Wave Basics

Something moves, but nothing goes anywhere

Drop a stone in a pond and a ring of ripples spreads out. A leaf floating two metres away bobs up and down as the ripple passes, and then sits exactly where it was. The disturbance travelled. The water did not.

That is the whole trick of a wave, and it is why waves get their own set of tools. In the thermodynamics unit you tracked energy moving as heat and work through a gas, and every joule went with a lump of matter or a moving piston. A wave carries energy from here to there while the medium stays put. So the question "how fast does it go?" has a new kind of answer: not "as fast as you threw it", but "as fast as this medium allows".

A transverse wave moves the medium perpendicular to the direction of travel: a rope you flick, the S wave from an earthquake. A longitudinal wave moves the medium back and forth along the direction of travel: sound, the earthquake's faster P wave. Almost everything below applies to both. The pictures are easier to draw for a rope, so the rope is where you start.

Reading a snapshot

Take a photograph of a long rope while a wave runs along it. You get a curve: height $y$ against position $x$. Two numbers describe it.

The amplitude $A$ is the height of a crest above the resting line, in metres. Not crest-to-trough: half of that. The wavelength $\lambda$ is the distance from one crest to the next crest, in metres, which is the same as the distance over which the pattern repeats.

Now stop photographing and instead watch one single speck of the rope. It goes up, comes back, goes down, comes back. The time for one full round trip is the period $T$, in seconds, and the number of round trips per second is the frequency $f = 1/T$, in hertz. A 512 Hz tuning fork makes 512 round trips a second, so one vibration takes $1/512 = 0.001953$ s.

These two pictures lock together. In one period, the pattern slides forward by exactly one wavelength: a crest arrives where the previous crest was. Speed is distance over time, so

$v = \frac{\lambda}{T} = f\lambda$

Ripples at 4.0 Hz with crests 0.25 m apart travel at $4.0 \times 0.25 = 1.0$ m/s. A singer holding 686 Hz in air, where sound goes 343 m/s, makes a wavelength of $343/686 = 0.50$ m. The relation is not a law of nature so much as bookkeeping, but it is bookkeeping you will use in half the problems in this unit.

Writing the wave down

You want one formula that gives the height of the rope at any place and any time. Start with the snapshot at $t = 0$: a sine curve that repeats every $\lambda$ in $x$, which is $y = A\sin(kx)$ where $k = 2\pi/\lambda$. The constant $k$ is called the wave number, in radians per metre; it exists because sine repeats every $2\pi$ radians, and you need $k\lambda = 2\pi$ so the curve repeats every wavelength.

Now let it move. A shape moving right at speed $v$ is the same shape with $x$ replaced by $x - vt$, since the point of the curve that was at $x = 0$ is at $x = vt$ a time $t$ later. Substituting gives

$y(x,t) = A\sin\big(k(x - vt)\big) = A\sin(kx - \omega t)$

where $\omega = kv$ is the angular frequency in radians per second. Since $v = f\lambda$ and $k = 2\pi/\lambda$, you get $\omega = 2\pi f$. A minus sign between the terms means the wave moves in the $+x$ direction; a plus sign means it moves in $-x$.

So from any wave written this way you can read off everything by pattern matching. For $y = 0.040\sin(5.0x - 200t)$ in SI units: $A = 0.040$ m, which is 4.0 cm; $k = 5.0$ rad/m, so $\lambda = 2\pi/5.0 = 1.257$ m; $\omega = 200$ rad/s, so $f = 200/2\pi = 31.83$ Hz; and $v = \omega/k = 200/5.0 = 40.0$ m/s.

Worked example 1: how fast the rope moves

A rope carries $y(x,t) = 0.030\sin(2.0x - 8.0t)$ in SI units. Find the transverse velocity of the element at $x = 0.50$ m at $t = 0.10$ s.

The wave moves at $\omega/k = 8.0/2.0 = 4.0$ m/s, but that is not what is asked. A rope element does not move sideways at all. Its velocity is the rate of change of $y$ with the element held at fixed $x$, so differentiate with respect to $t$ treating $x$ as a constant:

$v_y = \frac{\partial y}{\partial t} = -\omega A\cos(kx - \omega t)$

Put the numbers in. The phase is $kx - \omega t = 2.0(0.50) - 8.0(0.10) = 1.00 - 0.80 = 0.20$ rad. Then $\omega A = 8.0 \times 0.030 = 0.240$ m/s, and $\cos(0.20) = 0.9801$, so

$v_y = -0.240 \times 0.9801 = -0.235 \text{ m/s}$

Negative: that bit of rope is heading downward at this instant. The largest $v_y$ can ever get is $\omega A$, when the cosine hits 1, which happens as the element passes through $y = 0$. Differentiate once more and you get $a_y = -\omega^2 A \sin(kx - \omega t)$, with maximum size $\omega^2 A$, reached at the crests and troughs. For the wave $0.040\sin(5.0x - 200t)$ that is a maximum speed of $200 \times 0.040 = 8.0$ m/s and a maximum acceleration of $200^2 \times 0.040 = 1600$ m/s².

