order, are the first, second, … overtones.

For a string fixed at both ends every harmonic is present, so the first overtone is the second harmonic, the second overtone the third harmonic, and so on. A plucked string vibrates in a mixture of many modes at once; the fundamental fixes the note we hear, and the mixture of harmonics (its timbre) is why a guitar, a sitar and a piano playing the same note sound different. The fundamental frequency $\dfrac1{2L}\sqrt{\dfrac T\mu}$ contains the three "laws of a vibrating string", tested on a sonometer: $\nu\propto\dfrac1L$ at fixed $T$ and $\mu$ (a shorter string gives a higher note — a guitarist's frets); $\nu\propto\sqrt T$ (tuning pegs); $\nu\propto\dfrac1{\sqrt\mu}$ (thick strings for low notes).

✏️ Example 4 — Tuning a guitar string

The vibrating length of a guitar string is $0.65\ \text{m}$ and its linear mass density is $5.0\ \text{g/m}$. It is to sound the note A$_2$, of fundamental frequency $110\ \text{Hz}$. (a) Find the wave speed needed and the tension. (b) Find the frequencies of its second and third harmonics. (c) Where must the string be pressed against a fret (shortening the vibrating length) to sound $165\ \text{Hz}$?

💡 Show solution▶

(a) $v=2L\nu_1=2\times0.65\times110=143\ \text{m/s}$; $T=\mu v^2=5.0\times10^{-3}\times143^2=1.0\times10^2\ \text{N}$ ($102$).

(b) $\nu_2=220\ \text{Hz}$, $\nu_3=330\ \text{Hz}$.

Check $\dfrac{L}{L'}=\dfrac{0.65}{0.433}=1.50=\dfrac{165}{110}$ ✓ — frequency inversely proportional to length.

✏️ Example 5 — Counting loops

A string $1.2\ \text{m}$ long, fixed at both ends, vibrates at $150\ \text{Hz}$ in a standing wave with three loops. Find (a) the wavelength and the wave speed; (b) the fundamental frequency; (c) the positions of the nodes and antinodes; (d) the next higher frequency at which it can vibrate.

💡 Show solution▶

(a) Three loops: $L=3\dfrac\lambda2$, $\lambda=0.80\ \text{m}$; $v=\nu\lambda=150\times0.80=120\ \text{m/s}$.

(b) This is the third harmonic: $\nu_1=\dfrac{150}{3}=50\ \text{Hz}$ $\left(=\dfrac{120}{2\times1.2}\ ✓\right)$.

(c) Nodes every $\dfrac\lambda2=0.40\ \text{m}$: at $0,\ 0.40,\ 0.80,\ 1.20\ \text{m}$. Antinodes midway: $0.20,\ 0.60,\ 1.00\ \text{m}$.

(d) The fourth harmonic, $4\times50=200\ \text{Hz}$.

Standing waves in air columns

The air in a pipe has normal modes too: sound waves travel along the pipe, reflect at its ends, and form standing longitudinal waves. Think in terms of the displacement of the air along the pipe. At a closed end the air cannot move into the wall, so the closed end is a displacement node. At an open end the air is free to surge in and out, and the open end is (very nearly) a displacement antinode. Where the displacement has a node the air is alternately squeezed from both sides and pulled apart, so a displacement node is a pressure antinode, and a displacement antinode is a pressure node (the pressure at an open end stays atmospheric).

Pipe closed at one end (length $L$): a node at the closed end and an antinode at the open end, a quarter-wavelength apart in the simplest case. In general $L$ must be an odd number of quarter-wavelengths, $L=(2n-1)\dfrac\lambda4$. Pipe open at both ends: an antinode at each end, so, as for the string (with nodes and antinodes swapped), $L=n\dfrac\lambda2$.

📐 Result 14.13 — Normal modes of air columns

For sound of speed $v$ in a pipe of length $L$ (ignoring the small end correction):

Closed at one end — only the odd harmonics are present:

$$\nu=\frac{(2n-1)\,v}{4L}=\frac{v}{4L},\ \frac{3v}{4L},\ \frac{5v}{4L},\dots$$

Open at both ends — all harmonics are present:

$$\nu=\frac{n\,v}{2L}=\frac{v}{2L},\ \frac{2v}{2L},\ \frac{3v}{2L},\dots$$

The fundamental of an open pipe is twice that of a closed pipe of the same length.

