Wave Basics

Beginner

Picture a stadium wave: each fan only stands up and sits down again, yet a bulge of raised arms races all the way round the stands. That is what a wave is: a disturbance that travels from place to place and carries energy, while the material itself stays put. A boat on a lake rises and falls as ripples pass, but they do not carry it to the shore.

In the animation a hand shakes the end of a rope. The gold dots ride the crests, the high points, steadily to the right, while the red bead only bobs up and down on its dashed line.

There are two kinds of wave. On the rope each bit moves across the direction of travel: a transverse wave. In the slinky at the bottom the coils move back and forth along that direction, crowding together and spreading apart: a longitudinal wave. Sound in air is of this kind.

Four words describe a regular wave. The wavelength is the distance from one crest to the next; the amplitude is how high a crest rises above the rest line; the frequency is how many crests pass a point each second, in hertz (Hz); the speed is how far a crest travels in one second. If 2 crests pass you every second and they are 3 metres apart, the wave covers $2\times 3=6$ metres each second: speed = frequency × wavelength.

The speed is fixed by the medium, not by the hand. Shake faster and more crests leave each second, so they have to sit closer together.

Check yourself: you shake the rope three times as often. What happens to the wavelength? (It becomes three times shorter; the speed does not change.)

Intermediate

What moves, and what we measure

It is tempting to think that a shaken rope travels along with its crests. It does not. A wave is a disturbance that travels through a medium and carries energy, while each part of the medium only oscillates about its resting place. On a shaken rope each bit copies its neighbour a moment later: the motion is handed on, the rope stays. In the animation the crests run to the right, yet the red bead only bobs along its dashed line.

On the rope each piece moves at right angles to the direction of travel: a transverse wave. In the slinky below, the coils move back and forth along that direction, crowding into compressions and spreading into rarefactions: a longitudinal wave. Sound in air is longitudinal, because a gas resists squeezing but not sideways shearing. Light is transverse too, with electric and magnetic fields oscillating instead of matter.

The displacement $y$ is how far a point is from its rest position (the dashed line), in metres; the amplitude $A$ is the largest displacement, from the rest line to a crest. The wavelength $\lambda$ is the length of one cycle in space, crest to crest. The period $T$ is the time of one full oscillation of a point, and the frequency $f$ the number of oscillations per second, in hertz (Hz).

One wavelength per period

Watch a single crest. While the point under it makes one full oscillation, the next crest arrives where the first one was: in one period $T$ the pattern moves forward by exactly one wavelength, so its speed is $\lambda /T$. Since the frequency is the number of periods per second, the two key relations of this lesson are:

• $f=\frac{1}{T}$
• $v=f\lambda$

They hold for every repeating wave, on a rope, in air or for light, and any two of $v$, $f$ and $\lambda$ give the third. The relation does not say which quantity is fixed, though: that is the job of the medium and of the source.

The medium sets the speed

How fast a disturbance is handed on depends on the medium alone: on how strongly it pulls back and how much mass must be set moving. For a string or rope with tension $F$, in newtons, and mass per metre $\mu$, in kilograms per metre,
$v=\sqrt{\frac{F}{\mu }}$

Pull harder and the wave runs faster; a heavier rope makes it slower, and four times the tension doubles the speed. Sound travels at about 343 m/s in air at 20 °C, 1480 m/s in water and 6000 m/s in steel, light at $3.00\times 10^{8}m/s$ in a vacuum.

The source sets the frequency: the hand or the loudspeaker decides how many cycles leave each second, and the wavelength follows, $\lambda =v/f$. In a new medium each point is driven by the old one at the same rate, so $f$ stays the same while $v$ and $\lambda$ change together. The energy carried grows with the square of the amplitude: doubling $A$ carries four times as much energy each second.

Worked example: the rope in the animation

The red bead makes one oscillation every 2 s, and the purple double arrow spans one wavelength, 2 m. So
$f=1/T=1/2=0.5\mathrm{Hz}$
$v=f\lambda =0.5\cdot 2=1m/s$
These are the values at the top of the animation. The purple arrow rides 2 m forward with the crests in each period, then jumps back to where the picture looks the same.