What sets the speed

Flick a rope harder and the pulse is taller, not faster. So what does control $v$?

Two things fight over a bit of rope. Tension $T$, in newtons, pulls the displaced bit back toward the line (more pull, faster return, faster wave. Mass resists that pull) a heavier rope responds sluggishly. The relevant mass is the mass per unit length, called the linear density $\mu = m/L$ in kilograms per metre. A dimensional check settles the combination: $T/\mu$ has units of N/(kg/m) $=$ m²/s², so its square root is a speed, and the full derivation from Newton's second law on a small curved segment gives exactly

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

A cord with $\mu = 0.010$ kg/m at 90 N carries waves at $\sqrt{90/0.010} = \sqrt{9000} = 94.87$ m/s. A guitar string of mass 2.5 g and length 1.25 m has $\mu = 0.0025/1.25 = 0.0020$ kg/m, so at 50 N it runs at $\sqrt{50/0.0020} = \sqrt{25000} = 158.1$ m/s.

The same logic shows up everywhere. In a solid rod, stiffness plays the role of tension and density plays the role of $\mu$: $v = \sqrt{Y/\rho}$ with $Y$ the Young's modulus. Steel with $Y = 2.0\times10^{11}$ Pa and $\rho = 7800$ kg/m³ gives $\sqrt{2.0\times10^{11}/7800} = 5064$ m/s, which is why a rail hums with the arriving train long before you hear it through the air.

Worked example 2: a pulse across a splice

A light rope 3.0 m long with $\mu_1 = 0.0020$ kg/m is spliced to a heavy rope 4.0 m long with $\mu_2 = 0.0080$ kg/m. The whole thing is pulled taut at 32 N. How long does a pulse take to go from the light end to the far end?

The thing to see first: tension is the same throughout, because the splice transmits whatever pull is applied. Density is not. So there are two speeds and you time each stretch separately.

$v_1 = \sqrt{\frac{32}{0.0020}} = \sqrt{16000} = 126.5 \text{ m/s}$

$v_2 = \sqrt{\frac{32}{0.0080}} = \sqrt{4000} = 63.25 \text{ m/s}$

Then $t_1 = 3.0/126.5 = 0.02372$ s and $t_2 = 4.0/63.25 = 0.06325$ s, for a total of $0.02372 + 0.06325 = 0.0870$ s. Averaging the two speeds and dividing once would have given 0.0737 s, which is wrong: time adds, speed does not.

Worked example 3: a rope that hangs

A uniform rope of length $L = 4.0$ m hangs from a ceiling with nothing on its lower end. How long does a pulse take to climb from the bottom to the top?

Here the tension is not constant. At a height $x$ above the bottom end, the rope below that point has mass $\mu x$ and hangs from it, so $T(x) = \mu x g$. The speed there is

$v(x) = \sqrt{\frac{\mu x g}{\mu}} = \sqrt{gx}$

The $\mu$ cancels, so the answer will not depend on how heavy the rope is. The speed is zero at the loose bottom end and grows as the pulse climbs. A changing speed means you cannot use distance over speed; you add up the little times $dt = dx/v(x)$:

$t = \int_0^L \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}}\Big[2\sqrt{x}\Big]_0^L = 2\sqrt{\frac{L}{g}}$

With $L = 4.0$ m and $g = 9.8$ m/s², $t = 2\sqrt{4.0/9.8} = 2\sqrt{0.4082} = 2(0.6389) = 1.278$ s. Worth noticing that this is twice the free-fall time from height $L$: a fact you can quote to check your algebra.

Energy, power, intensity

A wave carries energy because every element of the rope is oscillating like a little mass on a spring. An element of length $dx$ has mass $\mu\,dx$ and maximum speed $\omega A$, so its total energy is $\frac{1}{2}(\mu\,dx)(\omega A)^2$. Add up a whole wavelength:

$E_\lambda = \tfrac{1}{2}\mu\omega^2 A^2\lambda$

For $\mu = 0.020$ kg/m, $A = 0.010$ m, $\omega = 100$ rad/s and $\lambda = 0.40$ m: $\frac{1}{2}(0.020)(10000)(0.0001)(0.40) = 0.0040$ J.

One wavelength passes a point every period, so the average power is $E_\lambda/T = \frac{1}{2}\mu\omega^2A^2 (\lambda/T)$, that is

$P = \tfrac{1}{2}\mu\omega^2 A^2 v$

Take a string with $\mu = 0.0500$ kg/m at 80 N, shaken at 60 Hz with $A = 0.020$ m. First $v = \sqrt{80/0.0500} = \sqrt{1600} = 40.0$ m/s and $\omega = 2\pi(60) = 377.0$ rad/s. Then $P = 0.5(0.0500)(377.0^2)(0.020^2)(40.0)$. Step by step: $377.0^2 = 142{,}100$; $0.5 \times 0.0500 = 0.0250$; $0.0250 \times 142{,}100 = 3553$; $\times 0.000400 = 1.421$; $\times 40.0 = 56.9$ W.

