Waves

Yo ts is not done yet please edit and stuff

Wave definition

Definition (Wave). A wave is a disturbance that transports energy and momentum through a medium (or through space) without any net transport of matter. The particles of the medium oscillate about fixed equilibrium positions; it is the pattern of disturbance that travels.

In a transverse wave the medium oscillates perpendicular to the direction of propagation (a wave on a string, light).
In a longitudinal wave the medium oscillates along the direction of propagation (sound, compression waves in a spring).

Mechanical waves require a medium with two ingredients: something that provides a restoring force (tension, pressure, elasticity) and something that provides inertia (mass density). The wave speed is always set by the competition between these two.

The wave function and the wave equation

Consider a disturbance $y$ that depends on position $x$ and time $t$ . A pulse of any shape moving in the $+x$ direction at speed $v$ without changing shape must depend on $x$ and $t$ only through the combination $x - vt$ :

y(x,t) = f(x - vt).

A wave moving in the $-x$ direction is $f(x+vt)$ . A sinusoidal solution is

y(x,t) = A\sin(kx - \omega t + \phi),

where

k = \frac{2\pi}{\lambda} \quad(\text{wave number}), \qquad \omega = 2\pi f = \frac{2\pi}{T} \quad(\text{angular frequency}),

and the speed, wavelength, and frequency are linked by

v = \frac{\omega}{k} = \lambda f = \frac{\lambda}{T}.

Both forms satisfy the wave equation, the partial differential equation every nondispersive wave obeys:

Theorem (Wave equation). Every nondispersive wave $y(x,t)$ obeys

\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 y}{\partial t^2}.

USAPhO will not require you to solve these types of equations, but it is good to know the general solutions.

A slick way to see what those solutions are: factor the wave equation as a difference of squares,

\left(\partial_t^2 - v^2\partial_x^2\right)y = (\partial_t - v\partial_x)(\partial_t + v\partial_x)\,y = 0.

Anything killed by either factor solves the equation, and those are exactly the right- and left-movers $f(x-vt)$ and $g(x+vt)$ . Because the equation is linear, the general solution is their superposition,

y(x,t) = f(x-vt) + g(x+vt),

for arbitrary shapes $f$ and $g$ . Every wave problem on a string is ultimately about choosing $f$ and $g$ to match the initial conditions and boundaries.

Proof (Wave definition). Let $u = x - vt$ and $y = f(u)$ . By the chain rule,

\frac{\partial y}{\partial x} = f'(u), \qquad \frac{\partial y}{\partial t} = -v f'(u),

and differentiating again,

\frac{\partial^2 y}{\partial x^2} = f''(u), \qquad \frac{\partial^2 y}{\partial t^2} = v^2 f''(u).

Dividing, $\partial^2 y/\partial x^2 = (1/v^2)\,\partial^2 y/\partial t^2$ for any twice-differentiable shape $f$ . So the wave equation does not care about the shape of the pulse — only that it translates rigidly at speed $v$ .

A useful distinction: the wave speed $v$ is how fast the pattern moves, while the transverse velocity of a point on the string is

u_y = \frac{\partial y}{\partial t} = -\omega A\cos(kx - \omega t + \phi),

with maximum magnitude $\omega A$ .

Waves on a string

For a string with tension $T$ and linear mass density $\mu$ (mass per length), the wave speed is

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

This is the prototype of “restoring force over inertia”: more tension means a faster wave, more mass per length means a slower one.

When the tension varies along the string, so does the wave speed. The classic example is a rope hanging under its own weight: at a height $x$ above the bottom, the tension equals the weight of rope below it, $T = \mu g x$ , so

v(x) = \sqrt{\frac{T}{\mu}} = \sqrt{gx}.

A pulse therefore speeds up as it climbs. The time to travel the full length $L$ is

\tau = \int_0^L \frac{dx}{\sqrt{gx}} = 2\sqrt{\frac{L}{g}},

which is — pleasingly — twice the time an object would take to fall the length of the rope.

Proof (Wave velocity equation). Consider a small arc of string of length $\Delta x$ carrying a wave. The tension pulls tangentially at both ends; if the slope is small, the net upward (transverse) force is

F_y = T\sin\theta_2 - T\sin\theta_1 \approx T\left(\frac{\partial y}{\partial x}\Big|_{x+\Delta x} - \frac{\partial y}{\partial x}\Big|_{x}\right) \approx T\frac{\partial^2 y}{\partial x^2}\Delta x.