A closed pipe's first overtone is therefore its third harmonic, its second overtone the fifth, and so on; it lacks the even harmonics, which gives the clarinet (which acts roughly as a closed pipe) its dark, hollow sound compared with the flute (an open pipe). Blowing harder makes a flute jump from its fundamental to its second harmonic — an octave up — while a clarinet jumps to its third harmonic, a twelfth up. The bottle you blow across is a closed pipe: pouring in water shortens the air column and raises the note.

✏️ Example 6 — Closed and open organ pipes

An organ pipe is $0.85\ \text{m}$ long and the speed of sound is $340\ \text{m/s}$. Find the three lowest frequencies it produces (a) when it is closed at one end; (b) when it is open at both ends. (c) What length of open pipe has the same fundamental as the closed $0.85\ \text{m}$ pipe?

💡 Show solution▶

(a) $\dfrac{v}{4L}=\dfrac{340}{3.4}=100\ \text{Hz}$; then $300\ \text{Hz}$ and $500\ \text{Hz}$ (odd harmonics only).

(b) $\dfrac{v}{2L}=\dfrac{340}{1.7}=200\ \text{Hz}$; then $400\ \text{Hz}$ and $600\ \text{Hz}$.

Check Closed, $100\ \text{Hz}$: $\lambda=3.4\ \text{m}$, and $\dfrac\lambda4=0.85\ \text{m}=L$ ✓. Interactive 2 starts with this pipe.

Pick a string fixed at both ends, a pipe open at both ends, or a pipe closed at the left end, and step through its modes. For the string the curve is the string itself; for the pipes the dots are slices of air moving along the pipe, and the dashed curves show their displacement amplitude. N marks displacement nodes and A antinodes. (The motion is slowed down to the same pace for every mode.) Which system never shows an even harmonic?

system:mode:1length $L$:0.85 mspeed $v$:340 m/s

harmonic—

wavelength $\lambda$—

frequency $\nu$—

fundamental $\nu_1$—

⚠️ Common mistakes

1. Node spacing $=\lambda$. Neighbouring nodes are $\dfrac\lambda2$ apart; a node and the next antinode are $\dfrac\lambda4$ apart.

2. Overtone number $=$ harmonic number. For a string or open pipe, the $n$th overtone is the $(n+1)$th harmonic. For a closed pipe, the first overtone is the third harmonic, the second overtone the fifth.

3. Even harmonics in a closed pipe. There are none: $L$ must be an odd number of quarter-wavelengths.

4. Using the wavelength of the standing wave as the distance between "humps". One loop of a string is half a wavelength: a string with three loops holds $1\tfrac12$ wavelengths.

5. Reflection at a fixed end. A pulse comes back inverted from a rigid end (phase change $\pi$), upright from a free end. Its speed, wavelength and frequency do not change.

6. Displacement and pressure. A closed end is a displacement node but a pressure antinode; an open end is a displacement antinode and a pressure node.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — end correction, the resonance tube, and unequal amplitudes

In JEE Main, NEET and JEE Advanced (standing waves in air columns, superposition, and the resonance-tube experiment); not required by the CBSE theory syllabus, though the resonance tube is a CBSE practical.

End correction. The antinode at an open end lies slightly outside the pipe, by about $e\approx0.6r$ for a pipe of radius $r$. The effective length is $L+e$ for a closed pipe and $L+2e$ for an open one, so a closed pipe's fundamental is really $\dfrac{v}{4(L+e)}$.

