Bouncing Ball in a Cavity II

I thought today I'd review the bouncing ball in a cavity problem I looked at a couple days ago. In the original post, I had calculated the number of bounces that it would take the ball to hit the ground. I was thinking today that I'd just finish off the problem by calculating the trajectory. Nothing really fancy here, I'm afraid. I put everything into a normalized coordinate system, with normalized X given by $X^\prime = \frac{x}{L}$, $Y^\prime = \frac{y}{\phi^2}$ and $T^\prime = \frac{t}{\frac{L}{V_0}}$. I then just worked out the positions vs time by brute force--because X'(T') is not differentiable, there isn't really an elegant way to make these plots. I only plotted for $\alpha = 0.9$ and $\phi = 6$, which gives $N = 4.84$, so four bounces total. Here are the results.

X'(T') is pretty much what you'd expect. Linear and non-differentiable as I said. You'll notice that the time between each direction change is growing as you'd expect in this situation.

Y'(T') is perfectly parabolic, since in the original problem I had the loss in velocity only along X'. The proper way to do this is probably to have the loss be a loss in kinetic energy rather than X velocity, but this complicates matters substantially, since the x and y components of velocity change at different rates. I don't know that there's a really nice way to do this.

And here's the trajectory of the bouncing ball. Y' depends on $X^'2$, of course, although looking at the plot it is a bit hard to tell. I expect if I had chosen a different value of $\phi$ and $\alpha$, then this would probably be more pronounced.

Bouncing Ball in a Cavity

Here's a quick, fun problem I just thought up. Suppose you have a rectangular cavity like the one I've drawn below. A ball is fired vertically across the cavity with initial velocity $V_0$. When it hits the wall, its direction is reversed and it bounces back the other way. Suppose that at each bounce, its x-velocity is decreased by $\alpha$. How many bounces will the ball make during the trip?

I decided to do this problem with only the x-velocity changing on bounces since it gets pretty complicated if both x and y (or kinetic energy, say) is changing, but it should be doable in principle either way. To start off, let's calculate the time it takes the ball to hit the ground.

$h = \frac{1}{2}gt^2$

$t = \sqrt{\frac{2h}{g}$

Easy enough. Now what about the velocity in the x direction? Well, for the first pass, we have:

$L = V_0 t_0$

But the next pass takes longer...

$L = V_0 \alpha t_1$

If we add up all of the times that it travels along x, we should get the total y time...

$\sqrt{\frac{2h}{g}} = \frac{L}{V_0} + \frac{L}{V_0\alpha} + \frac{L}{V_0\alpha^2} + \ldots$

Let's define $\phi = \frac{\sqrt{\frac{2h}{g}}}{\frac{L}{V_0}}$. This is the ratio of the total travel time to the single pass time. If $\alpha$ were equal to zero, this value alone would give us the number of passes N. As it is, we've still got some more work to do:

$\phi = \Sigma_{i=0}^{N-1} \alpha^{-i}$

This is a geometric series and can be computed analytically:

$\phi = \frac{\alpha^{-N} - 1}{\alpha^{-1}-1}$

Rearrange a bit:

$\phi (\alpha^{-1}-1) + 1 = \alpha^{-N}$

And finally

$N = \frac{\ln \big( \phi (\alpha^{-1} - 1) \big)}{\ln (\alpha^{-1})}$

Here's some graphs of N vs $\phi$ for different values of $\alpha$. Intuitively, this is pretty much what we might expect: as $\phi$ (that is, the ratio between the total time and the single-pass time) grows, we get progressively more bounces, but the effect is moderated significantly by the energy lost per bounce. For relatively small $\alpha$ like the pink graph, we get many bounces, although we still end up with only about half as many bounces as the $\alpha = 1$ case where no energy is lost.

I'll look at the trajectories in part II

Why do rainbows form?

I thought I'd start off with a classic problem in optics: Why do rainbows appear?

