I'm convinced that water waves for example:
are a combination of longitudinal and transverse. Any references or proofs of this or otherwise?
Classical Mechanics – Are Water Waves Longitudinal or Transverse?
classical-mechanicswaterwaves
Related Solutions
For example, in solids, the velocities of transverse and the longitudinal waves depend on the shear modulus and compressive modulus, respectively, and shear modulus is less than compression modulus (the velocities also depend on the density, which is the same for both types of waves).
A fake derivation
We can rather easily compute a horizontal velocity for the string fi we assume that the total velocity vector is everywhere normal to the string (this assumption is not always valid, see below). The following picture then illustrates the computation:
Take two infinitesimally separated points $x$ and $x+\mathrm{d}x$ and let the wave motion be $\varphi(x,t)$. The vertical/transverse velocity is $v_\text{vert} = \partial_t \varphi(x,t)$, and the horizontal component is $v_\text{hor} = -v_\text{vert}\tan(\vartheta)$, where $\vartheta$ is the angle between the normal and the vertical, and the minus sign is because if we measure $\vartheta$ in the usual counterclockwise direction then the horizonal velocity points to $-x$ for small $\vartheta$. Now $\tan(\vartheta)$ is $\frac{\varphi(x+\mathrm{d}x) - \varphi(x)}{\mathrm{d}x} = \partial_x\varphi(x)$ , so we get $$ v_\text{hor} = -\partial_t\varphi\partial_x\varphi$$ and if you plug in the sinusoidal solution and take the time average you get exactly the same result as for longitudinal waves. However, you might protext - the transverse wave equation was derived assuming no longitudinal motion, and this computation just blatantly assumes something different.
A Lagrangian derivation
Oddly enough, the result of the above computation is the correct momentum for a pure transverse wave. The Lagrangian of a transverse wave is $$ L = \frac{1}{2}\rho (\partial_t\varphi)^2 - \frac{1}{2}\tau(\partial_x\varphi)^2$$ and translation invariance gives us a momentum density $$ T_{xt} = \partial_x L \partial_t \varphi = - \rho\partial_x\varphi\partial_t\varphi$$ which is conserved by Noether's theorem.
The actual answer
In reality, there are no purely transverse waves on a string, there will always be secondary longitudinal waves generated when trying to excite it purely transversely. The "true" momentum of a realistic "transverse" wave is rather half of the theoretical prediction, i.e. $\frac{1}{2}\rho\partial_t\varphi\partial_x\varphi$, for more on this see "The missing wave momentum mystery"[pdf link] by Rowland and Pask.
Best Answer
Each point is moving according to:
$x(t) = x_0 + a e^{-y_0/l} \cos(k x_0+\omega t)$
$y(t) = y_0 + a e^{-y_0/l} \sin(k x_0+\omega t)$
With $x_0,y_0$ -- "motion centre" for each particle, $a$ -- the amplitude, $l$ -- decay length with depth.
So you have exact "circular" superposition of longitudinal and transverse waves.