The bead sits 3 m from the hand, so the hand's motion reaches it $3/1=3\mathrm{s}$ later, one and a half periods: the bead is half a cycle out of step. Whenever the hand is at its lowest, $-0.3\mathrm{m}$, the bead is at its highest, $+0.3\mathrm{m}$; the red readout swings between these two values, the amplitude marked by the gold arrow $A$.

Compare how far the bead and a crest travel in one period. In each period the bead travels $4A=1.2\mathrm{m}$ up and down, an average of 0.6 m/s across the rope, while each crest moves 2 m along it. The slinky's compressions, too, are 2 m apart.

Worked example: two ropes tied together

A light rope of 0.040 kg per metre is tied to a heavy one of 0.160 kg per metre, each 6 m long, under a tension of 36 N. You shake the light rope 5 times per second. On each rope,
$v_1=\sqrt{36/0.040}=\sqrt{900}=30m/s,{\lambda }_1=30/5=6\mathrm{m}$
$v_2=\sqrt{36/0.160}=\sqrt{225}=15m/s,{\lambda }_2=15/5=3\mathrm{m}$
The frequency is 5 Hz on both, because the joint shakes the heavy rope as often as it is shaken itself. Four times the mass per metre halves the speed, and with it the wavelength.

A pulse crosses the light rope in $6/30=0.2\mathrm{s}$ and the heavy one in $6/15=0.4\mathrm{s}$, 0.6 s in all. Part of each wave also bounces back from the joint: that is where reflection begins.

Snapshots and histories

A snapshot plots $y$ against the position $x$ at one instant, like the rope itself, and shows the wavelength and the amplitude. A history plots $y$ against the time $t$ for one point, like the red bead's graph in the advanced view, and shows the period. Both are sines of the same height, one repeating every 2 m and the other every 2 s; their ratio is the 1 m/s.

A longitudinal wave is drawn the same way, with the displacement along the slinky plotted upward, so the curve looks transverse. Read it with care: a compression is not where the displacement is largest, but at a coil sitting exactly at its rest position with both neighbours pushed towards it. Half a wavelength on, they are pulled away: a rarefaction.

Common mistakes

• Shaking faster does not make the wave faster: on the 36 N light rope, 10 shakes per second still travel at 30 m/s; only the wavelength shrinks, from 6 m to 3 m.
• The wave speed is not the speed of the medium: the red bead averages 0.6 m/s, and more if the amplitude grows, while the crests travel at a steady 1 m/s.
• The amplitude is measured from the rest line: the rope spans 0.6 m from its lowest to its highest point, but $A=0.3\mathrm{m}$.
• From crest to trough is half a wavelength, and from top to bottom half a period: a point that takes 0.25 s to go from top to bottom has $T=0.5\mathrm{s}$ and $f=2\mathrm{Hz}$.

Where you meet it

Sound needs about 3 s per kilometre, so the delay between lightning and thunder gives the distance of a storm. Audible sound, 20 Hz to 20 kHz, has wavelengths from 17 m down to 1.7 cm; medical ultrasound at 5 MHz, at about 1540 m/s in tissue, has 0.3 mm waves, fine enough to show a beating heart. Earthquakes send both kinds of wave through the ground, and the transverse ones cannot cross the Earth's outer core, which is how we know that it is liquid.

Next come Sound and the Doppler Effect, Standing Waves, Interference and Superposition, Reflection and Refraction, and Electromagnetic Waves.

Advanced

A shape that slides

A wave is a disturbance that travels while the medium stays where it is: each point of the rope only moves up and down, yet a shape runs along it. To work with waves we need one function that gives the displacement of every point at every instant, and we need to know what form that function takes when a shape moves without changing.

