I took this photograph at dusk recently from the beach at Portobello, where Edinburgh meets the sea. As sunset pictures go, it’s not much to look at. But what caught my attention was the faint radiating pattern of light and dark in the sky. The light areas are where the sun’s rays are illuminating suspended particles in the air. The dark areas are where the air is unlit, because a cloud is casting a shadow. You may have seen similar crepuscular rays when the sun has disappeared behind the skyline and the landscape features on the skyline cast shadows in the air.
The rays in my picture appear to radiate from a point below the horizon, because that’s where the sun is…isn’t it?
No! Portobello beach faces north-east, not west. The sun is actually just about to set behind me! So why do the rays appear to come from a point in front of me? Shouldn’t they appear to diverge from the unseen sun behind me?
To understand why, we need to realise that the rays aren’t really diverging at all. The Sun is a very long way away (about 150 million kilometres), so its rays are to all intents and purposes parallel. But just as a pair of parallel railway tracks appear to diverge from a point in the distance, so the parallel rays of light appear to diverge from a point near the horizon.
The point from which the rays seem to diverge is the antisolar point, the point in the sky exactly opposite the sun, from my point of view. It’s where the shadow of my head would be. When I took the photograph, the sun was just above the horizon in the sky behind me, so the antisolar point, and hence the point of apparent divergence, is just below the horizon in the sky ahead of me.
For normal crepuscular rays, the (obscured) sun is ahead, and the light is travelling generally* towards the observer. The rays in the picture are anticrepuscular rays, because the light is generally travelling away from me. This was the first time that I had knowingly seen anticrepuscular rays.
*I say “generally” because the almost all of the rays aren’t travelling directly towards the observer. An analogy would be standing on a railway station platform as a train approaches: you’d say that it was travelling generally towards you even though it isn’t actually going to hit you.
The distressed man on the right is Garry McDougall. Garry’s just found out that his colour vision is not the standard-issue colour vision that most of us have. He made this discovery while watching my talk on the science of colour vision, in Kirkwall as part of the Orkney International Science Festival 2018.
Garry and I were part of a team funded by the Institute of Physics to perform at the festival. Also on the team were Siân Hickson (IOP Public Engagement Manager for Scotland) and Beth Godfrey.
Garry needn’t look quite so woebegone: he’s not colour blind, and he’s in plentiful company – about 1 in 20 men have colour vision like his.
How did Garry’s unusual colour vision come to light? In one of the demos in my talk, I compare two coloured lights. One (at the bottom in the picture on the right) is made only of light from the yellow part of the spectrum. The other (at the top) is made of a mixture of light from the red and green parts of the spectrum. If I adjust the proportions of red and green correctly, the red/green mixture at the top appears identical to the “pure” yellow light at the bottom.
Except that to Garry it didn’t. The mixture (the top light) looked far too red. By turning the red light down, I could get a mixture that matched the “pure” yellow light as far as Garry was concerned. But it no longer matched for the rest of us! To us, the mixture looked much greener than the “pure” yellow
light; the lower picture on the right shows roughly how big the difference was. This gives us an insight into how different the original pair of lights (that we saw as identical) may have appeared to Garry. It’s not a subtle difference.
We can learn a lot from this experiment.
Firstly, we’re all colour blind. The red/green mixture and the “pure” yellow light are physically very different, but we can’t tell them apart. “Colour normal” people are just one step less colour blind than the people we call colour blind.
Secondly, it shows that there’s no objective reality to colour. People can disagree about how to adjust two lights to look the same colour, and there’s no reason to say who’s right.
Thirdly, it shows that Garry has unusual colour vision. Our colour vision is based on three kinds of light-sensitive cell in our eyes. They’re called cones. The three kinds of cone are sensitive to light from three (overlapping) bands of the spectrum. Comparison of the strengths of the signals from the three cone types is the basis of our ability to tell colours apart. Garry is unusual in that the sensitivity band of one of his three cones is slightly shifted along the spectrum compared to the “normal” version of the cone. This makes him less sensitive to green than the rest of us, which is why the red/green mixture that matches the “pure” yellow to Garry looks distinctly green to nearly everyone else.
Garry isn’t colour blind. He’s colour anomalous. A truly red-green colour blind person has only two types of cone in their eyes. Garry’s kind of colour anomaly is quite common, affecting about 6% of men and 0.4% of women. It’s called deuteranomaly, the deuter- indicating that it’s the second of the three cone types that’s affected, ie the middle one if you think of their sensitivity bands arranged along the spectrum.
My thanks to Siân Hickson for the photographs.