The resonance tube. A tuning fork of frequency $\nu$ is held over a tube whose air column (closed below by water) is lengthened until the sound swells. The first and second resonances occur at lengths $L_1$ and $L_2$ with $L_1+e=\dfrac\lambda4$ and $L_2+e=\dfrac{3\lambda}4$. Subtracting removes the unknown end correction:

$$\lambda=2(L_2-L_1),\qquad v=2\nu(L_2-L_1),\qquad e=\frac{L_2-3L_1}{2}.$$

For example, with a $512\ \text{Hz}$ fork, $L_1=15.8\ \text{cm}$ and $L_2=49.2\ \text{cm}$: $v=2\times512\times0.334=342\ \text{m/s}$ and $e=\dfrac{49.2-47.4}{2}=0.9\ \text{cm}$.

Unequal amplitudes. Two waves of amplitudes $a_1$, $a_2$ and phase difference $\phi$ add like the two SHMs of Section 13.1's extension box:

$$A^2=a_1^2+a_2^2+2a_1a_2\cos\phi .$$

The intensity of a wave goes as the square of its amplitude, so $I_\text{max}:I_\text{min}=(a_1+a_2)^2:(a_1-a_2)^2$ — for $a_1=3a_2$ this is $16:4=4:1$.

Exercises

Take the speed of sound in air as $340\ \text{m/s}$ unless told otherwise, and ignore end corrections. Give answers to 2–3 significant figures.

Superposition and interference.

1 Two pulses of the same shape, of heights $+2.0\ \text{cm}$ and $+1.5\ \text{cm}$, overlap exactly. Find the resultant height. What is it if the second pulse is inverted?
2 Two identical pulses, one upright and one inverted, overlap exactly and the string is momentarily straight. Where is the energy at that instant?
3 Two waves of amplitude $3.0\ \text{mm}$, the same frequency and direction, differ in phase by $\dfrac\pi2$. Find the resultant amplitude.
4 What phase difference between two equal-amplitude waves gives a resultant amplitude equal to the amplitude of each?
5 Find the resultant of $y_1=0.020\sin(kx-\omega t)$ and $y_2=0.020\sin(kx-\omega t+\pi)$.
6 Repeat part 3 with a phase difference of $2\pi$.
7 Noise-cancelling headphones add a sound wave to the incoming noise. What phase difference must it have? Why do they work best on steady, low-pitched noise?

💡 Show solutions▶

1 $2.0+1.5=3.5\ \text{cm}$; with one inverted, $2.0-1.5=0.5\ \text{cm}$ (upwards).

2 All kinetic: the elements of the straight part are moving fast (one pulse's motion adds to the other's), even though no element is displaced.

3 $A=2\times3.0\cos\dfrac\pi4=4.2\ \text{mm}$ ($4.24$).

4 $2a\cos\dfrac\phi2=a$: $\phi=\dfrac{2\pi}{3}$ ($120^\circ$).

5 $y=0$ everywhere at all times: complete destructive interference.

6 $\cos\pi=-1$, $|{-1}|=1$: $A=6.0\ \text{mm}$ — a phase difference of $2\pi$ is the same as none (constructive).

7 A phase difference of $\pi$ (antiphase). The electronics must predict the noise waveform a little in advance; steady noise is predictable, and low-pitched noise has long wavelengths, so small timing and position errors are a small fraction of a cycle.

Reflection and standing waves on strings.

1 Describe the reflected pulse when an upright pulse on a string reaches (a) a fixed end, (b) a free end, (c) a junction with a much heavier string.
2 A standing wave on a string is $y=0.040\sin(5.0\pi x)\cos(200\pi t)$ (SI). Find the amplitude and wavelength of the two travelling waves that form it, their frequency and speed, and the distance between neighbouring nodes.
3 For the standing wave of part 2, find the amplitude of oscillation of the elements at $x=0.050\ \text{m}$ and at $x=0.10\ \text{m}$.
4 A string $0.90\ \text{m}$ long is fixed at both ends; waves travel along it at $180\ \text{m/s}$. Find the frequencies of its first three harmonics.
5 A string fixed at both ends vibrates with four loops at $240\ \text{Hz}$. Find its fundamental frequency.
6 Two successive natural frequencies of a string fixed at both ends are $420\ \text{Hz}$ and $490\ \text{Hz}$. Find the fundamental, say which harmonics these are, and find the length of the string if waves travel along it at $140\ \text{m/s}$.
7 A sonometer wire $1.0\ \text{m}$ long with $\mu=1.0\ \text{g/m}$ is under a tension of $100\ \text{N}$. Find its fundamental frequency, and the tension needed to raise it to $200\ \text{Hz}$.
8 By what factor does the fundamental frequency of a string change if (a) its vibrating length is halved, (b) its tension is made four times as large, (c) it is replaced by a string of the same material with twice the diameter (same tension)?