Legend

$y$	Displacement of the rope from its rest line.	$\mathrm{m}$
$x$	Position along the rope; the dot on the dashed pulse is at $x$.	$\mathrm{m}$
$t$	Time: the solid pulse is the rope at $t=0$, the dashed one a time $t$ later.	$\mathrm{s}$
$v$	Speed of the wave, towards $+x$.	$m/s$
$vt$	Distance the shape has moved in the time $t$.	$\mathrm{m}$
$x-vt$	Position of the dot on the solid pulse, whose height is now found at $x$.	$\mathrm{m}$

Write $y(x,t)$ for the sideways displacement of the rope at position $x$ and time $t$. Figure 1 follows a lopsided pulse, steeper on its left side than on its right, so that you can see the whole shape being carried along unchanged. The solid curve is the rope at $t=0$; call its shape $y=f(x)$, where for now $f$ names a shape, not a frequency. The dashed curve is the same rope a time $t$ later. In that time the shape has moved towards $+x$ at speed $v$, so it has shifted by the distance $vt$ marked under the axis.

Now look at the two dots. The right one is on the dashed curve at position $x$. The left one is on the solid curve a distance $vt$ further back, at $x-vt$. The dashed horizontal line joining them shows that they are at the same height: the moving shape has simply carried that height forward by $vt$. The same is true for every point, so the height found at $x$ at time $t$ is the height that was at $x-vt$ at time zero:
$y(x,t)=f(x-vt)$

Everything depends on the single combination $x-vt$. Holding it fixed, $x-vt=x_1$, picks out one feature of the shape, a peak say, and that feature is at $x=x_1+vt$: it moves forward at speed $v$, as it should. By the same argument a shape moving towards $-x$ has the form $g(x+vt)$.

The most important shape is a sine. Give it a wavelength $\lambda$, the distance between neighbouring crests, and a period $T$, the time one point takes to go through a full cycle. The argument of the sine, called the phase, must then grow by $2\pi$ when $x$ increases by $\lambda$, and also when $t$ increases by $T$. Both requirements are met by
$y(x,t)=A\sin (kx-\omega t)$
$k=\frac{2\pi }{\lambda }$
$\omega =\frac{2\pi }{T}=2\pi f$
where $A$ is the amplitude and $f=1/T$ is now the frequency. The wave number $k$ counts radians of phase per metre of rope, and the angular frequency $\omega$ counts radians of phase per second. To find the speed, take $k$ out of the phase: $kx-\omega t=k(x-(\omega /k)t)$. The sine therefore depends on $x$ and $t$ only through $x-vt$, with
$v=\frac{\omega }{k}=f\lambda$

In the animation $k=\pi rad/m$ and $\omega =\pi rad/s$, so the wavelength is 2 m, the period 2 s and the speed 1 m/s. With $x$ measured from the hand, the rope is $y=0.3\sin (\pi x-\pi t)$, in metres and seconds. Adding a phase constant, $A\sin (kx-\omega t+\phi )$, only shifts the point of its cycle at which the wave starts.

Newton's law for a piece of string

So far we have only described a travelling shape. Now we ask why a disturbance on a string travels at all, and what sets its speed. The answer comes from Newton's second law applied to one tiny piece of string.

Legend

$x$	Position of the left end of the piece.	$\mathrm{m}$
$x+dx$	Position of the right end of the piece.	$\mathrm{m}$
$dx$	Length of the piece along $x$.	$\mathrm{m}$
$\mu dx$	Mass of the piece, with $\mu$ the mass per unit length.	$\mathrm{kg}$
$F$	Tension: the pull of the rest of the string on the piece, along the string, at each end.	$\mathrm{N}$
${\theta }_1$	Angle between the pull at the left end and the horizontal.	$\mathrm{rad}$
${\theta }_2$	Angle between the pull at the right end and the horizontal.	$\mathrm{rad}$
$F_y$	Net sideways force on the piece: the sum of the sideways parts of the two pulls.	$\mathrm{N}$

Figure 2 shows a short piece of string between $x$ and $x+dx$, at a place where the string is bent. If $\mu$ is the mass per unit length, the piece has mass $\mu dx$. The rest of the string pulls on it at both ends with the tension $F$, always along the string, at the angles ${\theta }_1$ and ${\theta }_2$ to the horizontal.