A note to deuteranomalous readers
Please don’t expect the illustrations of the colour matches/mismatches above to work for you as they would have done if you’d seen them live. A computer monitor provides only one way to produce any particular colour, so the lights that appear identical to colour “normal” people (image duplicated on the right) will also appear identical to you, because, in this illustration, they are physically identical.
I was rather taken by the video above, which I first saw on Core77. I started wondering how many times you have to put the roll of silicone material through the machine to get satisfactory mixing of the two colours of material. The people in the video consider the job done after four passes. What does that mean in terms of the thickness of the red and white layers within the material?
The roll is a rather complicated object, so I worked with an idealised version of the real process, where the sheet emerging from the rollers isn’t rolled up, but cut into several pieces which are stacked up before being passed through the rollers again. I came up with the following:
After only 2 passes, the layers in the slab are too thin to see with the naked eye. And by some margin, too: there are over 600 of them and they’re only a fortieth of a millimetre thick. If you made a perpendicular cut through the slab, it wouldn’t appear to have red and white layers in it.
After only 4 passes, a standard compound microscope operating in visible light wouldn’t be able to resolve the layers in the slab.
After only 6 passes, the layers would be thinner than the width of the molecules of the silicone material. At this stage the concept of red and white layers no longer makes sense.
These results will only apply to material near the centre of the roll. It’s easy to see from the video that material near the edges is not mixed so well.
From the video, it looks like there are about 9 turns in the roll. Each time the roll is flattened by the rollers, those 9 turns are converted into 18 layers. The resulting sheet is rolled up and passed through the rollers again, multiplying the number of layers by 18, and so on.
This doesn’t work at the sides of the roll. We’ll ignore that complication, and work with a flat analogue of the actual situation. We’ll assume that we start with two long rectangular flat sheets of material, a white one and a red one, laid on top of each other. We’ll cut this assembly into 18 identical pieces, and make a stack of them; this stack will have 36 layers. We now flatten this stack in the rollers, cut it into 18 pieces, stack them up (giving us 648 layers), and repeat.
On emerging from the roller, the sheet appears, by eye, about 1.5 cm thick. We’ll assume that we start with two layers of half this thickness. The table below shows the number of layers and the thickness of each layer after 0, 1, 2, 3… passes through the rollers.
Number of passes
Number of layers
Layer thickness (m)
7.50 × 10-3
4.17 × 10-4
2.31 × 10-5
1.29 × 10-6
7.14 × 10-8
3 779 136
3.97 × 10-9
68 024 448
2.21 × 10-10
We can identify various milestones, as follows:
Limit of visual acuity. A person with clinically normal vision can resolve detail that subtends roughly 1 minute of arc at the eye. At a viewing distance of 30 cm, this corresponds to about 0.1 mm (10-4 m). The layers of material are much thinner than this after only 2 passes. If you made a perpendicular cut through the slab of material, after two passes you wouldn’t be able to see the layered structure. (This might not be true if the cut was oblique.)
Limit of standard light microscopy. A compound microscope working in visible light can resolve detail down to about 200 nm (2 × 10-7 m). The layers become thinner than this after only 4 passes.
Single-molecule layers. The question here is the number of passes needed before the layers are less than a molecule thick (at which point the idea of layers fails). The difficulty is that molecules of silicones are long chains, and these chains are almost certainly bent, so their size is ill-defined. This part of the calculation will be hugely approximate. We’ll be as pessimistic as possible, assuming that the molecules are roughly straight and that they lie parallel to the layers in the slab of material.
A common silicone material is polydimethylsiloxane or PDMS. This consists of a silicon-oxygen backbone with methyl groups attached. The lengths of carbon-silicon and carbon-hydrogen bonds are 1.86 × 10-10 m and 1.09 × 10-10 m respectively. So the width of the molecule is going to be, very, very approximately, of the order of 4 × 10-10 m. The layers are thinner than this after only 6 passes.
In the morning, as the sun climbs in the sky, my shadow gets shorter and shorter. In the afternoon, as sun gets lower and lower in the sky, my shadow grows again. At local noon, when the sun is at its highest, my shadow is as short as it’s going to be that day. My noon-time shadow is my shortest shadow.
How high the sun gets at local noon depends on the time of year. In the summer, it gets much higher than in the winter. On the day of the winter solstice, the noon-time sun is lower than the noon-time sun on any other day of the year. Therefore, of all of my shortest (noon-time) shadows, the one on the winter solstice will be the longest. It’s my longest shortest shadow.