💡 Show solutions▶

1 (a) Inverted (phase change $\pi$). (b) Upright. (c) Inverted, almost as at a fixed end, with a small upright pulse transmitted into the heavy string.

2 $2a=0.040$: $a=0.020\ \text{m}$; $k=5.0\pi$: $\lambda=0.40\ \text{m}$; $\nu=100\ \text{Hz}$; $v=\nu\lambda=40\ \text{m/s}$; nodes $\dfrac\lambda2=0.20\ \text{m}$ apart.

3 $x=0.050$: $0.040\left|\sin\dfrac\pi4\right|=0.028\ \text{m}$; $x=0.10$: $0.040\left|\sin\dfrac\pi2\right|=0.040\ \text{m}$ (an antinode).

4 $\nu_1=\dfrac{180}{1.8}=100\ \text{Hz}$; $200\ \text{Hz}$; $300\ \text{Hz}$.

5 Four loops is the fourth harmonic: $\nu_1=60\ \text{Hz}$.

6 Successive harmonics differ by $\nu_1$: $\nu_1=70\ \text{Hz}$; $420=6\times70$, $490=7\times70$: the 6th and 7th harmonics. $L=\dfrac{v}{2\nu_1}=\dfrac{140}{140}=1.0\ \text{m}$.

7 $\nu_1=\dfrac1{2.0}\sqrt{\dfrac{100}{0.0010}}=1.6\times10^2\ \text{Hz}$ ($158$). $T=\mu(2L\nu)^2=0.0010\times400^2=1.6\times10^2\ \text{N}$.

8 (a) $\times2$; (b) $\times2$; (c) $\mu\times4$, so $\nu\times\dfrac12$.

III

Air columns.

1 Find the three lowest natural frequencies of a pipe $0.50\ \text{m}$ long, closed at one end.
2 Repeat part 1 for the same pipe open at both ends.
3 A pipe closed at one end has a fundamental frequency of $250\ \text{Hz}$. How long is it?
4 Explain, with a sketch of the displacement nodes and antinodes, why a pipe closed at one end has only odd harmonics.
5 An open pipe and a closed pipe have the same fundamental frequency. Find the ratio of their lengths.
6 A pipe $0.75\ \text{m}$ long, closed at one end, sounds its third harmonic. Find the wavelength and the positions of the displacement nodes and antinodes, measured from the closed end.
7 Three successive natural frequencies of a pipe are $425\ \text{Hz}$, $595\ \text{Hz}$ and $765\ \text{Hz}$. Is the pipe open or closed? Find its fundamental and its length.
8 Where in a pipe closed at one end does the pressure vary most as it sounds its fundamental? Where does it vary least?

💡 Show solutions▶

1 $\dfrac{340}{4\times0.50}=170\ \text{Hz}$; $510\ \text{Hz}$; $850\ \text{Hz}$.

2 $\dfrac{340}{2\times0.50}=340\ \text{Hz}$; $680\ \text{Hz}$; $1020\ \text{Hz}$ ($1.02\ \text{kHz}$).

3 $L=\dfrac{v}{4\nu}=\dfrac{340}{1000}=0.34\ \text{m}$.

4 There must be a node (N) at the closed end and an antinode (A) at the open end. N to the next A is $\dfrac\lambda4$, and each further step to the next A adds $\dfrac\lambda2$, so $L=\dfrac\lambda4,\dfrac{3\lambda}4,\dfrac{5\lambda}4,\dots$ and $\nu=\dfrac{v}{4L}\times1,3,5,\dots$ — an even multiple would need an antinode or a node at both ends.

