Why does frequency remain the same when waves travel from one medium to another? I was reading about reflection and refraction on BBC Bitesize and I can't understand why frequency is a constant in the wave speed equation. I can't visualise the idea of it. I know that wave speed and wavelength are proportional to each other but how can I tell the speed of a wave by looking at a random oscillation? 
Here's where I got confused: https://www.bbc.co.uk/bitesize/guides/zw42ng8/revision/2 (the bottom of the page about the water)
 A: Imagine a rope attached to the bottom of a swimming pool. You, above the surface, grab it and shake it back and forth to create waves along the rope. The part of the rope just above the surface has to be moving back and forth at the same frequency as the part of the rope just below the surface. Otherwise the rope would have a break at the surface.
A: Loosely speaking, you can think of it this way: a wave propagates in space and time, but it encounters a change in the spatial properties of the medium at the interface.  What should the time part change? 
A: Instead of thinking of a travelling wave, it is better to think of a field, where each point in space and instant of time is associated to an electric and a magnetic field. The EM fields are normal to the direction of propagation. Let's choose a field in the $z$ direction and propagating in the $x$ direction, and suppose the boundary normal to $x$:
$$E_z(x,t) = E_0cos(kx - \omega t)$$
If $k$ is different for each media, it is possible to write for a point in the boundary both equations below, for any given $t$:  
$E_z(x_b,t) = E_0cos(k_1x_b - \omega t)$
$E_z(x_b,t) = E_0cos(k_2x_b - \omega t)$
I can always choose the boundary as the origin, $x_b = 0$ and continuity is assured. The idea is that it is possible to have the 2 cosines in phase at the boundary, independent of the time.
But if $\omega$ is different:
$E_z(x_b,t) = E_0cos(kx_b - \omega_1 t)$
$E_z(x_b,t) = E_0cos(kx_b - \omega_2 t)$
It is not possible to assure continuity all the time, only for some t's:
$kx_b - \omega_1 t = kx_b - \omega_2 t + 2\pi n$ =>
$t = 2\pi n / \Delta \omega$
