Standing Waves

Beginner

Two friends turning a long skipping rope make it swing as one big loop: the middle flies highest while the ends stay in their hands. Shake one end faster and the rope can settle into two loops with a still point in the middle. Neither shape travels along the rope: this is a standing wave.

Where does it come from? Waves run along the string, bounce off the fixed ends and come back. The outgoing and returning waves overlap, and together they make a pattern that stays in place. On the middle string of the animation the dashed waves run in opposite directions, and the solid string is what they add up to.

Some points never move at all. They are the nodes, marked by gold dots; the fixed ends are nodes too. Halfway between two nodes the string swings the most: that is an antinode.

Because both ends must stay still, only a whole number of loops fits: one, two, three and so on. The one-loop pattern is the fundamental and gives the lowest note. Two loops vibrate twice as fast as one and three loops three times as fast, giving higher notes called harmonics. A real string vibrates in several of these patterns at once, and that mix is why a guitar and a violin playing the same note still sound different.

Pressing a string against a fret shortens the vibrating part, so every loop is shorter and the note is higher. Tightening a string, or using a thinner one, also raises the note. Flutes and organ pipes do the same with standing sound waves in a tube: longer pipes, lower notes.

Check yourself: a string vibrates in three loops. How many nodes does it have, counting both ends? Answer: four.

Intermediate

Two travelling waves make a standing wave

At a given length and tension, a guitar string held fixed at both ends vibrates only at certain frequencies. Waves on a string clamped at both ends bounce off the ends and run through each other. When two waves with the same amplitude $A$, wavelength $\lambda$ and frequency $f$ travel in opposite directions, their sum does not travel at all: it is a standing wave. On the middle string of the animation, the dashed curves are the two waves and the solid string is their sum.

Where the two waves always arrive with opposite displacements, they cancel: these nodes, marked by gold dots, never move. Halfway between two nodes the waves always agree, and the string swings with amplitude $2A$: these are the antinodes. Every other point oscillates at the same frequency with a smaller amplitude, and neighbouring loops move in opposite directions.

Besides $\lambda$, $f$ in hertz and the wave speed $v=f\lambda$ from Wave Basics, a string has its length $L$ between the clamps, its tension $F$ in newtons and its linear density $\mu$, the mass of one metre of string, in kg/m. The harmonic number $n$ counts the loops.

Fitting half-wavelengths between the ends

Neighbouring nodes are half a wavelength apart, and each antinode lies a quarter-wavelength from the nearest node. A clamped end cannot move, so it must be a node, and the string holds a whole number $n$ of loops, each $\lambda /2$ long:
$L=\frac{n\lambda }{2}$
so the only wavelengths it allows are
${\lambda }_n=\frac{2L}{n}$
with $n=1,2,3,\dots$ The longest is twice the length of the string.

A wave that does not fit returns from each round trip out of step with the wave still arriving, and its reflections largely cancel. Only the fitting wavelengths reinforce themselves, so every confined wave has a definite set of natural frequencies.

The harmonic series and the wave speed

All waves on one string travel at the same speed $v$, so each allowed wavelength has its own frequency, $f_n=v/{\lambda }_n$:
$f_n=\frac{nv}{2L}=n{f}_1$
The lowest, $f_1=v/(2L)$, is the fundamental; the others, $2f_1$, $3f_1$ and so on, are its harmonics, equally spaced by $f_1$. The second harmonic sounds an octave above the fundamental.

The wave speed depends on the tension and the linear density:
$v=\sqrt{\frac{F}{\mu }}$
A tight string pulls a displaced piece back hard, so waves run fast; a heavy string has more inertia, so they run slowly.

There are therefore three ways to raise the pitch: shorten the vibrating part (a fret), tighten the string (the peg) or use a lighter one. The frequency is inversely proportional to $L$ but grows only as $\sqrt{F}$: halving the length doubles it, while doing the same with tension alone takes four times the force.