The mass of the arc is $\mu\,\Delta x$ and its transverse acceleration is $\partial^2 y/\partial t^2$ , so Newton’s second law gives

T\frac{\partial^2 y}{\partial x^2}\Delta x = \mu\,\Delta x\,\frac{\partial^2 y}{\partial t^2} \quad\Longrightarrow\quad \frac{\partial^2 y}{\partial x^2} = \frac{\mu}{T}\frac{\partial^2 y}{\partial t^2}.

Comparing with the wave equation, $v^2 = T/\mu$ .

Energy and power

A sinusoidal wave carries energy past a point at an average rate

P_{\text{avg}} = \tfrac12 \mu\, \omega^2 A^2 v.

The key scalings to remember: power (and intensity) go as the square of both amplitude and frequency. Doubling the frequency at fixed amplitude quadruples the power transmitted.

The energy is split between kinetic energy (transverse motion) and potential energy (stretching of the string against tension). A useful fact for any traveling wave $y = f(x - vt)$ : the kinetic and potential energy densities are equal at every point and every instant, because $\partial y/\partial t = -v\,\partial y/\partial x$ ties the two together. This equipartition fails for a standing wave, where energy sloshes back and forth between purely kinetic (string flat, moving fast) and purely potential (string maximally bent, momentarily at rest).

Superposition and interference

Theorem (Principle of superposition). When two or more waves overlap in a linear medium, the net disturbance is the sum of the individual disturbances:

y_{\text{net}} = y_1 + y_2 + \cdots

This is what makes interference possible. For two waves of equal amplitude and frequency differing in phase by $\delta$ ,

y = A\sin(kx-\omega t) + A\sin(kx-\omega t+\delta) = \underbrace{2A\cos\tfrac{\delta}{2}}_{\text{net amplitude}}\sin\!\left(kx-\omega t+\tfrac{\delta}{2}\right).

The interference is

Constructive (amplifies the wave) when $\delta = 0, 2\pi, 4\pi, \dots$ (amplitude $2A$ ),
Destructive (de-amplifies the wave) when $\delta = \pi, 3\pi, \dots$ (amplitude $0$ ).

For two sources, a phase difference usually arises from a path-length difference $\Delta r$ :

\delta = \frac{2\pi}{\lambda}\,\Delta r.

So constructive interference is $\Delta r = m\lambda$ and destructive is $\Delta r = (m+\tfrac12)\lambda$ for integer $m$ — the backbone of all interference problems (and the double-slit on the Optics page).

Standing waves and normal modes

Add two identical waves traveling in opposite directions:

y = A\sin(kx-\omega t) + A\sin(kx+\omega t) = 2A\sin(kx)\cos(\omega t).

The result does not travel — it is a standing wave. The space and time parts have separated: every point oscillates at the same frequency $\omega$ , but with a position-dependent amplitude $2A\sin(kx)$ .

Nodes (always at rest) occur where $\sin(kx)=0$ , i.e. $x = 0, \tfrac{\lambda}{2}, \lambda, \dots$ — spaced half a wavelength apart.
Antinodes (maximum swing) sit halfway between nodes.

Confining a wave between boundaries selects a discrete set of allowed wavelengths — the normal modes or harmonics.

String fixed at both ends (nodes at each end), length $L$ :

\lambda_n = \frac{2L}{n}, \qquad f_n = \frac{nv}{2L}, \qquad n = 1, 2, 3, \dots

The lowest mode ( $n=1$ ) is the fundamental mode, while the rest are higher harmonics, and here all integer harmonics are present.

In a pipe, a closed end forces a displacement node (pressure antinode); an open end is a displacement antinode (pressure node).

Open–open pipe: $\;f_n = \dfrac{nv}{2L},\quad n=1,2,3,\dots$ (all harmonics)
Open–closed pipe: $\;f_n = \dfrac{(2n-1)v}{4L},\quad n=1,2,3,\dots$ (odd harmonics only)

Example. A guitar string of length $L = 0.65\text{ m}$ and linear density $\mu = 5.0\times 10^{-3}\text{ kg/m}$ is tuned to a fundamental of $f_1 = 110\text{ Hz}$ . What tension is required?

