Travelling and Standing Waves
Pluck a guitar string and something strange happens. The disturbance you make does not fly off the
end of the string like a thrown ball — it sits there, humming, the string bulging up and down in a
fixed shape while the whole room fills with a single steady note. Yet the very same string, tapped at
one end, will send a visible pulse racing to the other end. One string, two utterly
different behaviours: a wave that travels, and a wave that stands
still. This page is about both, and — the punchline worth holding onto from the start —
the standing one is secretly built out of two travelling ones running in opposite
directions.
We will write down the tidy piece of mathematics that describes a travelling wave, learn to read
every symbol in it, and use it to compute real numbers. Then we will add two such waves nose to nose
and watch a standing wave fall out of the algebra. Finally we will see why a string clamped at both
ends can only sing certain notes — the harmonics — and why that is the reason a
guitar sounds like a guitar.
The sinusoidal travelling wave
The simplest travelling wave is a single pure sine shape sliding along the x
axis without changing form. If y(x,t) is the displacement of the medium at
position x and time t, that wave is
y(x,t) = A\,\sin(kx - \omega t).
Four symbols carry all the meaning, and each one deserves a moment:
-
Amplitude A — the maximum displacement from rest, how
"tall" the wave is. Louder sound, bigger ripple: larger A.
-
Wavenumber k — how much phase the wave packs into each
metre of space. It is set by the wavelength \lambda (the distance
between two crests) through
k = \dfrac{2\pi}{\lambda}, measured in radians per metre. Short
wavelength means large k: many wiggles crammed into a short distance.
-
Angular frequency \omega — how much phase the wave
sweeps through each second. It is set by the frequency f (oscillations
per second) through \omega = 2\pi f, in radians per second.
-
Phase \varphi = kx - \omega t — the angle fed to the
sine. Everything that matters about "where in its cycle" a point is lives here.
Why does the combination kx - \omega t make the shape move? Pick
out one crest — a point of fixed phase, say kx - \omega t = \tfrac{\pi}{2}.
As time t ticks up, x must grow to keep the
phase constant, so that crest slides steadily towards larger x: the wave
moves in the +x direction. Flip the sign to
kx + \omega t and the same argument sends the crest the other way, towards
−x. Sign of the \omega t term
= direction of travel. Remember that; the standing wave depends on it.
Watch it move
Drag the Time slider and the whole shape marches to the right, exactly as the
constant-phase argument promised — but its form never changes. Push the
Wavenumber slider up and the crests crowd together (shorter wavelength); push
Amplitude and it grows taller. This one curve is the entire behaviour of an ideal
travelling wave in a single picture.
The relations that tie it together
A crest of fixed phase advances at the wave speed v.
Setting kx - \omega t = \text{const} and differentiating gives
k\,\dfrac{dx}{dt} = \omega, so
-
Speed.
v = \dfrac{\omega}{k} = \dfrac{2\pi f}{2\pi/\lambda} = f\lambda.
The wave advances one wavelength every period.
-
Period.
T = \dfrac{2\pi}{\omega} = \dfrac{1}{f} — the time for one full
oscillation at a fixed point.
-
Space and time echo each other.
k = \dfrac{2\pi}{\lambda} counts radians per metre;
\omega = \dfrac{2\pi}{T} counts radians per second. Same idea, one for
space, one for time.
Example 1 — from wavelength and frequency to everything. A wave on a rope has
wavelength \lambda = 0.50\ \text{m} and frequency
f = 10\ \text{Hz}. Then
k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.50} \approx 12.6\ \text{rad/m}, \qquad
\omega = 2\pi f = 2\pi(10) \approx 62.8\ \text{rad/s},
v = f\lambda = (10)(0.50) = 5.0\ \text{m/s}, \qquad
T = \frac{1}{f} = 0.10\ \text{s}.
Example 2 — reading a formula. A wave is written as
y = (0.02\ \text{m})\sin\!\big(4x - 200t\big) in SI units. Match term by
term: A = 0.02\ \text{m}, k = 4\ \text{rad/m},
\omega = 200\ \text{rad/s}. Hence
\lambda = \frac{2\pi}{k} = \frac{2\pi}{4} \approx 1.57\ \text{m}, \qquad
f = \frac{\omega}{2\pi} = \frac{200}{2\pi} \approx 31.8\ \text{Hz}, \qquad
v = \frac{\omega}{k} = \frac{200}{4} = 50\ \text{m/s}.
Notice you never needed the wavelength to get the speed — v = \omega/k
reads it straight off the two coefficients.
They are cousins, not twins, and mixing them up is the classic first-week slip.
Frequency lives in time: how many oscillations pass a fixed point each
second (f, in Hz), or in radians, \omega = 2\pi f.
Wavenumber lives in space: how many radians of the wave fit into one metre
(k = 2\pi/\lambda, in rad/m). One answers "how fast does it wiggle in
time?", the other "how fast does it wiggle in space?". They are only linked through the medium, by
v = \omega/k. A useful sanity check: \omega has
units of 1/second, k has units of 1/metre, and their ratio is a speed.