Split each pull into a horizontal part and a sideways part; the dashed lines at each end show the split. When the angles are small, both horizontal parts are very nearly $F$ and they point in opposite directions, so they cancel. The sideways part of each pull is $F\sin \theta$. For a small angle $\sin \theta \approx \tan \theta$, and $\tan \theta$ is just the slope of the string, $\partial y/\partial x$. So the sideways part at each end is $F$ times the slope of the string at that end.

In the figure the slope is negative at the left end and positive at the right end, so both sideways parts point up and add. In general the left end is pulled towards $-x$ and the right end towards $+x$, so their sideways parts enter with opposite signs: the net sideways force is $F$ times the slope at $x+dx$ minus $F$ times the slope at $x$. That difference of slopes across a length $dx$ is $({\partial }^{2}y/\partial x^2)dx$, so
$F_y=F\frac{{\partial }^{2}y}{\partial x^2}dx$

A straight piece therefore feels no net force, and a curved piece is pulled towards the inside of its curve, as the arrow $F_y$ shows. Newton's second law for the piece reads $\mu dx{\partial }^{2}y/\partial t^2=F_y$, and dividing by $\mu dx$ gives
$\frac{{\partial }^{2}y}{\partial t^2}=\frac{F}{\mu }\frac{{\partial }^{2}y}{\partial x^2}$

This is the wave equation. In words: each bit of string accelerates in proportion to how sharply the string is curved there. Does the sliding shape of the first section, $y=f(x-vt)$ with $f$ again naming a shape, obey it? Each derivative with respect to $t$ brings out a factor $-v$ by the chain rule, and each derivative with respect to $x$ a factor 1, so ${\partial }^{2}y/\partial t^2=v^2{f}^{\prime \prime }$ and ${\partial }^{2}y/\partial x^2=f^{\prime \prime }$, where $f^{\prime \prime }$ is the second derivative of the shape. The equation then reads $v^2{f}^{\prime \prime }=(F/\mu )f^{\prime \prime }$. It holds for every shape, provided that
$v=\sqrt{\frac{F}{\mu }}$

The tension supplies the restoring force and the mass per unit length the inertia, just as the stiffness and the mass do for an oscillator on a spring: a tighter string carries waves faster, a heavier one slower. Two approximations are hidden in the derivation. The horizontal parts of the pulls, $F\cos \theta \approx F$, were taken to cancel, so the piece moves only sideways; and the tension was taken to be the same everywhere. Both need small slopes, $|\partial y/\partial x|\ll 1$.

Sound: the same law in a gas

Sound is mechanics too. In air the restoring force comes from pressure instead of tension, and the gas moves along the direction of travel instead of across it. This section derives the wave equation for a gas, and from it the speed of sound.

Legend

$x$	Rest position of the left face of the layer.	$\mathrm{m}$
$x+dx$	Rest position of the right face of the layer.	$\mathrm{m}$
$dx$	Thickness of the layer at rest.	$\mathrm{m}$
$s$	Displacement of the left face from its rest position, along $x$.	$\mathrm{m}$
$s+ds$	Displacement of the right face; $ds$ is how much more it moved than the left face.	$\mathrm{m}$
$dx+ds$	Thickness of the layer once displaced.	$\mathrm{m}$

In a sound wave each thin layer of gas moves back and forth along $x$, like the coils of the slinky in the animation. Let $s(x,t)$ be the displacement along $x$ of the layer that rests at $x$. Figure 3 follows one layer, between $x$ and $x+dx$ while the gas is at rest (top row). In the bottom row a sound wave has displaced the gas. The left face of the layer has moved by $s$ and the right face by $s+ds$, a different amount, so the layer is now $dx+ds$ thick. In the figure $ds$ is negative: the right face has moved back, the layer is squeezed and the gas in it is denser.

Three links lead from this picture to a wave equation.

First, mass is conserved. The two faces have moved by amounts that differ by $ds=(\partial s/\partial x)dx$, so the same mass now fills a thickness $(1+\partial s/\partial x)dx$. The density is inversely proportional to the thickness, so for a small $\partial s/\partial x$ it changes by $\delta \rho \approx -{\rho }_0\partial s/\partial x$, where ${\rho }_0$ is the density at rest.