The fundamental of a string fixed at both ends is $f_1 = v/2L$ , so the required wave speed is

v = 2Lf_1 = 2(0.65)(110) = 143\text{ m/s}.

Since $v = \sqrt{T/\mu}$ ,

T = \mu v^2 = (5.0\times 10^{-3})(143)^2 \approx 102\text{ N}.

Roughly $100\text{ N}$ — about the weight of a $10\text{ kg}$ mass, which is why guitar necks must be sturdy.

Reflection and boundary conditions

When a pulse reaches a boundary it is partly reflected and partly transmitted.

At a fixed end (string tied to a wall, or a denser medium), the reflected pulse is inverted — it picks up a phase shift of $\pi$ .
At a free end (or a lighter medium), the reflected pulse is upright — no phase shift.

More generally, for a wave going from a string of density $\mu_1$ to one of density $\mu_2$ (same tension, so wave speeds $v_1, v_2$ ), the amplitude reflection and transmission coefficients are

r = \frac{v_2 - v_1}{v_2 + v_1}, \qquad t = \frac{2v_2}{v_2 + v_1}.

When the second medium is denser ( $v_2 < v_1$ ), $r<0$ — the reflection is inverted, recovering the fixed-end rule as the limiting case $v_2 \to 0$ .

The cleaner way to package this is the impedance of the medium,

Z = \sqrt{\mu T} = \mu v,

in terms of which the reflection and transmission coefficients are

r = \frac{Z_1 - Z_2}{Z_1 + Z_2}, \qquad t = \frac{2Z_1}{Z_1 + Z_2}.

Reflection happens whenever the impedances mismatch, and the reflection vanishes ( $r=0$ , perfect transmission) when $Z_1 = Z_2$ even if the media are otherwise different. This is exactly the same idea as matching impedances on a transmission line, and it is why engineers insert gradual “impedance-matching” devices (a tapered horn, an anti-reflection coating) to suppress unwanted reflections by softening the discontinuity.

Sound waves

Sound is a longitudinal pressure wave. Its speed in a fluid of bulk modulus $B$ and density $\rho$ is

v = \sqrt{\frac{B}{\rho}},

and for an ideal gas this becomes

v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma R T}{M}},

where $\gamma$ is the ratio of specific heats (see the Thermodynamics page). Note that sound speed depends on temperature but not on pressure for an ideal gas (since $P/\rho \propto T$ ).

A sound wave can be described either by the displacement $\xi(x,t)$ of the gas parcels or by the pressure variation $\delta P \propto -\partial\xi/\partial x$ . These are a quarter-wavelength out of step, which makes the boundary rules subtle. The reliable principle: whatever quantity the boundary forces to zero is the one that gets a node and flips sign on reflection.

A hard wall pins the displacement, $\xi = 0$ : it is a displacement node and a pressure antinode.
An open end of a tube sits at atmospheric pressure, $\delta P = 0$ : it is a pressure node and a displacement antinode.

This is why an open–closed pipe has the open end as a displacement antinode and the closed end as a displacement node, giving the odd-harmonic series quoted above. (Plenty of textbooks botch this by claiming “hard boundaries flip transverse waves but not longitudinal ones” — not true; track which quantity is fixed.)

Intensity is power per unit area, $I = P/A$ . For a point source radiating uniformly into spheres,

I = \frac{P}{4\pi r^2} \propto \frac{1}{r^2},

so amplitude falls as $1/r$ . Because human hearing spans many orders of magnitude, loudness is measured on a logarithmic decibel scale:

\beta = 10\log_{10}\frac{I}{I_0}, \qquad I_0 = 10^{-12}\text{ W/m}^2.

Every factor of $10$ in intensity adds $10\text{ dB}$ ; a factor of $2$ adds about $3\text{ dB}$ .

Beats

Two waves of nearly equal frequencies $f_1$ and $f_2$ superpose to give a slow amplitude modulation, called beats, heard at the difference frequency:

f_{\text{beat}} = \lvert f_1 - f_2\rvert.

Musicians tune by listening for the beats to slow to zero.