5 $\dfrac{v}{2L_\text{o}}=\dfrac{v}{4L_\text{c}}$: $L_\text{o}:L_\text{c}=2:1$.

6 $L=\dfrac{3\lambda}4$: $\lambda=1.0\ \text{m}$. Nodes at $0$ and $0.50\ \text{m}$; antinodes at $0.25\ \text{m}$ and $0.75\ \text{m}$ (the open end).

7 Differences $170\ \text{Hz}$, and $425:595:765=5:7:9$ — odd multiples of $85\ \text{Hz}$, so it is closed at one end with $\nu_1=85\ \text{Hz}$ (if it were open, the frequencies would be whole multiples of $170\ \text{Hz}$). $L=\dfrac{340}{4\times85}=1.0\ \text{m}$.

8 Most at the closed end (a displacement node, so a pressure antinode); least at the open end, where the pressure stays at atmospheric (a pressure node).

Applications and harder problems.

1 (Modelling) The human ear canal is roughly a tube $2.5\ \text{cm}$ long, open at the outside and closed by the eardrum. Find its fundamental frequency ($v=343\ \text{m/s}$). Why are we especially sensitive to sounds of about $3$–$4\ \text{kHz}$?
2 (Modelling) A flute behaves as a pipe open at both ends, $0.66\ \text{m}$ long. Find its lowest note ($v=343\ \text{m/s}$).
3 (Modelling) A violin's A string is $0.33\ \text{m}$ long, has $\mu=0.60\ \text{g/m}$ and is tuned to $440\ \text{Hz}$. Find its tension.
4 A guitarist touches a string lightly at one-third of its length from the bridge and plucks it, creating a node at that point. If the open string's fundamental is $110\ \text{Hz}$, what is the lowest note that can now sound?
5 A string $0.50\ \text{m}$ long, fixed at both ends, vibrating in its fundamental mode, is in resonance with the fundamental of a $0.25\ \text{m}$ pipe closed at one end. Find the speed of waves on the string.
6 Show that the points $x=\dfrac\lambda4$ and $x=\dfrac{3\lambda}4$ of the standing wave $y=2a\sin kx\cos\omega t$ always move in antiphase, and that the points $x=\dfrac\lambda8$ and $x=\dfrac{3\lambda}8$ always move in phase with equal amplitudes.

💡 Show solutions▶

1 $\nu=\dfrac{343}{4\times0.025}=3.4\times10^3\ \text{Hz}$. The air in the canal resonates near this frequency, boosting the pressure at the eardrum for sounds of $3$–$4\ \text{kHz}$.

2 $\nu=\dfrac{343}{2\times0.66}=2.6\times10^2\ \text{Hz}$ ($260$, close to middle C).

3 $v=2L\nu=2\times0.33\times440=290\ \text{m/s}$; $T=\mu v^2=6.0\times10^{-4}\times290.4^2=51\ \text{N}$.

4 Only modes with a node at $\dfrac L3$ survive: $n=3,6,9,\dots$ The lowest is the third harmonic, $330\ \text{Hz}$.

5 Pipe: $\nu=\dfrac{340}{4\times0.25}=340\ \text{Hz}$. String: $v=2L\nu=2\times0.50\times340=340\ \text{m/s}$.

6 $\sin\dfrac\pi2=1$ and $\sin\dfrac{3\pi}2=-1$: displacements $2a\cos\omega t$ and $-2a\cos\omega t$ — antiphase. $\sin\dfrac\pi4=\sin\dfrac{3\pi}4=\dfrac1{\sqrt2}$: both $\sqrt2\,a\cos\omega t$ — in phase (same loop), equal amplitudes.

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Two identical waves travel in opposite directions on a string fixed at $x=0$ and $x=L$. Show that their superposition is a standing wave, and deduce that the allowed frequencies are $\nu_n=\dfrac{nv}{2L}$. Sketch the first two modes.