Today is the winter solstice, and I’d hoped to photograph my longest shortest shadow on the beach at Portobello in Edinburgh. But implicit in all of the above is that the sun will be visible in the sky at the critical moment in order to cast the shadow. In Scotland in December this can be a tricky condition to meet, and it wasn’t quite met today. My shadow was barely visible. Since this idea came to me, I’ve only had one chance to photograph my longest shortest shadow, back in 2010.
In the previous post, we discovered that the kinetic energy of a drifting continent is of the same general magnitude as that of a moving bicycle and its rider – 1500 joules would be a typical figure.
I went on to calculate that, whereas it takes me only about 10 seconds to get my bike up to full speed, it would take me hundreds of years to get the continent up to its tiny full speed were I to put my shoulder against it and push (assuming that it was perfectly free to move). How can this be, when the amount of energy that I’m giving each of these objects is the same?
The problem is that when I push the continent, I am, effectively, in the wrong gear.
On a bike with gears, you’ve got a range of choices about how you power it: you can ride in a high gear, pedalling slowly but pushing hard on the pedals, or ride in a low gear, pedalling more quickly but pushing less hard on the pedals. There’s a simple tradeoff: if you want to pedal half as fast, you’ve got to push twice as hard for the same effect.
But there’s a limit to how hard you can push on the pedals, which means that if you move up too far up through the gears, there comes a point where you can no longer make up for the decreased pedalling rate by pushing harder on the pedals, and the power that you can supply to the bicycle falls.
Anyone who’s tried to accelerate a bicycle when they are in too high a gear will have experienced this problem, and it’s what I experience when I try to push the continent directly. Because the top speed of the continent is extremely low (about the speed of a growing fingernail), I’m necessarily pushing it very slowly as I accelerate it. This means that to give it energy at the rate that I want to (1500 joules in 10 seconds, like the bike) I would have to push it impossibly hard – the force needed is about the same as the weight of a 300-metre cube of solid rock.
Is there a way that we can put me into a lower gear, so that I can push with a force that suits me, over a longer distance, and still apply the very high force over a short distance to the continent?
Yes. Just as we’ve all used a screwdriver as a lever to get the lid of a tin of paint off, so I could use a lever to move the continent. Similarly to the bike gears, the lever allows me to exchange pushing hard over a small distance with pushing less hard over a longer distance. To do the job, the lever would need to be long enough to allow me to push, with all my might, through a distance of about 2.5 metres, with the short arm of the lever pushing the continent. We’d need an imaginary immoveable place for me to stand, and we could use the edge of the neighbouring continent as the pivot (just as we use the rim of a paint tin as the pivot). The catch is the length of the lever: if the short arm was 1 metre long, the long arm would be about 1.5 million kilometres long.
Simon Gage of Edinburgh International Science Festival suggested a more compact arrangement: a bicycle with an extremely low gear ratio, with the front wheel immobilised on the neighbouring continent (assumed immoveable), and the back wheel resting on the continent we’re trying to accelerate. A transmission giving 17 successive 4:1 speed reductions would do the job nicely. Ten seconds of hard pedalling would get the continent up to full speed. To me on the saddle, it shouldn’t feel any different to accelerating my bike away from the lights.
A wee caveat. This is a thought experiment, and we’ve swept some fairly significant engineering issues under the carpet. The rearmost parts of the power train would be moving at speeds that are literally geological, so in reality it would take me years of pedalling to take all of the slack and stretch out of the system. These parts would also be transmitting mountainous forces, and so they’d need to be supernaturally strong. There will be frictional losses. And then there’s the issue of transmitting a gigantic force to the continent through the contact of a bike tyre on the ground.
What force is required to accelerate the Eurasian plate to top speed in 10 seconds?
The top speed of the plate is 3.2 × 10-10 ms-1. If I accelerate it uniformly, its average speed will be half of this, and so in the 10 seconds over which I hope to accelerate it, it will travel 1.6 × 10-9 m.
Now W = fd
where W is the work that I do on the plate (ie the kinetic energy that I give it), f is the force that I apply to it, and d is the distance through which I push the plate. Rearranging gives us
f = W/d
We know W from the previous post (it’s 1500 joules) and we’ve just calculated d. Thus f works out at about 9.4 × 1011 newtons.
For comparison, a 300-metre cube of rock of density 2700 kg m-3 will have a weight of (300 m)3 × 2700 kg m-3 × 9.81 m s-2 = 7 × 1011 newtons roughly.