Worked example: the guitar string in the animation

The animation's strings are 0.65 m long, like a classical guitar's, and their waves travel at 143 m/s. Which notes do they play? In the fundamental the string holds half a wavelength:
${\lambda }_1=2L=2\cdot 0.65=1.30\mathrm{m}$
$f_1=\frac{v}{{\lambda }_1}=\frac{143}{1.30}=110\mathrm{Hz}$
Then $f_2=220\mathrm{Hz}$ and $f_3=330\mathrm{Hz}$, the purple labels beside the strings; 110 Hz is the A of a guitar's fifth string. A real period of $1/110\mathrm{s}\approx 9.1\mathrm{ms}$ is far too short to watch, so the animation stretches it to 2 s, 220 times slower.

On the third string ${\lambda }_3=2L/3\approx 0.433\mathrm{m}$, so the nodes are ${\lambda }_3/2\approx 0.217\mathrm{m}$ apart and each antinode lies ${\lambda }_3/4\approx 0.108\mathrm{m}$ from its neighbours: the spacing of the letters N A N A N A N under the string, four nodes counting both ends and three antinodes.

To raise the fundamental an octave, to 220 Hz, at the same length, $v$ must double to 286 m/s; since $v$ grows as $\sqrt{F}$, the tension must be multiplied by four, not two.

Worked example: tuning a violin string

A violin's A string vibrates over 0.325 m and has a linear density of 0.60 g/m, that is $6.0\times 10^{-4}kg/m$. What tension tunes it to 440 Hz? Find the wave speed first:
$v=2L{f}_1=2\cdot 0.325\cdot 440=286m/s$
$F=\mu v^2=6.0\times 10^{-4}\cdot 286^2\approx 49\mathrm{N}$
That is about the weight of a 5 kg mass; leaving $\mu$ in g/m would give a tension a thousand times too large.

If the string slackens and sounds 436 Hz, the tension must grow by the factor $(440/436{)}^2\approx 1.018$, less than 2 % or about 0.9 N, because $f_1$ is proportional to $\sqrt{F}$. To play the B at 494 Hz the violinist shortens the string instead, to $L^{\prime }=v/(2f)=286/988\approx 0.289\mathrm{m}$: the finger goes down about 3.5 cm from the nut.

Reading the pattern: loops, phases and pipes

To read a standing-wave picture, count the loops: there are $n$ of them, each $\lambda /2$ long, and $n+1$ nodes including the ends. Points within one loop move in step; neighbouring loops are half a cycle apart. Twice per cycle the whole string passes through its straight position, moving fastest, and at the extremes it stops for an instant.

In a pipe the air moves back and forth along the pipe. A closed end stops it and is a displacement node; an open end is close to a displacement antinode. Open at both ends, a pipe has the same series as a string, $f_n=nv/(2L)$. Closed at one end, it holds an odd number of quarter-wavelengths, so $f_n=nv/(4L)$ with odd $n$ only. With sound at 343 m/s, a 0.65 m pipe sounds 132, 396 and 660 Hz when closed at one end, and 264, 528 and 792 Hz when open at both.

Common mistakes

• A standing wave is not frozen: every point except the nodes keeps oscillating, all at one frequency, 330 Hz on the bottom string of the animation.
• The wavelength is not the length of the string: in the fundamental the string holds half a wavelength, so the 0.65 m string has ${\lambda }_1=1.30\mathrm{m}$.
• Nodes are half a wavelength apart, not a whole one: with $\lambda =0.8\mathrm{m}$ they are 0.4 m apart, and a node lies 0.2 m from the nearest antinode.
• Doubling the tension does not double the pitch: since $f$ grows as $\sqrt{F}$, the 110 Hz string would rise only to about 156 Hz.
• Harmonics are not louder copies of one note but different frequencies, here 110, 220 and 330 Hz; the mix that a pluck excites gives each instrument its timbre.

Where you meet it

Every string instrument is a set of standing waves: frets change $L$, pegs change $F$, and the bass strings are wound with wire to add mass. A violinist who touches a string lightly at its midpoint forces a node there; the fundamental and the odd harmonics, which need motion at the centre, die out, and the note jumps an octave. Flutes, clarinets and organ pipes hold standing sound waves. In a microwave oven the hot spots lie half a wavelength, about 6 cm, apart.