Rainbows are the result of a property of matter called dispersion. What dispersion means is that, when light passes through a material, the refractive index of the light will depend upon its wavelength (or frequency, if you prefer). If you recall Snell's law, you'll recognize that the angle that light is refracted through a material depends on the refractive index. Therefore, if the refractive index itself depends on wavelength, this will cause different colours to deflect at different angles and split apart. The splitting that we see gives rise to rainbows.

The dispersion relation for water is fairly complicated. I will be using the relation found here, where I've taken a temperature of 293.15K (20C), and density of $1000 kg/m^3$. Solving this for n, we get a relation that looks like this:

$n = \sqrt{\big(\frac{1+2C(\lambda)}{1-C(\lambda)}\big)}$

Where $C(\lambda)$ is the numerical relation listed in the reference. Graphically, the dispersion relation looks like this, which probably tells us all we really need to know

The refractive index doesn't change by much, but it changes enough for it to make a big difference, as we'll see.

Let's now calculate the refraction of light through a spherical droplet. You can see the setup that I'll be using in the figure.

I've labelled the incident angle $a$, the refracted angle $b$. The normals are all in blue, and I've shown the paths of a red and green ray to illustrate the differences this will likely make. Note that, by definition, the outgoing ray a will be the same input and output angles $a$ and $b$, respectively. We're interested in the angle $q$, which is the change between the entry ray and the exit ray. Note that we don't know any of the angles at this point. We don't actually need them at this point, because the angles are all related.

The angles $a$ and $b$ are related by Snell's Law. That is,

$\sin a = n(\lambda) \sin b$

I've set the refractive index of water to 1. I note here that $n$ is dependent on $\lambda$ as above. I won't include it beyond this point, however, for brevity's sake. Now, you'll note that all of the normal line converge in the middle, by definition of the way circles work. The angles $c$ form isoceles triangles with the angles b. In a total circle the angles add up to 360 degrees, hence.

$360 = 2a + 2c + q$

But $b$ and $c$ are related as well.

$360 = 2a + 2(180 - 2b) + q$

This simplifies to a relation between q, b, and a.

$q = 4b - 2a$

Let's take the derivative of Snell's law and the derivative of the equation above, with respect to $b$.

$\frac{dq}{db} = 4 - 2\frac{da}{db}$
$\cos a \frac{da}{db} = n\cos b$

Combining these two equations, we can form a third relationship between the angles.

$\frac{dq}{db} = 4 - 2n\frac{\cos b}{\cos a}$

Let's set the derivative $\frac{dq}{db} = 0$, which will maximize/minimize the angle $q$ in terms of $a$ and $b$.

$4\cos a = 2n\cos b$

Let's multiple Snell's law by 4, square it, and add it to the square of this equation.

$16 \cos^2 a + 16\sin^2 a = 16n^2\sin^2 b + 4n^2\cos^2 b$

Noting that $\sin^2 b = 1 - \cos^2 b$ and $\sin^2 a + \cos^2a = 1$, we can solve for $b$.

$cos^2 b = \frac{4}{3}\big(1-\frac{1}{n^2}\big)$

We can then go back and find $a$.

$cos^2 a = \frac{n^2 - 1}{3}$

Now we can go back and find $q$. Unfortunately, there doesn't seem to be a particularly elegant way to write this out; it's a bit of a mess. So rather than writing the whole thing out, I'll just plot it for you here, including the dispersion relation for n.

You can see that we observe a deflection of about 3 degrees across the visible light, and that at the 600 nm, we get the "rainbow angle" of 42 degrees. At a rough estimate, if the rainbow appears to be 1 km away, then the red and blue light would be separated by 26m, more than enough to create the effect that we call a rainbow. Note that the example I'm doing here is a first-order rainbow--we can also make rainbows be counting further reflections within the droplet, and we'll see rainbows showing up at various other points. The method of calculation is pretty much the same.

And that is how we get a rainbow.