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Sum $a\sin(kx-\omega t)+a\sin(kx+\omega t)=2a\sin kx\cos\omega t$: each point oscillates with amplitude $2a|\sin kx|$; the pattern does not travel. [1]

Modes $x=0$ is a node automatically; $x=L$ is a node if $\sin kL=0$, $kL=n\pi$, so $\lambda_n=\dfrac{2L}n$ and $\nu_n=\dfrac v{\lambda_n}=\dfrac{nv}{2L}$. [1]

Sketch $n=1$: one loop, nodes at the ends, antinode at the middle. $n=2$: two loops, an extra node at the middle. [1]

Exam tip State both boundary conditions ($y=0$ at $x=0$ and $x=L$ at every $t$) — that is where the marks are.

E2. Board · Case study A music teacher has a metal pipe $0.85\ \text{m}$ long. With both ends open it sounds one note; with a cap on one end, a lower one. Take $v=340\ \text{m/s}$. (i) Find the fundamental frequency in each case. [2] (ii) Which harmonics are missing when the pipe is capped? [1] (iii) What is the frequency of the first overtone of the capped pipe? [1]

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(i) Open: $\dfrac{v}{2L}=200\ \text{Hz}$; capped: $\dfrac{v}{4L}=100\ \text{Hz}$ — an octave lower. [2]

(ii) All the even harmonics ($200,\ 400,\ 600\ \text{Hz},\dots$). [1]

(iii) The next mode is the third harmonic: $300\ \text{Hz}$. [1]

Exam tip "First overtone" of a closed pipe is the third harmonic, not the second.

E3. NEET A string fixed at both ends has two successive resonant frequencies of $375\ \text{Hz}$ and $450\ \text{Hz}$. Its fundamental frequency is (a) $75\ \text{Hz}$ (b) $37.5\ \text{Hz}$ (c) $150\ \text{Hz}$ (d) $25\ \text{Hz}$

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Idea Successive harmonics of a string differ by $\nu_1$: $450-375=75\ \text{Hz}$. Check: $375=5\times75$, $450=6\times75$ ✓.

Answer (a).

Exam tip (b) is the closed-pipe reasoning (successive modes differ by $2\nu_1$) applied to a string. For a closed pipe, also check that the frequencies are odd multiples.

E4. JEE Main (Numerical value) The second overtone of a pipe closed at one end is in unison with the first overtone of a pipe $60\ \text{cm}$ long open at both ends. Find the length of the closed pipe in centimetres.

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Closed Second overtone $=$ fifth harmonic $=\dfrac{5v}{4L_\text{c}}$.

Open First overtone $=$ second harmonic $=\dfrac{2v}{2L_\text{o}}=\dfrac{v}{L_\text{o}}$.

Answer $\dfrac{5}{4L_\text{c}}=\dfrac1{L_\text{o}}$: $L_\text{c}=\dfrac54\times60=75\ \text{cm}$.

Exam tip Translate overtones into harmonics first: closed pipe $1,3,5,\dots$; open pipe $1,2,3,\dots$ The speed of sound cancels.

E5. JEE Advanced (One or more correct) A standing wave on a string is $y=0.060\sin(2.5\pi x)\cos(500\pi t)$ in SI units. Which statements are correct? (A) It is formed by two waves of amplitude $0.030\ \text{m}$ travelling at $200\ \text{m/s}$. (B) Neighbouring nodes are $0.40\ \text{m}$ apart. (C) The element at $x=0.10\ \text{m}$ oscillates with amplitude $0.060\ \text{m}$. (D) The elements at $x=0.10\ \text{m}$ and $x=0.50\ \text{m}$ always move in antiphase.

💡 Show solution▶

(A) $2a=0.060$, $a=0.030\ \text{m}$; $v=\dfrac\omega k=\dfrac{500\pi}{2.5\pi}=200\ \text{m/s}$ ✓.

(B) $\lambda=\dfrac{2\pi}{2.5\pi}=0.80\ \text{m}$; nodes $\dfrac\lambda2=0.40\ \text{m}$ apart ✓.

(C) $0.060\left|\sin\dfrac\pi4\right|=0.042\ \text{m}$, not $0.060$ ✗ (the antinodes are at $x=0.20,\ 0.60\ \text{m},\dots$).

