0
$\begingroup$

I have a series of wave-readings which show wave amplitudes pr. time unit for different events. So on the $x $-axis we have seconds, and on the $y$-axis, wave height.

If I want to convert this to a time-independent plot, and go therefore from:

$$\frac{A}{t}\longrightarrow \lambda/\kappa,$$

where $A$ is the amplitude, $t$ is time (in sec), $\lambda$ is the wavelength, and $\kappa$ is the wavenumber, how do I do that?

Any suggestions appreciated.

$\endgroup$
2
  • 1
    $\begingroup$ Are the waves pure sine waves? Or are they more messy? Could you show a picture? Edit: based on the Fourier transform tag I suspect that they are more messy $\endgroup$ Commented Feb 15, 2023 at 14:00
  • $\begingroup$ very messy! However, some average velocity of the wave train can be assumed, i.e. $25 m/s$ $\endgroup$ Commented Feb 15, 2023 at 15:40

2 Answers 2

1
$\begingroup$

Your graph, amplitude as a function of time, does not include any factor of distance. It is only an oscillation at the one point being measured for amplitude. Wave number and wavelength are distance-based relationships. In a standard sine wave, wave number is $\kappa=1/\lambda$. Wave number and wavelength do not relate to amplitude. Unless you know the wave speed, they do not relate to time.

If the variation of amplitude with respect to time shows the summation of various frequencies, then you could use a Fourier transformation to find amplitude as a function of frequency. For this to work, however, the wave would need a repeating cycle. The period of that repeating cycle would give you the fundamental frequency on which to build your Fourier analysis.

$\endgroup$
2
  • $\begingroup$ I have the velocity. It is an average velocity for waves at sea during storms, it can be set to $20-25m/s$ for instance. $\endgroup$ Commented Feb 15, 2023 at 19:18
  • $\begingroup$ Find the period of the wave, the time for one cycle to occur. Multiply this by the wave speed to yield the wavelength. If period and therefore wavelength vary, then a graph of A vs. $\lambda$ will yield how amplitude varies with wavelength. If the period and thus wavelength are uniform, then all data will have the same value for the horizontal axis. $\endgroup$ Commented Feb 15, 2023 at 20:45
0
$\begingroup$

The plot of the function is generated from a series of $(x,y)$-coordinates, $\{(x_1,y_1), (x_2,y_2),\cdots(x_n,y_n)\}$, where $x$=time (t), $y$=Amplitude (A). In order to convert this plot into a stattionary plot, where $x$ is the position, we have to convert the set of $x$ coordinates from time to position.

We use the conversion operator:

$$Q:\frac{A}{t}\longrightarrow \frac{A}{x}$$

which is an operation on the set of coordinates of the plot:

\begin{equation} Q: \{(x_1,y_1), (x_2,y_2),\cdots(x_n,y_n)\}\longrightarrow \{v,1\}\cdot\{(x_1,y_1), (x_2,y_2),\cdots,(x_n,y_n)\} \end{equation}

Where $v$ is the average velocity of the wavetrain. This yields a new set of coordinates:

$$\{(v\cdot x_1, y_1), (v\cdot x_2, y_2),\cdots,(v\cdot x_n, y_n)\}$$

In this way, we have converted the time-axis (x-axis) into a position-axis, and when we perform an interpolation we obtain a stationary plot of the waves (waveprofile), where $\lambda$ is readable on the $x$-axis. This plot is in effect a "stretched" out version of the time-dependent plot, stretched on the x-axis.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.