The Doppler effect

Theorem (Doppler effect). When source and observer move relative to the medium, the observed frequency shifts. With all speeds measured relative to the medium,

f' = f\,\frac{v \pm v_{\text{obs}}}{v \mp v_{\text{src}}}.

The sign rule that never fails: choose signs so that approach raises the pitch and recession lowers it. Concretely, the top sign (numerator $+$ , denominator $-$ ) applies when the motion is toward the other party.

Example. An ambulance emits a $700\text{ Hz}$ siren and drives toward a stationary listener at $35\text{ m/s}$ . Take the speed of sound as $v = 343\text{ m/s}$ . What frequency does the listener hear, and what do they hear after the ambulance passes?

The observer is stationary ( $v_{\text{obs}}=0$ ) and the source approaches, so use the $-$ sign in the denominator:

f' = f\,\frac{v}{v - v_{\text{src}}} = 700\,\frac{343}{343 - 35} \approx 780\text{ Hz}.

After it passes, the source recedes, so the denominator sign flips to $+$ :

f' = 700\,\frac{343}{343 + 35} \approx 635\text{ Hz}.

The pitch drops by about $145\text{ Hz}$ as the ambulance goes by — the familiar falling “neeeow.”

When the source itself moves faster than the wave speed ( $v_{\text{src}} > v$ ), the wavefronts pile into a shock wave (a sonic boom for sound). The Mach cone half-angle is

\sin\theta = \frac{v}{v_{\text{src}}} = \frac{1}{\text{Mach number}}.

Water waves

Water waves are the most familiar waves in daily life and also the most complicated — the restoring force is gravity, and the result is dispersive, so the speed depends on wavelength. Two limiting cases are worth knowing.

Shallow water ( $\lambda \gg D$ , where $D$ is the depth). The wave speed depends only on the depth,

v = \sqrt{gD},

independent of wavelength — so shallow water waves are nondispersive. Two consequences: a tsunami in the deep ocean ( $D \sim 4\text{ km}$ ) travels at jet-airliner speeds, and as waves approach shore the dropping depth slows and steepens them. The speed change also explains why waves always arrive nearly parallel to the shoreline (the part of a crest in deeper water outruns the part in shallow water, swinging the crest around — the same refraction that bends light toward slower media).

Deep water ( $\lambda \ll D$ ). Now the depth drops out and gravity competes with wavelength,

v_p = \sqrt{\frac{g}{k}} = \sqrt{\frac{g\lambda}{2\pi}},

so longer swells travel faster and a storm sorts its waves by wavelength as they spread out. The dispersion relation $\omega = \sqrt{gk}$ gives a group velocity that is exactly half the phase velocity, $v_g = \tfrac12 v_p$ — individual crests appear to run forward through a wave group and vanish at its leading edge.

(For very short ripples, $\lambda \lesssim 1\text{ cm}$ , surface tension takes over from gravity and the trend reverses: shorter ripples go faster.)

Phase velocity and group velocity

In a dispersive medium the wave speed depends on frequency, so we distinguish two speeds. The phase velocity is how fast a single wave crest moves,

v_p = \frac{\omega}{k},

while the group velocity is how fast a wave packet (and its energy/information) moves,

v_g = \frac{d\omega}{dk}.

For a nondispersive wave ( $\omega = vk$ ) the two coincide. The relation $\omega(k)$ is called the dispersion relation, and computing $v_g = d\omega/dk$ from it is a recurring olympiad task.

The group velocity is the speed of a wave packet — a localized burst built by superposing waves over a band of wavenumbers $\Delta k$ . A single sinusoid has infinite extent; to make a packet of finite size $\Delta x$ you must combine a range of wavenumbers, and the two are inversely related:

\Delta x\,\Delta k \gtrsim 1, \qquad\text{equivalently}\qquad \Delta t\,\Delta \omega \gtrsim 1.

This is a purely classical statement about waves — a narrow pulse needs a broad spectrum — but feeding in the de Broglie relation $p = \hbar k$ turns it directly into the Heisenberg uncertainty principle $\Delta x\,\Delta p \gtrsim \hbar$ . In a dispersive medium the components of a packet travel at different speeds, so the packet gradually spreads out, or disperses.

However, not every dispersion relation passes through the origin. If a string is tied down by a bed of springs, its equation of motion picks up an extra restoring term and the dispersion relation becomes

\omega^2 = \omega_0^2 + v^2 k^2.

