4.Harmonic Waves

Consider a simplest wave profile in form of sinusoidal, the so-called harmonic wave. At t = 0, the disturbance is given by: 

$\psi {{(x,t)}_{t=0}}=\psi (x)=A\sin (kx)=f(x)$                                           (4.1)

In Equation (4.1), we introduced a positive constant k, as the propagation number, since the sine of a quantity is dimensionless. kx is thus in radians, which is not a real physical unit. The sine varies between -1 and +1 so that the maximum values of the disturbance, i.e. the amplitude, of the wave $\psi (x)=A$.

For a progressive wave traveling at speed v in the positive x-direction, at time t, Equation (4.1) can be rewritten as follows:

$\psi (x,t)=A\sin k(x-vt)=f(x-vt)$                                          (4.2)

Clearly, Equation (4.2) is a solution of differential wave equation.

We introduce a derived physical quantity: wavelength λ into Equation (4.2). The wavelength λ, is also known as the spatial period, i. e. the number of units of length per wave. An increase or decrease of x by the amount of λ should leave the wave function ψ unaltered:

$\psi (x,t)=\psi (x\pm \lambda ,t)$                                                      (4.3)

Equivalently, for a sinusoidal wave, this is the same as altering the argument of sine-function by $\pm 2\pi $, therefore:

$\sin k(x-vt)=\sin k[(x\pm \lambda )-vt]=\sin [k(x-vt)\pm 2\pi ]$                             (4.4)

From Equation (4.4), we conclude that: |kλ| = 2π. Since both k and λ are positive number, we obtain:

$k=\frac{2\pi }{\lambda }$                                                   (4.5)

Figure 4.1. A profile of harmonic wave. One wavelength corresponds to a change in phase φ of 2π rad. 

Figure 4.1 illustrates the harmonic wave equation (see Equation (4.1)) in terms of λ. In this figure, φ is the phase, i.e. the argument of the sine-function. From this figure, we notice that $\psi (x)=0$ whenever $\sin \varphi =0$, i.e. when $\varphi =0,\pi ,2\pi ,...$That occurs at $x=0,\frac{\lambda }{2},\lambda ,...$ respectively.

We now examine the temporal period τ, which is the amount of time it takes for one complete wave to pass a stationary observer. In an analogous fashion to Equation (4.3), an increase or decrease of t by the amount of τ should leave the wavefunction ψ unaltered:

$\psi (x,t)=\psi (x,t\pm \tau )$                                                             (4.6)

Therefore,

$\sin k(x-vt)=\sin k[x-v(t\pm \tau )]=\sin [k(x-vt)\pm 2\pi ]$                              (4.7)

From Equation (4.7), we conclude that: |kvτ| = 2π ; hence

$kv\tau =2\pi $                                                                 (4.8)

Or $\frac{2\pi }{\lambda }v\tau =2\pi $ from which it follows that

$\tau =\frac{\lambda }{v}$                                                                       (4.9)

The inverse of Equation (4.9) is the temporal frequency $\nu \equiv {}^{1}/{}_{\tau }$ , which is the number of waves per unit of time, i.e. per second. The unit of temporal frequency is cycles per second or Hertz. Equation (4.9) can then be rewritten as: $v=\nu \lambda $.

Two other quantities are often used in practice: the angular temporal frequency: $\omega \equiv {}^{2\pi }/{}_{\tau }=2\pi \nu $ and the wave number or the spatial frequency $\kappa \equiv {}^{1}/{}_{\lambda }$ which is measure in inverse meters. κ is thus the number of waves per unit length i.e. per meter.