Driving a system at a natural frequency builds up a large motion, the resonance of Damping and Resonance. Standing waves are a special case of Interference and Superposition, pipes use the speed of sound from Sound and the Doppler Effect, and fitting whole wavelengths around an orbit returns for electrons in Atomic Models: the Bohr Atom.

Advanced

Two travelling waves on a string

Why does a string held at both ends vibrate only at certain frequencies? The starting point is the simplest question: how does a single wave move along a stretched string?

Let $y(x,t)$ be the sideways displacement of the point of the string at position $x$, at time $t$. For small slopes, a string with tension $F$ and mass per metre $\mu$ obeys the wave equation
$\frac{{\partial }^{2}y}{\partial t^2}=v^2\frac{{\partial }^{2}y}{\partial x^2}$
where $v^2=F/\mu$ and $v$ is the wave speed. Its general solution is any shape running right plus any shape running left:
$y(x,t)=u(x-vt)+q(x+vt)$
The value of $u$ stays the same wherever $x-vt$ stays the same, so the shape $u$ slides towards larger $x$ at speed $v$ without changing; $q$ slides the other way. An end of the string ties these two waves together, and that is where standing waves come from.

Reflection at a clamped end

What happens when a pulse reaches a point that cannot move? The answer, an upside-down echo, is the first half of every standing wave.

Legend

$q(x+vt)$	The incoming pulse, running towards the wall. Where it is behind the wall, off the string, it is dashed.	$\mathrm{m}$
$q(vt-x)$	Its twin $-q(vt-x)$: the pulse mirrored in the wall and turned upside down. It runs away from the wall and is dashed while behind it.	$\mathrm{m}$
$v$	Speed of the pulse and its twin.	$m/s$
$x$	Position along the string, measured from the wall.	$\mathrm{m}$
$t_1$	An instant before the pulse reaches the wall (upper picture).	$\mathrm{s}$
$t_2$	A later instant, after the pulse and its twin have passed through each other (lower picture).	$\mathrm{s}$

Figure 1 clamps the string at $x=0$; the real string is the part to the right of the wall, $x>0$. A pulse $q(x+vt)$ arrives from the right, running left. The clamp means $y(0,t)=0$ at every instant. Putting $x=0$ into the general solution gives $u(-vt)+q(vt)=0$ for every $t$. Writing $s=-vt$, this says $u(s)=-q(-s)$ for every $s$: the returning wave is fixed completely by the incoming one, and
$y(x,t)=q(x+vt)-q(vt-x)$

Read the second term as a picture. Replacing $x$ by $-x$ in $q(x+vt)$ gives $q(vt-x)$, the incoming pulse mirrored in the clamp; the minus sign turns it upside down. In the upper picture of Figure 1, at the instant $t_1$, look at the two shapes. The purple pulse $q(x+vt)$ is on the real string and runs left. Its gold, dashed twin $-q(vt-x)$ hangs downwards at the mirror position behind the wall and runs right. Both move at speed $v$, so they reach the wall at the same moment, and at $x=0$ they cancel at every instant: the clamp never moves.

In the lower picture, at the later instant $t_2$, the two have passed through each other. The solid gold twin is now the only shape on the real string, upside down and running away from the wall, while the purple original carries on, dashed, behind the wall, where there is no string. The result: at a clamped end a pulse comes back inverted, with its shape and speed unchanged.

A free end, such as a ring sliding on a smooth rod, behaves differently. The ring can move but feels no sideways force from the rod, so the string must meet the rod at right angles: $\partial y/\partial x=0$ at $x=0$. The same steps then give $u(s)=+q(-s)$, and the pulse returns upright.

From two travelling waves to a standing wave

Now send in a sine wave instead of a single pulse. The incoming wave and its reflection overlap all along the string, and their sum is a standing wave.