Second, the pressure depends on the density. For a small change, $\delta P=(dP/d\rho {)}_0\delta \rho$, where the subscript 0 means that the derivative is taken in the gas at rest.

Third, Newton's law for the layer. Per unit area it has mass ${\rho }_{0}dx$. The pressure on its left face pushes it towards $+x$ and the pressure on its right face towards $-x$, so the net force per unit area is the drop in pressure across it, $-(\partial \delta P/\partial x)dx$. Dividing by $dx$, ${\rho }_0{\partial }^{2}s/\partial t^2=-\partial \delta P/\partial x$: the layer is pushed away from the side of higher pressure.

Chaining the three, put $\delta \rho$ into $\delta P$ and $\delta P$ into Newton's law, then divide by ${\rho }_0$:
$\frac{{\partial }^{2}s}{\partial t^2}=c^2\frac{{\partial }^{2}s}{\partial x^2},c^2={\left(\frac{dP}{d\rho }\right)}_0$

This is the same wave equation as for the string, and $c$ is the speed of sound. Which derivative $dP/d\rho$ is the right one depends on what the temperature does while the gas is squeezed. Newton assumed it stays constant, so that $P\propto \rho$; that gives $c=\sqrt{P_0/{\rho }_0}\approx 290m/s$ for air at 20 °C, some 15 % below the measured 343 m/s.

Laplace found the flaw: heat has no time to flow from the compressions, which warm up, to the rarefactions, which cool down. At 1 kHz heat diffuses about 0.1 mm in half a period, while a compression and the next rarefaction are 17 cm apart. Each layer is therefore squeezed without exchanging heat, an adiabatic process, in which $P\propto {\rho }^{\gamma }$ with $\gamma =1.40$ for air. Then $dP/d\rho =\gamma P/\rho$, and
$c=\sqrt{\frac{\gamma P_0}{{\rho }_0}}=\sqrt{\frac{\gamma k_{B}T}{m}}$

The second form uses the ideal gas law, $P_0={\rho }_0{k}_{B}T/m$. Here $T$ is the absolute temperature, not the period, $m$ the mass of one molecule and $k_B$ Boltzmann's constant. The pressure drops out, because at a fixed temperature a higher pressure means a proportionally higher density. The speed of sound depends only on the gas and its temperature: 331 m/s at 0 °C and 343 m/s at 20 °C. Compare the molecules themselves: their root-mean-square speed, $\sqrt{3k_{B}T/m}$ (see Kinetic Theory of Gases), is 502 m/s at 20 °C, so $c$ is $\sqrt{\gamma /3}\approx 0.68$ of it. A pressure signal, passed on from molecule to molecule by collisions, travels a little slower than the molecules that carry it.

Snapshot, history and the graph

A wave depends on two variables, so it can be graphed in two ways: the whole rope at one instant, or one point of the rope over time. This section shows that for a travelling wave the two graphs are the same curve, and what that says about how each point moves.

Legend

$y$	Displacement of the rope from its rest line.	$\mathrm{m}$
$x$	Position along the rope, measured from the hand.	$\mathrm{m}$
$t$	Time.	$\mathrm{s}$
$x_0$	Position of the red bead.	$\mathrm{m}$
$v$	Speed of the wave, towards $+x$.	$m/s$
$\partial y/\partial x$	Slope of the rope at the bead in the snapshot: the upper tangent line.
$\partial y/\partial t$	Velocity of the bead: the slope of its history, the lower tangent line.	$m/s$

Think of $y(x,t)$ as a surface over the $(x,t)$ plane; Figure 4 cuts it in two ways. The upper curve is a snapshot, a slice at one instant: the rope at $t=0$, $y=0.3\sin (\pi x)$. The lower curve is a history, a slice at one place: the red bead at $x_0=3\mathrm{m}$, the graph of the advanced view. Since $\sin (3\pi -a)=\sin a$, the bead moves as $y=0.3\sin (3\pi -\pi t)=0.3\sin (\pi t)$: a sine of period 2 s that starts at zero and rises. The hand at $x=0$ moves as $-0.3\sin (\pi t)$, the same curve upside down. The dashed line joins the two appearances of the bead at $t=0$.