If your answer for k comes out in hertz, something has gone wrong.
Two waves collide: the standing wave
Now send two identical waves at each other — same amplitude, same
k, same \omega — one going right, one going
left. This is exactly what happens when a wave on a string reflects off a fixed end and runs back
into the wave still arriving. By the principle of superposition, the total
displacement is just their sum:
y(x,t) = A\,\sin(kx - \omega t) + A\,\sin(kx + \omega t).
Use the identity \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 averages come out clean: \tfrac{P+Q}{2} = kx and
\tfrac{P-Q}{2} = -\omega t. Since cosine is even,
y(x,t) = 2A\,\sin(kx)\,\cos(\omega t).
Stare at what just happened. The x and the t
have separated. There is no longer any kx - \omega t
travelling combination — nothing moves along. Instead the wave has a fixed spatial shape,
\sin(kx), whose height is simply scaled up and down in time by the factor
\cos(\omega t). Every point oscillates in place. This is a
standing wave.
-
Nodes — points that never move, where
\sin(kx) = 0, i.e. kx = n\pi, so
x = \dfrac{n\pi}{k} = \dfrac{n\lambda}{2} for
n = 0, 1, 2, \dots Nodes sit half a wavelength apart.
-
Antinodes — points of maximum swing, where
|\sin(kx)| = 1, i.e. kx = \big(n+\tfrac12\big)\pi.
They fall exactly halfway between the nodes.
In the picture below, drag Time. The whole curve breathes up and down between the
two dashed envelope curves \pm 2A\sin(kx) — but the points
where it crosses zero, the nodes, never budge. That is the signature of a standing wave: fixed nodes,
pulsing antinodes.
A standing wave carries no net energy along the string — it does not "go" anywhere.
This is the misconception to stamp out. Because the pattern has no kx - \omega t
travelling term, there is no forward flow: each little segment of string just sloshes energy back and
forth with its neighbours, endlessly trading kinetic for potential, but on average zero
power crosses any point. (The two travelling waves it is built from do each carry energy —
one to the right, one to the left — and those flows exactly cancel.)
A second trap: it is tempting to think a node is "where the wave is weak" and an antinode "where it is
strong", as if the wave were fading. Not so. A node is a point that is held still by
destructive interference at all times, and an antinode is where the two waves always reinforce.
Both are permanent features of the pattern, not a wave running down. Don't picture a standing wave as
a travelling wave that got tired — picture it as two travelling waves locked in a standoff.
A string fixed at both ends: the normal modes
Clamp a string of length L at both ends — a guitar, a violin, a piano
wire. The ends can never move, so they are forced to be nodes. That single
boundary condition is enormously restrictive: only standing waves whose shape
\sin(kx) is zero at both x = 0 and
x = L can survive. The first is automatic; the second demands
\sin(kL) = 0, i.e. kL = n\pi for a whole number
n = 1, 2, 3, \dots
Translating k = n\pi/L back into wavelength and frequency (using
k = 2\pi/\lambda and f = v/\lambda) gives the
allowed normal modes:
-
Wavelengths. \lambda_n = \dfrac{2L}{n}. The string
length holds a whole number of half-wavelengths.
-
Frequencies. f_n = \dfrac{v}{\lambda_n} = \dfrac{n\,v}{2L} = n f_1.
-
Fundamental and harmonics. n = 1 is the
fundamental f_1 = \dfrac{v}{2L} (one antinode, a
single hump). n = 2, 3, \dots are the harmonics, whose
frequencies are exact integer multiples of the fundamental.
Step through the modes with the selector. Mode n has exactly
n humps (antinodes) and n-1 interior nodes,
with both clamped ends forced to zero. These are the only pure shapes the string will hold; every
real pluck is a mixture of them.
Example 3 — tuning a guitar string. A guitar's high E string has vibrating length
L = 0.65\ \text{m} and the wave speed along it is
v = 429\ \text{m/s}. Its fundamental is
f_1 = \frac{v}{2L} = \frac{429}{2(0.65)} = \frac{429}{1.30} \approx 330\ \text{Hz},
which is indeed the pitch of E4. The harmonics ring at
f_2 \approx 660\ \text{Hz},
f_3 \approx 990\ \text{Hz}, and so on — the integer stack that gives the
string its rich, musical timbre. Press the string down at a fret to shorten
L, and every f_n rises: shorter string, higher
note.
When you pluck a string you do not excite one clean mode — you jerk it into a triangular kink, which
is a superposition of the fundamental and a whole ladder of harmonics, each with its own
amplitude. The ear hears the fundamental f_1 as the pitch, but the
blend of harmonics above it — strong second, weak third, and so on — is what your brain reads as
timbre: the reason a plucked guitar, a bowed violin and a struck piano playing the
same note sound so different. It is all the same physics on this page: each instrument is just a
different recipe of standing-wave modes stacked on top of one another. Fourier's great insight was
that any shape the string can take is some sum of these sine modes — so the humble
\sin(n\pi x/L) family is a complete alphabet for everything a string can
do.