Take the incoming wave $A\sin (kx+\omega t)$, with amplitude $A$, wavenumber $k=2\pi /\lambda$ for wavelength $\lambda$, and angular frequency $\omega =vk$; it is a wave of the form $q(x+vt)$. Its inverted mirror image $-q(vt-x)$ is $-A\sin (\omega t-kx)=A\sin (kx-\omega t)$. Adding the two with $\sin a+\sin b=2\sin \frac{1}{2}(a+b)\cos \frac{1}{2}(a-b)$ gives
$y(x,t)=2A\sin (kx)\cos (\omega t)$

Space and time have separated: a fixed shape $\sin kx$ multiplied by one oscillation $\cos \omega t$ shared by every point. The point at $x$ moves up and down sinusoidally with amplitude $2A|\sin kx|$. Where $\sin kx=0$, that is where $kx=m\pi$ with whole $m$, the string never moves: these points are the nodes, and they are $\lambda /2$ apart because $k=2\pi /\lambda$. Halfway between nodes, where $|\sin kx|=1$, lie the antinodes, which swing with the full amplitude $2A$.

Because $\sin kx$ changes sign at each node, neighbouring loops are always on opposite sides of the line: they swing in antiphase. A motion in which every point oscillates at one frequency with a fixed shape is called a normal mode.

The animation shows this on a string in its second mode, $n=2$. The dashed gold wave $A\sin (kx-\omega t)$ runs right, the dashed purple one runs left, and the solid string is their sum. The graph follows the red antinode at $x=L/4$, where $\sin kx=1$: its motion is a pure cosine, $2A\cos \omega t$, which starts on the dashed $+2A$ line.

Two clamped ends: only certain wavenumbers

A second clamp turns a continuum of possible waves into a set of notes. This is the step that gives a string its pitch.

The standing wave above already has a node at $x=0$. A second clamp at $x=L$ also demands $y(L,t)=0$ at all times, so $\sin kL=0$, and $kL$ must be a whole multiple of $\pi$:
$k_n=\frac{n\pi }{L}$
with $n=1,2,3,\dots$ counting the loops. In words, a whole number of half-wavelengths must fit between the clamps, $L=n{\lambda }_n/2$. Because $\omega =vk$ for every wave on the string, the allowed angular frequencies are ${\omega }_n=n\pi v/L$, and the frequencies are
$f_n=\frac{nv}{2L}$
They are whole multiples of the lowest one, the fundamental $f_1=v/(2L)$: this is the harmonic series.

For the animation's string, $L=0.65\mathrm{m}$ and $v=143m/s$, so $f_1=143/(2\cdot 0.65)=110\mathrm{Hz}$, with $k_1=\pi /0.65\approx 4.83rad/m$ and ${\omega }_1=2\pi \cdot 110\approx 691rad/s$. A string with only one clamp accepts every $k$: it is the second boundary that makes the spectrum discrete.

The round trip and the harmonic series

The same frequencies follow without a single sine, by tracking one pulse as it bounces between the clamps. This view shows why the harmonics are exact multiples.

Legend

$x$	Position along the string.	$\mathrm{m}$
$t$	Time, increasing upwards.	$\mathrm{s}$
$L$	Distance between the two clamps.	$\mathrm{m}$
$x_0$	Where the pulse starts, and where it is back after one round trip.	$\mathrm{m}$
$v$	Speed of the pulse; the slope of each leg.	$m/s$
$T_1$	Time of one round trip, $2L/v$.	$\mathrm{s}$

Figure 2 is a space–time diagram: position $x$ runs across, time $t$ runs upwards, and the two thick vertical lines are the clamps at $x=0$ and $x=L$, which stay put. A pulse moving at constant speed is a straight sloping line, and each bounce is a corner. Follow the blue path from the bottom. The pulse leaves $x_0$ at $t=0$, running right at speed $v$; the sign $+$ says it is upright. At $x=L$ it reflects and turns upside down ($-$), runs left across the whole string, then reflects at $x=0$ and turns upright again ($+$). Two reflections mean two inversions, so it reaches $x_0$ at the top dot upright and moving right, exactly as it started.

The length of the path does not depend on where the pulse started: $(L-x_0)+L+x_0=2L$. So the round trip takes
$T_1=\frac{2L}{v}=\frac{1.30}{143}\approx 9.09\mathrm{ms}$
which is exactly $1/f_1$ for 110 Hz. A disturbance of any shape is made of parts that each make this same round trip, so after $T_1$ the whole string is back in its starting state and the motion repeats with period $T_1$. A motion with period $T_1$ can contain only frequencies that fit a whole number of cycles into $T_1$: the multiples of $1/T_1=110\mathrm{Hz}$. The harmonic series is the round-trip time written as a spectrum.