The $A^2$ is the part to hold on to. Tripling the amplitude at fixed frequency on the same string multiplies the power by nine, so a shaker delivering 15 W would need $9 \times 15 = 135$ W.

For sound spreading into open air from a small source, the useful quantity is intensity $I$: power per unit area, in W/m². A point source sending power $P$ equally in all directions spreads it over a sphere of area $4\pi r^2$, so

$I = \frac{P}{4\pi r^2}$

Intensity falls with the square of distance. An alarm giving $I = 2.0\times10^{-5}$ W/m² at 15 m radiates $P = 4\pi(15)^2(2.0\times10^{-5}) = 4\pi(225)(2.0\times10^{-5}) = 0.0566$ W. Move from 2.0 m to 10.0 m, five times further, and intensity drops by $5^2 = 25$: $8.0\times10^{-3}/25 = 3.2\times10^{-4}$ W/m².

Ears cover an enormous range of intensity, so sound is quoted on a logarithmic scale. The sound level in decibels is

$\beta = 10\log_{10}\!\left(\frac{I}{I_0}\right), \qquad I_0 = 1.0\times10^{-12}\ \text{W/m}^2$

A vacuum cleaner at $I = 1.0\times10^{-6}$ W/m² sits at $10\log_{10}(10^{6}) = 60.0$ dB. Running it backwards, city traffic at 85 dB has $I = I_0 10^{8.5} = 10^{-12} \times 3.162\times10^{8} = 3.16\times10^{-4}$ W/m².

Worked example 4: where the siren fades

A siren radiates 0.80 W uniformly over flat open ground. At what distance is the level 60 dB?

Go from decibels to intensity first, then from intensity to distance. 60 dB means $I/I_0 = 10^{6}$, so $I = 10^{-6}$ W/m². Then

$r = \sqrt{\frac{P}{4\pi I}} = \sqrt{\frac{0.80}{4\pi \times 10^{-6}}} = \sqrt{\frac{0.80}{1.2566\times10^{-5}}} = \sqrt{63{,}662} = 252 \text{ m}$

Two shortcuts save time on problems like this. Ten identical machines have ten times the intensity of one, and $10\log_{10}10 = 10$, so ten machines at 70 dB each give 80.0 dB, not 700. And every factor of ten in distance costs 20 dB, since intensity drops by 100. A jackhammer at 100 dB from 1.0 m is at 80 dB at 10 m and 60 dB at 100 m.

Two effects of relative motion

If the source moves toward you, it chases its own wavefronts and packs them closer, shortening the wavelength in the air. The speed of sound is unchanged, so the frequency you receive goes up. That is the Doppler effect:

$f' = f\,\frac{v \pm v_o}{v \mp v_s}$

Rather than memorising the signs, fix them by the direction. An 800 Hz siren approaching at 30 m/s, with sound at 340 m/s, must sound higher, so the denominator has to shrink: $f' = 800 \times 340/(340 - 30) = 800 \times 1.0968 = 877$ Hz.

Second effect: two tones of nearly equal frequency sounding together drift in and out of step, so the loudness pulses at the difference frequency. Forks at 440 Hz and 444 Hz beat at $|444 - 440| = 4$ Hz, four surges a second. That is how a piano is tuned by ear.

How to recognise which tool a problem wants

Sort by what the problem hands you.

Given two of speed, frequency and wavelength, and asking for the third, it is $v = f\lambda$ and nothing more. Given a rope, a tension and a mass per unit length, reach for $v = \sqrt{T/\mu}$, and if only a total mass and length are quoted, compute $\mu$ first. If the tension varies along the rope, as it does for anything hanging under its own weight, the speed varies too and you have to integrate $dx/v(x)$ instead of dividing.

Given an explicit $y(x,t)$, pattern match to $A\sin(kx - \omega t)$ and read the constants; if the question asks about a point on the rope rather than the wave, differentiate with respect to $t$. Any mention of watts, amplitude and frequency together points at $P = \frac{1}{2}\mu\omega^2A^2v$, and the giveaway is that changing the amplitude changes the answer at all. Distances from a source, W/m², or the word decibel mean $I = P/4\pi r^2$ and $\beta = 10\log(I/I_0)$. Anything moving relative to anything else is Doppler. Two close frequencies at once is beats.

One caution on the exam: a wave problem often runs two of these in series, as worked example 4 did. Convert decibels to intensity, intensity to distance. Convert mass and length to $\mu$, $\mu$ and $T$ to $v$, $v$ and $f$ to $\lambda$. Write down the chain before you touch the calculator.

Everything so far has followed one wave at a time. The interesting things start when two waves occupy the same rope at once and their displacements add, which is where superposition and standing waves picks up.