Wavelength and Angular Wave Number

Displacement Relation in a Progressive Wave

A wave travelling through a medium carries energy and momentum, but what exactly do we mean by the "displacement" of a wave? For a wave on a string, the displacement is simply how far a particular point on the string has moved from its equilibrium (rest) position. This displacement depends on two things: which point on the string you look at (its position $x$), and when you look at it (the time $t$).

The fundamental idea is that the shape of the wave moves. If you take a snapshot of the wave at time $t = 0$, you get a certain shape described by some function $f(x)$. As time passes, this entire shape shifts to the right (for a wave travelling in the positive $x$-direction) or to the left (for a wave travelling in the negative $x$-direction). The displacement of any point on the string at any time is therefore given by a function of both $x$ and $t$.

For a wave travelling in the positive $x$-direction with constant speed $v$, the displacement $y(x, t)$ is:

For a wave travelling in the negative $x$-direction:

The functions $f$ and $g$ can be any well-behaved functions — they determine the shape of the wave. The simplest and most important case is when the wave has a sinusoidal shape, which is what we now examine in detail.

Wavelength and Angular Wave Number

Consider a sinusoidal wave travelling along the positive $x$-direction. At a fixed time, say $t = 0$, the displacement as a function of position is:

Here $a$ is the amplitude — the maximum displacement from equilibrium. The constant $k$ is called the angular wave number (often just "wave number"). Its physical meaning becomes clear when we ask: after what distance does the wave pattern repeat itself? That distance is the wavelength $\lambda$.

Since the sine function repeats when its argument increases by $2\pi$, we require:

This gives:

The angular wave number $k$ tells you how many radians of phase the wave accumulates per unit distance along the $x$-axis. A larger $k$ means a shorter wavelength — the wave oscillates more rapidly in space.

The Complete Displacement Relation

For a sinusoidal wave travelling in the positive $x$-direction, the displacement at any position $x$ and any time $t$ is:

The term $\omega$ is the angular frequency (radians per second), and $\phi$ is the initial phase (or phase constant) — it determines the displacement at $x = 0$ and $t = 0$.

The argument of the sine function, $(kx - \omega t + \phi)$, is called the phase of the wave. The minus sign between $kx$ and $\omega t$ indicates a wave travelling to the right (positive $x$-direction). For a wave travelling to the left, the phase would be $(kx + \omega t + \phi)$.

Properties of the Sinusoidal Wave Function

The textbook lists three important properties of the function $y(x, t) = a \sin(kx - \omega t + \phi)$. Each one follows directly from the definitions of $k$, $\omega$, and the wave speed $v$.

›Proof

Property (I): Periodicity in space (wavelength)

At any fixed time $t$, the displacement repeats after a distance $\lambda$:

$y(x + \lambda, t) = y(x, t)$

Proof:

$y(x + \lambda, t) = a \sin[k(x + \lambda) - \omega t + \phi]$

$= a \sin(kx + k\lambda - \omega t + \phi)$

Since $k\lambda = 2\pi$:

$= a \sin(kx + 2\pi - \omega t + \phi)$

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

The sine function's periodicity of $2\pi$ ensures the equality.

›Proof

Property (II): Periodicity in time (time period)

At any fixed position $x$, the displacement repeats after a time interval $T$:

$y(x, t + T) = y(x, t)$

Proof:

$y(x, t + T) = a \sin[kx - \omega(t + T) + \phi]$

$= a \sin(kx - \omega t - \omega T + \phi)$

The angular frequency $\omega$ is related to the time period $T$ by $\omega T = 2\pi$:

$= a \sin(kx - \omega t - 2\pi + \phi)$

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

›Proof

Property (III): The wave speed

The wave travels with a constant speed $v$ given by:

$v = \frac{\omega}{k} = \frac{\lambda}{T}$

Proof:

Consider a point of constant phase on the wave. For a fixed value of the phase $(kx - \omega t + \phi)$, as time increases, the position $x$ must also increase to keep the phase constant. Differentiate the phase with respect to time:

$\frac{d}{dt}(kx - \omega t + \phi) = 0$

$k\frac{dx}{dt} - \omega = 0$

$\frac{dx}{dt} = \frac{\omega}{k}$

This speed $dx/dt$ is the wave speed $v$. Using $\omega = 2\pi/T$ and $k = 2\pi/\lambda$:

$v = \frac{2\pi/T}{2\pi/\lambda} = \frac{\lambda}{T}$

The Initial Phase $\phi$

The phase constant $\phi$ allows us to describe waves that do not start at zero displacement at $x = 0$, $t = 0$. For example:

• If $\phi = 0$: $y(0, 0) = a \sin(0) = 0$, and the wave is rising (positive slope) at the origin.
• If $\phi = \pi/2$: $y(0, 0) = a \sin(\pi/2) = a$, so the displacement starts at its maximum value. …