How to calculate Standing Waves in Electrical Cables? I have a 20 metre Coaxial Cable. I send digital signals down the cable ranging from 5 KHz to 50 KHz. 
I have noticed a pattern in the noise ratio, an oscillating wave. I predict this is to do with standing waves in the cable.
How can I calculate at what Frequencies these standing waves would occur in the cable?
 A: Your oscillatory pattern can't be owing to standing waves. The speed of light in such cables is of the order of $0.5\,c$ to $0.7\,c$, so your first standing wave resonant frequency (depending on the impedance connected to the cable's end) will be at least about 1MHz. This statement depends of course on how much of the "oscillating noise" period you saw - I presume you mean the noise varies sinusoidally with frequency. If the reflexion co-efficient from the cable's far end is $\Gamma$ (set by the impedance connected to the far end), then the impedance seen at the cable's input is:
$$Z_{in} = \frac{1 + \Gamma \exp\left(i\frac{4\,\pi\,\nu}{c_c} \ell\right)}{1 - \Gamma \exp\left(i\frac{4\,\pi\,\nu}{c_c} \ell\right)}$$
where $c_c\approx 0.5\,c$ is the speed of light in the cable, $\nu$ the source frequency and $\ell$ the cable's length. Your varying noise arises because the load $Z_{in}$ presented to the source varies with frequency, and thus affects the source's noise performance. If $Z_{in}$ is indeed modelled by the above equation, I plot the variation in this quantity below for frequency between 0Hz and 5MHz below for various values of $\Gamma$

and below, for a small value of $\Gamma = 0.1$ in the same frequency range:

As you can see, the variation in the range 0Hz to 50kHz is negligible. 
Therefore I agree with user1038377's answer: the coaxial cable is acting as a lumped impedance (theorised to be an LC circuit by user1038377) and you are maybe seeing resonances, although I can't explain how you would get a sinusoidal variation with frequency out of a lumped ciruit.
To calculate the standing wave frequencies for your particular values, note that you will get a peak in input impedance when $\arg(\Gamma) + \frac{4\,\pi\,\nu}{c_c} \ell = 2 j \pi;\;j\in\mathbb{Z}$ and a trough in impedance when $\arg(\Gamma) + \frac{4\,\pi\,\nu}{c_c} \ell = (2 j +1)\pi;\;j\in\mathbb{Z}$. Note than:
$$\Gamma = \frac{Z_0 - Z_{load}}{Z_0 + Z_{load}}$$
where $Z_0$ is the coax line's characteristic impedance (probably $50\Omega$) and $Z_{load}$ is the load connected to the line's far end. Since the latter is generally complex, note that $\arg(\Gamma)$ can be nonzero. And you don't actually get standing waves (i.e. waves with nodal points) unless the $\Gamma = \pm1$, you get instead constant points of minimum and maximum amplitude at points with axial co-ordinate $z$ measured backwards towards the source from the load on the end of the line as follows:
$$\arg(\Gamma) + \frac{4\,\pi\,\nu}{c_c} z_{max} = 2 j \pi;\;j\in\mathbb{Z}$$
defines the points where there is maximum voltage amplitude and minimum current amplitude and:
$$\arg(\Gamma) + \frac{4\,\pi\,\nu}{c_c} z_{min} = 2 j \pi + 1;\;j\in\mathbb{Z}$$
defines the points where there is minimum voltage amplitude and maximum current amplitude. A quantity often defined is the Standing wave ratio, which is the ratio of minimum voltage / current to maximum voltage / current and it is given by $SWR = (1-|\Gamma|)/(1+|\Gamma|)$;
More information can be found on the Wikepedia "Transmission Line page.
A: It's simply an LC circuit. The linear frequency is $f = 1/(2\pi\sqrt{LC})$. If you use the approximation of a thin long wire to compute L and C, then you may obtain $LC \propto l^2$ where $l$ is the length of the wire. So $f \propto 1/l$.
A: Sorry, can't comment the answer of Rod Vance so writing a new one. My previous offer regarding to the LC-circuit was wrong because accurate estimation of the eigenfrequency gives f ~ 1 MHz. The observed frequency ~50kHz may be formally connected with a speed ~(0.05-0.1)c but there is no such velocities in the experiment.
At the other side, the wavelength of the 50kHz signal is about tens times higher than the cable length. Maybe the phenomena observed is (a) seesaw of the electrons by the (b) force which is truncated in the frequency space. Fourier image of the meander has a series of diminishing maxima separated by frequency of the meander. The cable itself may act as a filter, so the smooth shifting of the input frequency leads to the periodical variation of the integrated power spectrum of the meander inside the filtering window of the cable. However, I'm not a specialist in this area, so I and can't make more detailed calculations.
