The Wave Function y(x, t)

In Physics 2 you read a wave two ways: a snapshot (the shape of the whole string at one instant) and a history (how one point moves as time passes). A wave function holds both at once. It is a single formula $y(x, t)$ that gives the sideways displacement of every point $x$ of the medium at every time $t$. Freeze $t$ and you get the snapshot; fix $x$ and you get the history.

Key idea

A sinusoidal wave moving toward $+x$ is $$y(x,t) = A\sin(kx - \omega t + \phi).$$ The wave number $k = 2\pi/\lambda$ (rad/m) says how fast the phase changes along the string; the angular frequency $\omega = 2\pi f$ (rad/s) says how fast it changes in time. The wave moves at $$v = \frac{\omega}{k} = f\lambda.$$ With $kx + \omega t$ instead, the same wave moves toward $-x$.

Why it works

Take any shape $y = f(x)$ and slide it to the right by a distance $vt$. Every feature that was at position $x_0$ is now at $x_0 + vt$, so the new curve is $y = f(x - vt)$. For a sine shape, $\sin[k(x - vt)] = \sin(kx - kvt)$, and $kv$ is the angular frequency: $\omega = kv$. So the minus sign means the same shape, a little later and a little farther to the right. To see it with calculus, ride on one crest, where the phase $kx - \omega t$ stays constant. Differentiate: $k\,\frac{dx}{dt} - \omega = 0$, so the crest moves at $\frac{dx}{dt} = +\frac{\omega}{k}$.

The parts of a sinusoidal wave function.
Symbol	Name	Unit	Connects to
$A$	amplitude	m	the largest displacement
$k$	wave number	rad/m	$k = 2\pi/\lambda$
$\omega$	angular frequency	rad/s	$\omega = 2\pi f = 2\pi/T$
$\phi$	phase constant	rad	sets $y$ at $x = 0,\ t = 0$
$v$	wave speed	m/s	$v = \omega/k = f\lambda$

Two waves with $k = 2$ rad/m and $\omega = 3$ rad/s. The solid one, $\sin(2x - 3t)$, slides to the right; the dashed one, $\sin(2x + 3t)$, slides to the left. Both move at $\omega/k = 1.5$ m/s.

Worked examples

Worked example: Read the numbers

A wave on a string is $$y = (0.030\ \text{m})\sin(2.0x - 6.0t)$$ with $x$ in meters and $t$ in seconds. Find its wavelength, frequency, speed and direction.

Compare with $A\sin(kx - \omega t)$: $A = 0.030$ m, $k = 2.0$ rad/m, $\omega = 6.0$ rad/s.
Wavelength: $\lambda = \frac{2\pi}{k} = \frac{2\pi}{2.0\ \text{rad/m}} = 3.14$ m.
Frequency: $f = \frac{\omega}{2\pi} = \frac{6.0\ \text{rad/s}}{2\pi} = 0.955$ Hz.
Speed: $v = \frac{\omega}{k} = \frac{6.0}{2.0} = 3.0$ m/s. Check: $f\lambda = 0.955 \times 3.14 = 3.0$ m/s.
The $kx$ and $\omega t$ terms have opposite signs, so the wave moves toward $+x$.

Answer: $\lambda = 3.14$ m, $f = 0.955$ Hz, $v = 3.0$ m/s toward $+x$

Worked example: The string moves, too

For the same wave, find the displacement and the velocity of the bit of string at $x = 0.50$ m when $t = 0.20$ s.

Phase: $kx - \omega t = (2.0)(0.50) - (6.0)(0.20) = 1.0 - 1.2 = -0.20$ rad.
Displacement: $y = 0.030\sin(-0.20) = -0.0060$ m.
The string's sideways velocity is a partial derivative, holding $x$ fixed: $$v_y = \frac{\partial y}{\partial t} = -A\omega\cos(kx - \omega t).$$
$v_y = -(0.030)(6.0)\cos(-0.20) = -0.18 \times 0.980 = -0.176$ m/s.

Answer: $y = -6.0$ mm, moving down at $0.176$ m/s

Notice the two very different speeds. The wave moves along the string at $\omega/k = 3.0$ m/s, set by the string's tension and mass. Each bit of string moves up and down with speeds up to $A\omega = 0.18$ m/s, set by how hard you shake it.

Worked example: Build a wave function

A wave on a rope has amplitude 5.0 cm, wavelength 0.80 m and frequency 2.5 Hz. It travels toward $-x$, and the rope at $x = 0$ is at $y = 0$ when $t = 0$. Write $y(x, t)$.

$k = \frac{2\pi}{0.80\ \text{m}} = 7.85$ rad/m and $\omega = 2\pi(2.5\ \text{Hz}) = 15.7$ rad/s.
Toward $-x$ means the terms have the same sign: $kx + \omega t$.
$y(0, 0) = 0$ is satisfied by $\sin$ with $\phi = 0$.
Check the speed: $\omega/k = 15.7/7.85 = 2.0$ m/s, and $f\lambda = 2.5 \times 0.80 = 2.0$ m/s.

Answer: $y = (0.050\ \text{m})\sin(7.85x + 15.7t)$

Common mistake

Seeing the minus sign in $kx - \omega t$ and deciding the wave moves toward $-x$. It is the other way around: opposite signs mean the wave moves toward $+x$, and $\sin(\omega t - kx)$ moves toward $+x$ too. Check by following a crest: set the phase equal to a constant and solve for $x$. For $kx - \omega t = C$, $x = (C + \omega t)/k$ grows as $t$ grows.

Common mistake

Mixing up the wave speed $\omega/k$ with the speed of the medium. The bits of string never travel along with the wave; they swing about their own positions with speed up to $A\omega$.

A traveling wave with its snapshot $y(x)$ at one instant and the history $y(t)$ of the red particle. Raise the frequency: the wavelength shrinks, but the speed stays the same.

The Wave Function y(x, t)

Worked examples

Practice

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