Any motion is a sum of modes

A real string is never set going in one pure mode, yet every motion is still a mixture of modes, and there is a way to find the mixture.

The wave equation is linear: if two motions are solutions, so is their sum. Every mode is a solution, and together the modes are enough to build any motion of the string:
$y(x,t)={\sum }_n{b}_n\sin \frac{n\pi x}{L}\cos ({\omega }_{n}t+{\phi }_n)$
Mode $n$ enters with its own amplitude $b_n$ and phase ${\phi }_n$. Both are set by the string's initial shape and velocity, and finding them is called Fourier analysis.

The tool that does the finding is orthogonality: multiply the shapes of two different modes and integrate along the string, and the result is zero. For a string released from rest with the shape $y_0(x)$, every phase ${\phi }_n$ is zero, because each mode must start at rest. Multiply $y_0(x)$ by the shape of mode $n$ and integrate from 0 to $L$. Orthogonality removes every term except mode $n$'s own, which gives $b_n$ times ${\int }_0^L{\sin }^2(n\pi x/L)dx=L/2$. Therefore
$b_n=\frac{2}{L}{\int }_0^L{y}_0(x)\sin \frac{n\pi x}{L}dx$

Symmetry settles one case without any integral. A pluck at the centre gives a shape symmetric about $x=L/2$, while every even mode is antisymmetric about that point: one half goes up while the other goes down. A symmetric shape times an antisymmetric one integrates to zero, so the even coefficients vanish, and a string plucked at its centre sounds only odd harmonics. Going the other way, a few modes already make rich motions: the first two alone, a quarter-cycle out of step, add up to a bump that runs back and forth between the ends, and many modes together build the sharp corner of a real pluck.

Orthogonality also splits the energy. The total energy of the string is the sum of the mode energies, with no cross terms between modes:
$E_n=\frac{1}{4}\mu L{\omega }_n^2{b}_n^2$
This is the kinetic energy of mode $n$ at the instant it passes through straight: $\frac{1}{2}\mu {\omega }_n^2{b}_n^2{\sin }^2(n\pi x/L)$ per metre, and the ${\sin }^2$ averages to one half along the string. In this sense any linear vibrating system, from a bridge to a molecule, is a collection of independent oscillators, one per mode.

Where the energy sits

A standing wave carries no energy along the string, yet energy keeps moving inside it. Figure 3 shows where it is stored.

Legend

$n$	Mode number. The string is in its second mode, $n=2$, drawn at its two extreme shapes.
$\lambda /2$	Distance between neighbouring nodes (gold dots): half a wavelength.	$\mathrm{m}$
$x$	Position along the string.	$\mathrm{m}$
$u_K$	Kinetic energy per metre: the largest value it reaches at each point during a cycle.	$J/m$
$u_E$	Elastic (stretching) energy per metre: the largest value it reaches at each point during a cycle.	$J/m$

Figure 3 shows the $n=2$ string at its two extreme shapes, with its nodes as gold dots, $\lambda /2$ apart. Below it, two curves show how the energy per metre is spread along $x$. Each curve is the largest value that energy reaches at each point during a cycle; the two largest values occur at different instants.

The kinetic energy per metre is $u_K=\frac{1}{2}\mu (\partial y/\partial t{)}^2$. With $\partial y/\partial t=-2A\omega \sin kx\sin \omega t$, it varies as ${\sin }^{2}kx{\sin }^2\omega t$. Its largest value, the solid purple curve, peaks at the antinodes, where the string moves fastest, and vanishes at the nodes, which never move. It is reached at the instants when the string passes through straight.

