Superposition & Standing Waves

A guitar string is about 0.65 m long and you can pull it into any shape you like before letting go. Yet when you let go it sings one note, the same note every time, and the note only changes if you change the string's length or how hard it is stretched. A flute is a tube of air with holes in it, and it does the same trick: one length of air column, one pitch. Nothing about a plucked string or a blown pipe says in advance which frequency it should pick.

So where does the note come from? It comes from the ends. In the previous lesson a wave ran along the string forever and never met anything. Put two ends on the string and the wave comes back. What arrives at any point is now two waves at once, one going each way, and almost every frequency dies from that interference. A few survive. Those are the notes.

Two waves on one string

Start with the rule that makes all of this work. Superposition: when two waves overlap, the displacement at a point is the sum of what each wave would have given on its own. Waves do not collide or push each other aside. They add, pass through, and come out the other side unchanged.

Take the wave from the last lesson, $y = A\sin(kx - \omega t)$, moving in the $+x$ direction with amplitude $A$, wave number $k = 2\pi/\lambda$ and angular frequency $\omega = 2\pi f$. Its reflection off a far end is an identical wave moving the other way, $y = A\sin(kx + \omega t)$. Add them:

Use $\sin(P) + \sin(Q) = 2\sin\!\big(\tfrac{P+Q}{2}\big)\cos\!\big(\tfrac{P-Q}{2}\big)$, with $P = kx - \omega t$ and $Q = kx + \omega t$. The half-sum is $kx$ and the half-difference is $-\omega t$, and cosine does not care about that sign. So

Look hard at what just happened. In the original waves, $x$ and $t$ were locked together inside one bracket, which is what made the shape travel. Now they sit in separate factors. Nothing moves along the string any more. Every point oscillates in time as $\cos(\omega t)$, all in step, and the only thing that changes from place to place is the size of the swing, $2A\sin(kx)$. That is a standing wave, and the factorising is the whole of it.

Nodes, antinodes, and the half-wavelength rule

The amplitude factor $2A\sin(kx)$ does two special things.

Where $\sin(kx) = 0$, the string never moves at all, no matter what $t$ is. Those points are nodes. They happen when $kx = 0, \pi, 2\pi, \dots$, that is at $x = 0, \lambda/2, \lambda, \dots$, since $k = 2\pi/\lambda$. So adjacent nodes are half a wavelength apart. Not a whole one. This single fact does more work in this topic than any formula.

Halfway between two nodes, $\sin(kx) = \pm 1$ and the point swings with amplitude $2A$, twice what either travelling wave carried alone. Those are antinodes. A node and its neighbouring antinode are a quarter wavelength apart.

A string fixed at both ends

Now add the ends. A string clamped at $x = 0$ and $x = L$ cannot move at either clamp, so both ends have to be nodes. The $x = 0$ end is already a node in $2A\sin(kx)$. The other end costs you something: you need $\sin(kL) = 0$, so $kL = n\pi$ for some whole number $n = 1, 2, 3, \dots$

Put $k = 2\pi/\lambda$ back in and that says $2\pi L/\lambda = n\pi$, or

In words: the string holds a whole number of half-wavelengths. Every other wavelength is forbidden, because its reflection comes back out of step with itself and cancels. Solve for the wavelength, $\lambda_n = 2L/n$, and feed it through $f = v/\lambda$:

These allowed frequencies are the harmonics. The lowest, $n = 1$, is the fundamental $f_1 = v/2L$, and every other one is a whole-number multiple of it: $f_2 = 2f_1$, $f_3 = 3f_1$, and so on. The $n$ also counts things you can see. Mode $n$ has $n$ loops (each loop is one half-wavelength), $n - 1$ nodes in the middle, and $n + 1$ nodes if you count the two clamped ends.

One piece is still missing: what sets $v$? For a string it is the tension $T$ pulling it taut and the linear mass density $\mu$, the mass per metre:

Tighter means faster means higher. Heavier per metre means slower means lower, which is why the low strings on a guitar are the thick ones. Combining,

Pipes, and the harmonics a stopped pipe cannot play

Air in a pipe does the same thing, with one change in the boundary conditions. A sound wave is a longitudinal wave, so "displacement" means how far a slab of air slides back and forth along the pipe.

A closed end is a solid wall. Air there cannot slide, so a closed end is a displacement node. An open end is free to breathe in and out against the open air, so it is a displacement antinode. That is the whole rule.

A pipe open at both ends needs an antinode at each end. Antinode to antinode, like node to node, is half a wavelength, so the condition is identical to the string's: $L = n\lambda/2$, and

with $v$ now the speed of sound, about 343 m/s in room-temperature air.