Why are the two slices related? Because $y$ depends on $x$ and $t$ only through $x-vt$, the chain rule gives $\partial y/\partial x=f^{\prime }$ and $\partial y/\partial t=-v{f}^{\prime }$. Hence
$\frac{\partial y}{\partial t}=-v\frac{\partial y}{\partial x}$

A point of the rope moves at minus the wave speed times the local slope. This makes sense: where the rope slopes down towards $+x$, the higher part behind is arriving, so the point rises. In Figure 4 the $x$ and $t$ axes have equal scales and $v=1m/s$, so the lower curve is the upper one flipped left to right, a mirror image about the dashed line. Look at the two tangent lines at the bead: the snapshot slopes down, $\partial y/\partial x=-kA$, and the history slopes up, $\partial y/\partial t=+\omega A$.

Where the rope crosses its rest line, as at the bead, its slope is largest in size, $kA$, and so is the speed of the point, $vkA=\omega A$; at a crest both are zero. In the animation $kA=0.3\pi \approx 0.94$, so the bead peaks at 0.94 m/s, almost the wave speed, and its graph starts with that slope.

Applying the rule twice gives back the wave equation, ${\partial }^{2}y/\partial t^2=v^2{\partial }^{2}y/\partial x^2$: the acceleration of a point is $v^2$ times the curvature of the rope there. A crest, the most curved place, accelerates downward the hardest, at ${\omega }^{2}A=0.3{\pi }^2\approx 2.96m/{\mathrm{s}}^2$. For a sine ${\partial }^{2}y/\partial x^2=-k^{2}y$, so every point obeys ${\partial }^{2}y/\partial t^2=-{\omega }^{2}y$. The wave is a row of harmonic oscillators like those of Simple Harmonic Motion, each a little behind its neighbour.

Superposition: adding waves

The wave equation is linear: $y$ and its derivatives appear only to the first power. So if two functions satisfy it, their sum does too. Two pulses running towards each other add where they overlap and leave with their shapes intact, and the general solution on a long string is $f(x-vt)+g(x+vt)$, with $f$ and $g$ any two shapes, as d'Alembert showed in 1747. Interference, beats and standing waves are all superpositions.

Linearity also works the other way round: any shape can be built as a sum of sines. Each sine travels at its own $\omega /k$; on the ideal string that is the same $v$ for all of them, so the sum, and with it the shape, travels unchanged. Light obeys the same equation, derived from Maxwell's equations instead of Newton's law.

Energy carried by a wave

Energy travels with the shape. Per metre, a piece of string has kinetic energy $\frac{1}{2}\mu (\partial y/\partial t{)}^2$. It also has potential energy $\frac{1}{2}F(\partial y/\partial x{)}^2$, because a slanted piece is longer than $dx$ by about $\frac{1}{2}(\partial y/\partial x{)}^{2}dx$ and the tension did work stretching it. For a travelling wave, $\partial y/\partial t=-v\partial y/\partial x$ and $F=\mu v^2$ make the two equal at every point.

For a sine the kinetic energy per metre is $\frac{1}{2}\mu {\omega }^2{A}^2{\cos }^2(kx-\omega t)$, which averages to $\frac{1}{4}\mu {\omega }^2{A}^2$ over a cycle. Doubling it for the equal potential energy, the total is $\frac{1}{2}\mu {\omega }^2{A}^2$ per metre. This energy moves along at $v$, so the mean power passing any point is
$\bar{P}=\frac{1}{2}\mu {\omega }^2{A}^{2}v$

It grows as $A^2$ and as ${\omega }^2$. Where two ropes meet, a jump in the product $\mu v$ sends part of the wave back. For the ropes of the intermediate example $\mu v$ doubles, from 1.2 to 2.4 kg/s; the reflected amplitude is $(1.2-2.4)/(1.2+2.4)=-1/3$ of the incoming one, so a third of the amplitude returns upside down, and since power goes as amplitude squared, a ninth of the power is reflected.