The elastic energy per metre, stored by stretching the string, is $u_E=\frac{1}{2}F(\partial y/\partial x{)}^2$. With $\partial y/\partial x=2Ak\cos kx\cos \omega t$, it varies as ${\cos }^{2}kx{\cos }^2\omega t$. Its largest value, the dashed green curve, peaks at the nodes, where the string is most tilted, and is reached at the instants of maximum displacement. The two curves reach the same height, $2\mu A^2{\omega }^2$, because $F{k}^2=\mu {\omega }^2$ (from $v^2=F/\mu$ and $\omega =vk$). So during each cycle the energy sloshes from elastic energy near the nodes to kinetic energy near the antinodes and back.

Why does none of it travel along the string? Each travelling component carries a mean power $\frac{1}{2}\mu {\omega }^2{A}^{2}v$, but in a standing wave the two carry it in opposite directions, so the net flow is zero. Instant by instant, the power passing a point is the sideways force from the string on the left, $-F\partial y/\partial x$, times the velocity of that point, $\partial y/\partial t$:
$P=-F\frac{\partial y}{\partial x}\frac{\partial y}{\partial t}=Fk\omega A^2\sin 2kx\sin 2\omega t$
The factor $\sin 2kx$ is zero at every node and every antinode, at all times. Energy never crosses those points: it swings back and forth within each quarter-wavelength, at twice the frequency of the motion. On the animation's graph, the antinode passes through zero at its top speed, $2A\omega$; at that instant all the energy of its region is kinetic.

Drive the string at one of its $f_n$ and the energy fed in during each cycle adds up, so the amplitude grows until the losses balance the drive. This is resonance, and the height of its peak is set by the quality factor $Q$ met in Damping and Resonance.

Beyond the ideal string

The rule that whole numbers of waves must fit carries over to other ends, stiffer strings and surfaces, with changes you can hear.

A string clamped at $x=0$ but free at $x=L$ needs zero slope at the free end, $\cos kL=0$, so $kL=(m-\frac{1}{2})\pi$ with whole $m\ge 1$. An odd number of quarter-wavelengths fits, and $f=(2m-1)v/(4L)$: only odd multiples of the lowest frequency $v/(4L)$. This is the pipe closed at one end. Its closed end is a displacement node, like the clamp, and its open end is a pressure node, like the free end. The air just outside an open end also takes part, adding about 0.6 of the pipe's radius to the effective length.

A real string also resists bending. This extra restoring force grows faster with curvature than tension does, so short waves run faster and the overtones go sharp:
$f_n\approx n{f}_1\sqrt{1+B{n}^2}$
where the stiffness constant $B$ grows with the fourth power of the wire's diameter. For a piano string near middle C, $B$ is of order $4\times 10^{-4}$, so for the tenth mode $\sqrt{1+0.04}\approx 1.02$: it lies about 2 % above ten times the fundamental, and tuners stretch their octaves slightly to match.

In two dimensions the waves must fit both ways at once. A rectangular membrane with sides $L_x$ and $L_y$ has modes $f_{mn}=(v/2)\sqrt{(m/L_x{)}^2+(n/L_y{)}^2}$. A round drumhead has modes at about 1.59, 2.14 and 2.30 times its lowest frequency, not whole multiples, which is why a drum's pitch is less definite than a string's.

Worked problem: plucking the animation's string

This problem puts the pieces together: the Fourier coefficients of a real pluck, the harmonics it leaves out, and the share of energy each harmonic gets.

Legend

$y_0$	Shape of the string at the moment of release: a triangle.	$\mathrm{m}$
$H$	How far the string is pulled aside at the pluck point.	$\mathrm{m}$
$d$	Distance from the left end to the pluck point.	$\mathrm{m}$
$L$	Length of the string.	$\mathrm{m}$
$n$	Mode number: mode $n$ has $n$ loops. Shown: $\sin (n\pi x/L)$ for $n=1$ and $n=5$.

Figure 4 draws the string at the moment of release. It is the animation's string, $L=0.65\mathrm{m}$ and $v=143m/s$, pulled aside by $H=3\mathrm{mm}$ at $d=0.13\mathrm{m}$ from the left end, a fifth of its length, and let go from rest. The height $H$ is drawn far larger than it is. Which harmonics sound, and how is the energy shared between them?