When a lever is used to amplify a force, the ratio of the lengths of the arms of the lever needs to be the same as the ratio of the two forces. Suppose that I can push with a force equal to my own body weight, about 600 newtons. If I’m to use a lever to amplify my push of 600 N to a force of 9.4 × 1011 N, the ratio of the lengths of the arms needs to be (9.4 × 1011)/600, or roughly 1.5 × 109. So if the short arm of the lever is 1 metre long, the long arm needs to be about 1.5 × 109 metres long, which is 1.5 million kilometres. For comparison, the Moon is about 400,000 kilometres away.
To do 1500 joules of work with a force of 600 N, I’d need to push over a distance of 2.5 metres (because 600 × 2.5 = 1500).
The bicycle gearing
I estimated that it takes me 15 pedal revolutions to get my bike up to full speed. Knowing the length of the pedal cranks, I know the total distance that I have pushed the pedals through, and I know how much work I have done on the bicycle – 1500 joules. (I’m ignoring energy losses here, because they are small at low speeds on a bike and the calculation is highly approximate anyway). Using work done = force × distance, this gives an average force on the pedals of about 94 newtons.
The 17 stages of 4:1 reduction mean that the back wheel is rotating 417 = 1.7 × 1010 times slower than I’m pedalling. The pedalling force is amplified in the same ratio, to give a force on the teeth of the rearmost gear of 1.6 × 1012 newtons. We now have to allow for the fact that the radius of the rear wheel is about twice the length of the pedal crank. This roughly halves the force available at the rim of the rear wheel, giving a force of about 8 × 1011 newtons, which is close to what we need.
…is broadly similar to the kinetic energy of me and my bike as I pedal along.
According the the theory of plate tectonics, the outer layer of the Earth is divided into a number of separate plates, which very slowly drift around, opening and closing oceans, causing earthquakes, and thrusting up mountain ranges.
A moving body has energy by virtue of its motion: kinetic energy. Kinetic energy is proportional to a body’s mass and to the square of its speed.
Now tectonic plates move extremely slowly: the usual comparison is with a growing fingernail. But they are also extremely heavy: tens of millions of square kilometres in area, over 100 km thick, and made of rock. I wondered how the minute speed and colossal mass play out against each other: what’s the kinetic energy of a drifting tectonic plate?
There are so many variables, that vary such a lot, that this calculation is going to be extremely approximate. But the answer is delightfully small: the kinetic energy of the tectonic plate on which I live, as observed from one of the plates next door, is about the same as the kinetic energy of me and my bike when I’m going at a reasonable pace: about 1500 joules.
This is a fun calculation to do, but we shouldn’t get carried away thinking about the kinetic energy of tectonic plates. Plates are driven by huge forces, and their motion is resisted by equally large forces. The mechanical work done by and against these forces will dominate a plate’s energy budget in comparison to its kinetic energy.
But the calculation does provoke an interesting thought about forces and motion. I can get my bike up to full speed in, say, 10 seconds. If the Eurasian plate were as free to move as my bike, and I were to put my shoulder against it and shove as hard as I could, it would take me about 500 years to get it up to its (very tiny) full speed.
In both cases, I’m giving the moving object roughly 1500 joules of kinetic energy. How come I can give that energy to my bike in a few seconds, but to give it to the plate would take me centuries?
I’ll return to that thought in a later post.
Depending on how you count them, there are 6-7 major tectonic plates, 10 minor plates, and many more microplates. The plates vary hugely in size, from the giant Pacific Plate with an area of 100 million km2, to the dinky New Hebridean plate, which is a hundred times smaller. The microplates are smaller still. Plates also vary a lot in speed: 10-40 mm is typical.
I’m going to be parochial, and choose the Eurasian plate for this calculation.
Let’s call the area of the plate a and its mean thickness t. Its volume is then given by at, and if its mean density is ρ, then its mass m is ρat.
A body of mass m moving at a speed v has kinetic energy ½mv2. So our plate will have kinetic energy ½ρatv2.
The area of the Eurasian plate is 67,800,000 km2 or 6.78 × 1013 m2, and its speed relative to the African plate is (the only speed I have) is given as 7-14 mm per year. We’ll use 10 mm per year, which is 3.2 × 10-10 ms-1. The thickness of tectonic plates in general varies roughly in the range 100-200 km depending upon whether we are talking about oceanic or continental lithosphere; let’s call it 150 km or 1.5× 105 m. The density of lithospheric material varies in the range 2700-2900 kg m-3; we’ll use 2800 kg m-3.
Putting all of these numbers into our formula for kinetic energy, we get a value of 1500 joules (to 2 significant figures, which the precision of the input data certainly doesn’t warrant).
