While water moves up and down on the surface, does it move the same way below surface also? Or does it fade into a compression wave somewhere below the surface?
[Physics] How deep is a ocean wave (the extent of tranverse motion of the water below surface)
waves
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When you first put your finger into the water, the situation is pretty much as you describe; the water is not solid and so the molecules can move, unlike those of your finger or the bowl, so they must move (no two atoms can be in exactly the same place at exactly the same time). Liquids are classically defined as incompressible (technically most liquids including water can be compressed, but the ratio of force to volume is much greater than with a gas), so not only must the water molecules move, they must push all the other water molecules out of the way instead of simply getting closer to them. However, air, a gas, can move and can compress, so initially it is easier for the water to move up against the air than out against more water to get out of the way of your finger.
That's what happens in the first fractions of a second after your finger enters. What happens next involves some very complex math, but the concept is still simple; while water is incompressible, it is still a fluid, and so as gravity pulls on all the water molecules, those molecules move around each other trying to reach a new equilibrium between the force of gravity and the forces exerted by all the other molecules of water. In short, the surface of the water wants to settle down to a new "level". The water that moved up when you inserted your finger now tries to move back down, pulled by gravity. That pushes other water molecules aside, just as your finger did, and with a similar result; some of them find it easier to go up than out. This process repeats itself over time, with the result that a wave of water is generated as water molecules get out of the way of other molecules and then settle. When this wave hits the edges of the bowl, well, that's a solid surface that won't move, so the water can't move out, so it moves up again, and then when it settles there's only one way the water can push and that's back in the opposite direction.
Now, the second law of thermodynamics states that "you can't get the same energy out of a system that you put in"; even at the molecular level, there is "friction", where molecules will simply absorb the energy as heat instead of moving. In addition, as waves bounce around the edges of the bowl, they meet, and when the crest of the wave meets a trough, the two cancel out (the sum of the energies of the peak and trough of a waveform of given amplitude is zero). Eventually, the waves cancel out enough that the water molecules themselves simply absorb the energy and do not move (perceptibly) at all, and the system has then reached its new resting state.
Here is what I know:
A wave generated in water is said to be both, transverse as well as longitudinal. Hence vertical and horizontal motion of water molecules, both generate it.
If your stone is an ideal point sized mass then it will create a trough. However, a practical stone will create a trough at center and crest around the submerged part of stone simultaneously (approx).
It forms a circle because the wave propagates in all directions on the surface of water with same speed.
Best Answer
Water wave is combination of traverse and longitudinal waves with a dispersion relation:
$$c=\sqrt{\left( \frac{g\lambda}{2\pi}+\frac{2\pi \gamma}{\rho \lambda} \right) \tanh \frac{2\pi h}{\lambda}}$$
and displacement profile $$\xi+\eta i=\frac{ga}{kc^{2}} \frac{\cosh[k(y+h)+i(kx-\omega t)]}{\cosh kh}$$
In deep water, the particles move in more circular way, and the amplitude decays with depth. No perceptible motion deeper than $\dfrac{\lambda}{2}$.
In shallow water, the particles move in more elliptical manner, and they're pure longitudinal at the bottom.
See also another answer here