Let us imagine that the oscillatory motion of a source is described by the function

${\displaystyle y_S(t)=A\sin (\omega t+\varphi )}$

(1)

(For a quick introduction to trigonometric functions see the page sine function.) In this case we say that the source oscillates with harmonic motion, and therefore it produces a harmonic wave.

From the source to the wave

• Starting from expression (1), which describes only the source, let us try to derive an expression that describes the emitted wave, that is, the oscillation that takes place not only at the point where the source is located, but also at points far away from it.
• Let us therefore assume that the sinusoidal oscillation propagates to the right with constant speed v. The function ${\displaystyle y(x,t)}$ is then given by

${\displaystyle y(x,t)=\{\begin{array}{ll}0& 0\le t<t_{\mathrm{r}\mathrm{i}\mathrm{t}}\\ y_S(t-t_{\mathrm{r}\mathrm{i}\mathrm{t}})& t\ge t_{\mathrm{r}\mathrm{i}\mathrm{t}}\end{array}}$

where ${\displaystyle t_{\mathrm{r}\mathrm{i}\mathrm{t}}=\frac{x}{v}}$ denotes the delay time.

• Let us manipulate algebraically the expression obtained for ${\displaystyle t>t_{\mathrm{r}\mathrm{i}\mathrm{t}}}$, in order to obtain an equivalent expression

${\displaystyle y(x,t)=y_S(t-t_{\mathrm{r}\mathrm{i}\mathrm{t}})=A\sin [\omega (t-\frac{x}{v})+\varphi ]}$

from which, recalling that ${\displaystyle \omega =\frac{2\pi }{T}}$

${\displaystyle y(x,t)=A\sin [\frac{2\pi }{T}(t-\frac{x}{v})+\varphi ]}$;

then, bearing in mind that ${\displaystyle vT=\lambda }$, we obtain the particularly "symmetric" expression

${\displaystyle y(x,t)=A\sin [2\pi (\frac{t}{T}-\frac{x}{\lambda })+\varphi ]}$;

which clearly shows that the emitted wave has a "temporal" periodicity (represented by the period T) and a "spatial" periodicity (represented by the wavelength ${\displaystyle \lambda }$); indeed, if the time variable t is increased by one period T, or if the variable x is increased by one wavelength ${\displaystyle \lambda }$, the argument of the sine function changes by exactly ${\displaystyle 2\pi }$, and therefore the value of the function does not change.

• Introducing the wave number ${\displaystyle k=\frac{2\pi }{\lambda }}$, the expression of ${\displaystyle y(x,t)}$ becomes:

${\displaystyle y(x,t)=A\sin [(\omega t-kx)+\varphi ]}$;

• If instead the wave propagates to the left (towards negative x), the delay at the point x is ${\displaystyle -x/v}$ and one obtains

${\displaystyle y(x,t)=A\sin [(\omega t+kx)+\varphi ]}$.

Check your understanding

Further reading and links

• If you are interested in learning more about the sine function, visit the section on harmonic motion. You will be surprised to find that many of the oscillations found in nature are correctly described by this periodic function. As you can read in the pages on Fourier's theorem, it is even a sort of fundamental periodic oscillation from which all kinds of oscillation can be built.
• A useful summary of the formulas used on this page can also be found on the page fundamental quantities of waves.
• Related questions and answers:
  • Why do we measure the angles in radians?