(D) $\sin(0.25\pi)=+0.707$, $\sin(1.25\pi)=-0.707$: opposite signs, so antiphase (they are in neighbouring loops, either side of the node at $0.40\ \text{m}$) ✓.

Answer (A), (B), (D).

Exam tip Within one loop all elements are in phase; crossing a node flips the phase by $\pi$. Count the nodes between two points to decide.

E6. NEET A pipe open at both ends has a fundamental frequency of $300\ \text{Hz}$. If one end is closed, the fundamental frequency becomes (a) $150\ \text{Hz}$ (b) $300\ \text{Hz}$ (c) $600\ \text{Hz}$ (d) $75\ \text{Hz}$

💡 Show solution▶

Idea Open: $\dfrac v{2L}=300\ \text{Hz}$; closed: $\dfrac v{4L}=\tfrac12\times300$.

Answer $150\ \text{Hz}$: (a).

Exam tip Closing one end turns the fundamental from half a wavelength in the pipe into a quarter — the wavelength doubles and the frequency halves.

🔬 Research–Teaching NexusResearch-tutored · optional, about 60 minutes

Research question. How "harmonic" are real instruments? Do a string's overtones really sit at exactly $2$, $3$, $4$… times its fundamental, and does a bottle really lack its even harmonics?

Why it matters. Instrument makers — guitar luthiers in Spain, sitar and tanpura makers in Kolkata and Miraj, piano builders in Germany and Japan — shape the harmonic content of their instruments by hand. Piano tuners deliberately "stretch" the tuning because stiff steel strings have overtones slightly sharp of whole multiples, and the buzzing bridge (jawari) of a tanpura is designed to spread energy into many high harmonics. Acoustic engineers use the same spectra to design concert halls and to diagnose machines by their sound.

What to do. Use a free spectrum-analyser app (for example phyphox "audio spectrum" or Spectroid) close to the sound source.

Pluck an open guitar (or other) string. Read the frequencies of the first six peaks and divide each by the first. How close are the ratios to $1,2,3,4,5,6$?
Measure the vibrating length $L$ and find the wave speed $v=2L\nu_1$. If you know the string gauge, estimate the tension.
Pluck the string at its middle, then near the bridge. Which harmonics disappear when you pluck at the middle, and why? (Hint: where are the nodes of the even modes?)
Blow across a glass bottle (a closed pipe). Record its spectrum: are the peaks at odd multiples of the fundamental? Add water in steps and plot $\dfrac1{\nu_1}$ against the air-column length.
From the slope of that graph find the speed of sound, and from the intercept the end correction.

Tools: A smartphone with a spectrum app, a stringed instrument, a bottle, a ruler, water, a spreadsheet.

Hand in: Annotated spectra; a table of overtone ratios; your $\dfrac1{\nu_1}$–$L$ graph with $v$ and $e$; a paragraph explaining any departures from ideal behaviour.

Going further. For a piano or a thick guitar string, fit the overtones to $\nu_n=n\nu_1\sqrt{1+Bn^2}$ and find the "inharmonicity" $B$. Which strings are most inharmonic, and why do thick, short strings suffer most?

Instructor note: for a closed pipe $\dfrac1{\nu_1}=\dfrac{4(L+e)}{v}$, so the graph has slope $\dfrac4v$ ($\approx0.0117\ \text{s/m}$) and a small negative $L$-intercept of $-e$. Nylon strings give ratios within about $0.1\%$ of whole numbers; plucking at the midpoint suppresses the even harmonics (which have a node there). A bottle's neck makes it partly a "Helmholtz resonator", so its fundamental can sit below the simple-pipe value — a good discussion point.

🧪 Self-check quiz

0/5 correct

Q1. In a standing wave, the distance between two neighbouring nodes is

Q2. A pipe closed at one end can produce

Q3. A wave on a string is reflected from a rigidly fixed end. The reflected wave

Q4. A string fixed at both ends has a fundamental frequency of $100\ \text{Hz}$. Its second overtone has frequency

Q5. Two waves of equal amplitude $a$ and the same frequency travel the same way with a phase difference of $\dfrac{2\pi}3$. The amplitude of the resultant is

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