In my new apartment, I sleep on the lower bunk where the top bunk is held up by a grid base. So every night before I sleep (or every afternoon before I nap), I have a great view of that grid above me. I have pictures of it, but they suck, so I will simply draw this on PS:
The blue is the bed sheet, and the brown is the bottom of the bed itself. The gray lines form what is supposed to be a square grid… apologies for the bad artistic skills.
So, with that said, what can I DO with this?
1. Poke the bed in between grid lines. ‘Nuff said.
2. Draw Sine Waves.
If you don’t believe me, consider this: an entire square grid can be formed from specified tangent lines to a sinusoid and their normals. The converse also works.
Let’s use the simplest sine function with wavelength 2π to represent the sinuosidal wave. At any arbitrary point on the grid, let’s say x=0, the slope of the wave is
Let the wavelength be λ. Halfway through a period, at x=λ/2, the slope is -1 since cos(π) = -1. At x=λ, the slope is back to 1. The upshot here is that tangent lines at half-wavelengths, starting from a point where the slope is unity, oscillates between +/- 1. This diagram sums it up quite well:
Here comes the kicker. The slopes of these tangent lines are normal. The normals of these tangent lines are also normal. Everything is normal! Furthermore, by symmetry, the lengths of the both the tangent and the normal lines between where they cross each other and where they cross the x-axis are the same. They form squares.
Now let’s try to relate the wavelength of the wave to the length of each square. From the illustration above, a half-wavelength will cover the diagonal of the square. Let the side of each square have length L. Thus,
In this case of a 2π-periodic wave, L = π√2 / 2. But let’s generalize it for any L and any λ.
First we observe that for this to work, the slopes MUST be +/- 1 at the midpoints between extrema. Otherwise the square breaks down into some other ugly quadrilateral because tangents at half-wavelengths won’t be normal to each other. Mathematics wise, we observe that for
Since the wavenumber k = 2π/λ,
and so, after substituting L for λ and rearranging,
And so, given the side length of an arbitrary square in a regular grid, a sine wave may be fitted to the grid, as we can find A and k = 1/A from the above equations.
Since this is getting long, I will save a Part 2 for later. 🙂
[Note: Graphs generated with http://rechneronline.de/function-graphs/]