The initial shape $y_0(x)$ is the blue triangle: two straight lines from the clamps up to the height $H$ at $x=d$. Putting it into the formula for $b_n$ and integrating each straight piece gives
$b_n=\frac{2H{L}^2}{{\pi }^2{n}^{2}d(L-d)}\sin \frac{n\pi d}{L}=\frac{25H}{2{\pi }^2{n}^2}\sin \frac{n\pi }{5}$
where the second form uses $d=L/5$, so that $d(L-d)=4L^2/25$. This gives $b_1\approx 2.23\mathrm{mm}$, $b_2\approx 0.90\mathrm{mm}$ and $b_3\approx 0.40\mathrm{mm}$, at 110, 220 and 330 Hz.

The two lower pictures show why some harmonics are missing. Look along the dashed vertical line through the pluck point $x=d$. The shape of mode 1, $\sin (\pi x/L)$, is well above zero there, so pulling at $d$ feeds mode 1 strongly. The shape of mode 5, $\sin (5\pi x/L)$, crosses zero exactly there, because $d=L/5$. The pluck point is a node of mode 5, a place where that mode never moves, so no pull applied there can start it. The same holds for every mode with a node at $d$: $\sin (n\pi /5)=0$ for $n=5,10,\dots$, so 550 Hz and 1100 Hz are absent.

For the energy, $E_n\propto {\omega }_n^2{b}_n^2$, with ${\omega }_n\propto n$ and $b_n\propto \sin (n\pi d/L)/n^2$, so $E_n\propto {\sin }^2(n\pi d/L)/n^2$. Dividing each term by the sum over all modes, the fundamental gets 44 % of the energy, the second harmonic 29 % and the third 13 %. Plucked at the centre, the same string would put 81 % into the fundamental, 9 % into the third harmonic and nothing into the even ones. Plucking nearer the end feeds the upper harmonics, which is why a guitar sounds bright and thin near the bridge and round over the soundhole.

Limits of validity

The ideal string is a model. Knowing where it stops being right tells you when the results above can be trusted.

The model assumes small slopes, so that the string hardly stretches and its tension stays equal to $F$. Here the steepest slope is $H/d=3/130\approx 0.023$, well within that. A hard pluck does stretch the string, so the note starts slightly sharp and settles as it fades.

Nor are the ends perfectly rigid, and they must not be: a string alone pushes very little air, and an instrument sounds because the bridge moves and passes the vibration to the body. That leak, together with air drag and internal friction, damps the modes, the high ones fastest. If the fundamental's amplitude falls by a factor e in $\tau \approx 3\mathrm{s}$, its quality factor is $Q=\pi f_1\tau =\pi \cdot 110\cdot 3\approx 1000$.

A string is made of atoms about 0.3 nm apart, so the continuous wave equation holds only for wavelengths much longer than that spacing. A chain of $N$ atoms has exactly $N$ modes, the shortest wavelength being two spacings. In a crystal these modes reach about $10^{13}\mathrm{Hz}$, and their quanta, called phonons, carry sound and heat.

Where this leads

The idea that only waves which fit can survive reaches far beyond strings.

Light between two mirrors forms standing waves spaced in frequency by $c/(2L)$, where $c$ is the speed of light: about 500 MHz for a 30 cm laser cavity (Electromagnetic Waves). Adding waves in general leads on to the next lesson, Interference and Superposition.

The deepest echo is in quantum mechanics, where an electron is a wave and its frequency is an energy, $E=hf$, with $h$ the Planck constant. An electron confined to a box of length $L$ must also fit whole half-wavelengths, ${\lambda }_n=2L/n$. Its momentum is $p=h/\lambda$ and its energy is $p^2/(2m_e)$, with $m_e$ the electron mass, so
$E_n=\frac{n^2{h}^2}{8m_e{L}^2}$
which is about 0.38 eV for $n=1$ in a 1 nm box. The levels grow as $n^2$, not as $n$, because for matter waves the frequency goes as $k^2$, while on a string it goes as $k$. Fitting whole wavelengths around an orbit is the idea behind Atomic Models: the Bohr Atom; the box is solved properly in Schrödinger's Equation and Tunnelling, and Wave–Particle Duality explains why electrons are waves at all.