A pipe closed at one end (a stopped pipe) needs a node at the closed end and an antinode at the open one. The shortest fit is a quarter wavelength. Add a half wavelength and it still fits; add another and it still fits. So the pipe holds a quarter, or three quarters, or five quarters, never a half:

Two consequences, both of which problems test. First, the fundamental of a stopped pipe is $v/4L$, half that of an open pipe the same length: stopping one end drops the pitch an octave. Second, only the odd harmonics exist. After $f_1$ comes $3f_1$, then $5f_1$. There is no $2f_1$, because that mode would need a node at the open end.

There is one more boundary pair worth naming, because it turns up on strings too. Clamp a string at one end and tie the other to a light ring that slides freely on a frictionless rod. The free end cannot be held down, so it is a displacement antinode. Node at one end, antinode at the other: exactly the stopped-pipe geometry. So $f_n = (2n-1)v/4L$ there as well, odd harmonics only. Whenever the two ends are different in kind, you get quarter wavelengths and odd harmonics. Whenever they match, you get half wavelengths and all harmonics.

Three worked examples

A violin string. The vibrating length is $L = 0.65$ m and transverse waves travel along it at $v = 520$ m/s. Fixed at both ends, so $f_1 = v/2L$. The denominator is $2(0.65) = 1.30$ m, and $520/1.30 = 400$ Hz. Its third harmonic is just $3 \times 400 = 1200$ Hz, because a both-ends-fixed string keeps every multiple.

A pipe tuning a string. A pipe 0.32 m long, closed at one end, is sounded at its fundamental with $v_{\text{sound}} = 343$ m/s. A string 0.40 m long with $\mu = 2.5\times10^{-3}$ kg/m, fixed at both ends, resonates in its fundamental with that tone. Find the string's tension.

Nothing here can be plugged in directly, so set up first. The only thing the two objects share is a frequency, so get it from the pipe. Stopped pipe, fundamental, so $f = v/4L = 343/(4 \times 0.32) = 343/1.28 = 268.0$ Hz.

Now go into the string with that number. Its fundamental is $f = v/2L$, so the wave speed on the string must be

That is the speed of a wave on a wire, not the speed of sound; they are different numbers for the same note, and mixing them up is the main way this problem is lost. Finally $v = \sqrt{T/\mu}$ rearranges to $T = \mu v^2$:

Counting resonances as water drains. A tube 1.00 m long stands vertically, full of water. A 440 Hz fork sounds over the open top while the water drains out. Sound travels at 343 m/s. How many times do you hear it resonate?

The frequency is fixed by the fork, so the wavelength is fixed too: $\lambda = v/f = 343/440 = 0.7795$ m. What changes is the length of air column above the water. The water surface is a closed end (a node), the top is open (an antinode), so the column resonates when

Work through the list: $n = 1$ gives 0.195 m, $n = 2$ gives 0.585 m, $n = 3$ gives 0.974 m, and $n = 4$ would need 1.364 m, which is longer than the tube. The first three all fit inside 1.00 m, so you hear three resonances.

Notice the spacing of those positions: 0.390 m each time, exactly $\lambda/2$. That is a general result and a fast way to measure the speed of sound in a lab: find two neighbouring resonance lengths, double the gap to get $\lambda$, and multiply by the fork's frequency.

How to recognise one of these

The giveaway word is resonates, but the real signal is a length with two ends: a string between bridges, an air column above water, a pipe, a wire on a sonometer. Whenever a problem fixes a length and then asks for a frequency, or fixes a frequency and asks for a length, it is a standing-wave problem and the recipe never changes.

1. Decide what each end is. Fixed, clamped, closed, in water (that is a node. Open, free, sliding on a frictionless rod) that is an antinode.
2. Same at both ends, use $L = n\lambda/2$ and $f_n = nv/2L$, all harmonics. Different at the two ends, use $L = (2n-1)\lambda/4$ and $f_n = (2n-1)v/4L$, odd harmonics only.
3. Get $v$ from the right place. On a string, $v = \sqrt{T/\mu}$. In air, it is the speed of sound, and if a temperature is quoted, $v$ scales as $\sqrt{T_{\text{kelvin}}}$, so a warm pipe sounds sharp.
4. Read what mode you are in. "Second harmonic", "nodes 0.20 m apart", "the next resonance above": each of these pins down $n$ before you solve.

Where students actually lose the marks is step 1 and step 3: calling a stopped pipe an open one, or putting the speed of sound into a string formula. Write down the two boundary conditions before you write anything else.

So far the two waves you added had the same frequency. What if they do not quite? Then the interference pattern cannot stand still, and what you hear is a throb: that is beats and interference.