There is now a minimum frequency $\omega_0$ : drive the string below it and no traveling wave propagates — the disturbance decays exponentially instead. Reading $E = \hbar\omega$ and $p = \hbar k$ , this is exactly the energy–momentum relation $E^2 = (mc^2)^2 + (pc)^2$ of a relativistic massive particle, with the minimum frequency playing the role of rest mass. It is a toy model for how a field can acquire mass.

Problem-solving strategy

Match the situation to the right tool before reaching for algebra:

Problem-solving strategy

A pulse or wave travels through a medium and you want its speed: use the “restoring force over inertia” template — $v=\sqrt{T/\mu}$ on a string, $v=\sqrt{B/\rho}=\sqrt{\gamma RT/M}$ for sound, $v=\sqrt{gD}$ in shallow water. If the tension (hence speed) varies along the path, integrate $\tau=\int dx/v(x)$ , as for the hanging rope.
The medium is confined between boundaries (string or pipe): it is a normal-mode problem. Fit half-wavelengths to the boundary conditions — node at a fixed/closed end, antinode at a free/open end — to get $\lambda_n=2L/n$ , $f_n=nv/2L$ for symmetric ends, or the odd-only series $f_n=(2n-1)v/4L$ for open–closed.
Two waves overlap and you want loud/quiet spots: superpose. Convert any path-length difference to phase via $\delta=\tfrac{2\pi}{\lambda}\Delta r$ ; constructive at $\Delta r=m\lambda$ , destructive at $\Delta r=(m+\tfrac12)\lambda$ , net amplitude $2A\cos\tfrac{\delta}{2}$ .
Two nearly-equal frequencies are sounded together: beats, heard at $f_{\text{beat}}=\lvert f_1-f_2\rvert$ — not an interference-in-space question.
Source and/or observer move and the pitch changes: Doppler, $f'=f\,\dfrac{v\pm v_{\text{obs}}}{v\mp v_{\text{src}}}$ . Fix signs by the physics (approach raises pitch). If $v_{\text{src}}>v$ , switch to the shock-cone result $\sin\theta=v/v_{\text{src}}$ .
A wave hits an interface and you want reflection/transmission: use impedance $Z=\mu v=\sqrt{\mu T}$ with $r=\dfrac{Z_1-Z_2}{Z_1+Z_2}$ , $t=\dfrac{2Z_1}{Z_1+Z_2}$ ; inversion ( $r<0$ ) means a hard/denser boundary, and $Z_1=Z_2$ gives zero reflection.
Intensity, loudness, or how amplitude falls with distance: spread power over the wavefront, $I=P/4\pi r^2$ (so amplitude $\propto 1/r$ ), and convert to decibels with $\beta=10\log_{10}(I/I_0)$ .
The medium is dispersive (water waves, springs-loaded string, anything with $\omega(k)$ not $\propto k$ ): distinguish $v_p=\omega/k$ from $v_g=d\omega/dk$ . Differentiate the dispersion relation for the group (energy/packet) speed, and remember a pulse of width $\Delta x$ needs a band $\Delta k\sim 1/\Delta x$ .

Common mistakes

Wave speed vs. transverse speed. $v$ is how fast the pattern moves; a point on the string moves at $u_y=\partial y/\partial t$ with max $\omega A$ . They are unrelated quantities.
Pipe-end rules in longitudinal waves. Track which quantity the boundary pins: a hard wall is a displacement node but a pressure antinode; an open end is a pressure node but a displacement antinode. Do not assume “hard = inverted” the way you would for a string.
Open–closed pipe harmonics. Only odd harmonics exist ( $f_n=(2n-1)v/4L$ ); there is no second harmonic — a frequent source of wrong mode spacings.
Doppler sign confusion. Speeds are measured relative to the medium, and source vs. observer motion enter through different parts of the fraction. Sanity-check with “approach raises pitch” rather than memorizing $\pm$ .
Standing waves carry no net energy. They are not traveling waves; the kinetic–potential equipartition that holds for a traveling wave fails here, and power does not flow past a node.
Sound speed and pressure. For an ideal gas $v$ depends on temperature, not pressure, since $P/\rho\propto T$ — raising the pressure at fixed $T$ does nothing.