The above construction of a longitudinal wave out of a transversal wave has been encountered somewhere in an old physics textbook. There are several drawbacks with this construction.

- The maximum densities of the longitudinal wave are not where you would expect them, namely at the extrema of the transversal wave.
- More seriously. There is a restriction on the maximum amplitude $A$ of the transversal wave, namely $A < \lambda/(2\pi)$, where $\lambda$ is the wavelength.

After some thought, it seems that the comment by dfeuer is quite to the point: "what properties you want this construction to have?" A good question is half the answer. So here comes my wish list:

- the sequence of points $x_k$ must represent a longitudinal wave
- the sequence $x_k$ must must be monotonically increasing
- the construction must be independent of the discretization $\Delta$

Monotonically increasing means that $x_{k+1} > x_k$ , giving: $$(k+1)\Delta + A.\sin\left(\frac{2\pi}{\lambda}(k+1)\Delta\right) > k\Delta + A.\sin\left(\frac{2\pi}{\lambda}k\Delta\right)$$ With help of the formula $\;\sin(p)-\sin(q)=2.\cos\frac{1}{2}(p+q).\sin\frac{1}{2}(p-q)\;$ we find: $$\Delta + 2.A.\cos\left(\frac{2\pi}{\lambda}k\Delta + \frac{\pi}{\lambda}\Delta\right) .\sin\left(\frac{\pi}{\lambda}\Delta\right) > 0$$ This must be independent of $k$ in the first place. Worst case scenario: $\;\cos\left(\frac{2\pi}{\lambda}k\Delta+\frac{\pi}{\lambda}\Delta\right) = -1\;$. Giving, while anticipating a bit on the next requirement: $$\Delta - 2.A.\left[\frac{\sin\left(\frac{\pi}{\lambda}\Delta\right)}{\frac{\pi}{\lambda}\Delta}\right] \frac{\pi}{\lambda}\Delta > 0$$ Let's assume that the discretization $\Delta$ is reasonable with respect to the wavelength $\lambda$, say $\Delta < \lambda/4$. This is no serious restriction. Then the worst case scenario for $\Delta$ is that it becomes infinitesimally small. Because we know that $$\frac{\sin\left(\frac{\pi}{\lambda}\Delta\right)}{\frac{\pi}{\lambda}\Delta} < 1 \qquad \mbox{and} \qquad \lim_{\Delta \to 0} \frac{\sin\left(\frac{\pi}{\lambda}\Delta\right)}{\frac{\pi}{\lambda}\Delta} = 1$$ Consequently, if all of our wishes are fulfilled,