Worked problem: powering the rope

Suppose the rope of the animation, with $A=0.3\mathrm{m}$, $\omega =\pi rad/s$ and $v=1m/s$, has $\mu =0.2kg/m$ and lies on a smooth floor, shaken sideways so that gravity plays no part. What tension does it need, what power must the hand supply, and what changes if the hand shakes twice as often?

The tension follows from $v=\sqrt{F/\mu }$, and the power from the formula above:
$F=\mu v^2=0.2\cdot 1^2=0.2\mathrm{N}$
$\bar{P}=\frac{1}{2}\mu {\omega }^2{A}^{2}v=\frac{1}{2}\cdot 0.2\cdot {\pi }^2\cdot 0.3^2\cdot 1\approx 0.089\mathrm{W}$

Check the power by counting energy. The rope holds $\frac{1}{2}\mu {\omega }^2{A}^2\approx 0.089\mathrm{J}$ per metre, so one wavelength, 2 m of rope, holds about 0.18 J. In each 2 s period the hand delivers $0.089\cdot 2\approx 0.18\mathrm{J}$, exactly that amount: every period, the hand fills one more wavelength of rope with energy.

Now shake at 1 Hz with the same amplitude. The speed is set by $F$ and $\mu$ alone, so it is still 1 m/s, and the wavelength becomes $\lambda =v/f=1\mathrm{m}$. The power grows with ${\omega }^2$, four times, to about 0.36 W. To keep 89 mW, $\omega A$ must stay the same, so the amplitude must halve to 0.15 m. The largest slope $kA$ then stays at 0.94, and so does the bead's top speed $\omega A$. In a given medium the power is set by the peak speed of the moving points, $\omega A$, not by the amplitude alone.

Limits of validity

Three assumptions were made along the way: small slopes, a smooth continuous medium and a speed that is the same for all frequencies.

Small slopes. Every step used $\sin \theta \approx \tan \theta$ and $\cos \theta \approx 1$. At a slope of 0.1 the neglected terms are about half a per cent; at the animation's 0.94, drawn large to be visible, they would be tens of per cent. A real guitar string with a 1 mm amplitude and a 1.3 m wavelength has $kA\approx 0.005$. For sound, the same condition is $\delta P\ll P_0$: even a painful 120 dB sound has a pressure amplitude of only about 28 Pa, $3\times 10^{-4}$ of atmospheric pressure. In a blast the condition fails: the compressed, heated parts of the wave travel faster and outrun the parts ahead of them, and the front steepens into a shock wave.

A continuous medium. Treating the gas as smooth needs wavelengths far longer than the mean free path, the average distance a molecule travels between collisions, about 70 nm in air. Audible sound, with wavelengths from 1.7 cm to 17 m, is safe, but ultrasound near 1 GHz, with $\lambda \approx 0.34\mu \mathrm{m}$, is absorbed within a short distance.

One speed for all frequencies. A shape keeps only if $v$ does not depend on frequency. Deep-water waves have $v=\sqrt{g\lambda /(2\pi )}$, 1.8 m/s at a wavelength of 2 m and 3.5 m/s at 8 m, and light in glass is slower for blue than for red. Such waves are called dispersive: a pulse spreads out as it goes, and its energy moves at the group velocity $d\omega /dk$, half the crest speed for deep water.

Where this leads

Sound and the Doppler Effect follows moving sources and observers; Loudness, Intensity and Decibels turns $\bar{P}\propto A^2$ into decibels; Standing Waves and Interference and Superposition apply linearity; Reflection and Refraction, and Prisms, Dispersion and Colour, deal with speeds that change between media or with frequency. Electromagnetic Waves derives the speed of light from the same equation. In Wave–Particle Duality the frequency becomes proportional to energy and the wave number to momentum, and Schrödinger's Equation and Tunnelling describes waves that are dispersive by nature.