Now for me and my bike. I weigh about 57 kg, my bike is probably about 10 kg. Suppose I’m riding at 15 mph, which is 6.7 ms-1. My kinetic energy is almost exactly…
The closeness of these two values is unmitigated luck*, and we shouldn’t be seduced by the coincidence. Just varying the speed of the plate in the range 7-14 mm would cause a 4-fold change in kinetic energy, and there’s the variability in plate thickness and rock density to take into account as well. The choice of bike speed was arbitrary, I guessed the mass of the bike, and I’ve since realised that I didn’t account for the fact that the wheels of my bike rotate as well as translate.
However, what we can say is that the kinetic energy of a drifting continent is definitely on a human scale, which leads to a new question:
Suppose the Eurasian plate were as free to move as my bicycle, and that I put my shoulder against it and shoved, how long would it take me to get it up to speed?
From the figures above, the mass of the plate is 2.85 × 1022 kg. If I can push with a force equal to my own weight (about 560 newtons) then by Newton’s 2nd Law I can give it an acceleration of about 1.96 × 10-20 ms-2. Rearranging the equation of motion v = at, where v is the final speed, a is the acceleration, and t is the time, then t = v/a. Inserting the values for v and a, we get t = 1.6 × 1010 seconds, or about 500 years.
* I didn’t tweak my assumptions: what you see above really is the very first version of the calculation!
When you stir a cup of tea, the surface of the rotating liquid develops a dip in the middle. The faster you stir, the deeper the dip. But the liquid surface is quite uneven; to get a smooth surface, throw the spoon away and spin the whole cup continuously. Once the liquid inside has caught up with the cup, and everything is turning at the same speed, the liquid surface forms a beautiful smooth curve known as a paraboloid of revolution.
Sarah McLeary and I are applying this idea to make thin paraboloidal plaster shells. We spin a bucket on a potter’s wheel (Sarah is a potter), and pour plaster into it. The plaster rapidly flows until its surface forms a deep paraboloidal curve, and then sets. We now use this cast, still spinning, as a mould, and cast a thin layer of plaster inside it, to make a paraboloidal shell.
That’s our first attempt above. It’s about 20 cm across and 3 mm thick. It might look like a part of a sphere, but in a profile view it’s easy to see that the curvature is tightest at the base and gradually decreases up the sides, as you’d expect for a paraboloid.
I love the fact that we didn’t decide what shape this shell was going to be: physics did.
There’s lots of experimentation ahead. When we’ve got the hang of it, I’ll explain our methods in more detail. But as this picture of our third attempt shows, we haven’t quite cracked it yet.
This image is my version of Edward Adelson’s checkershadow illusion (with a little inspiration from Magritte). It’s a photograph of a real, physical scene.
Take a look at the central square of the checkerboard, and the square indicated by the arrow. Which is lighter? Quite clearly, it’s the central square, isn’t it?
Remarkably, the central square actually emits less light than the square indicated by the arrow! You could use a light meter to check this claim, but it’s easier to verify it directly by using a piece of card with two holes cut in it to mask off the rest of the image.
Some people will tell you that this image shows you how easy it is to fool your brain. But it does the exact opposite: it shows you what a marvellous piece of equipment your brain is.
Think about the checkerboard itself, and the materials it’s made of. The arrowed square is coated with dark grey paint, and the central square is coated with light grey paint—and that’s exactly what you perceive.
The shadow cast by the pipe means that the light-grey central square is more dimly lit than the dark-grey arrowed square, so much so that it actually reflects less light into your eye than the arrowed square. But your brain cleverly manages to determine the actual lightnesses of the physical surfaces, despite the uneven lighting. Isn’t that a good thing for your brain to do?
If you still don’t believe me, try this thought experiment. Imagine that you live in a forest where there are two kinds of fruit. One is light grey and poisonous, and the other is dark grey and nutritious. Two of these fruits hang next to each other, but in the dappled forest light the (light grey) poisonous fruit is in shadow, and the (dark grey) nutritious fruit is in bright light. Suppose that the depth of the shadow is such that the light-grey poisonous fruit actually reflects slightly less light into your eye than the dark-grey nutritious fruit, just as with the two squares in the picture above. Would you really want your vision to tell you that the poisonous fruit was the dark one and therefore the one to pick? Or would you want it to discount the irrelevant effect of the shadow and tell you which fruit was actually dark and which was actually light (and would kill you)? I know what I’d want.
I think that it is wrong to call this effect an illusion (and so does Adelson). There is nothing illusory about what you see. You perceive the useful truth about the scene in front of you.
How fast does a plucked guitar string move? It’s a complete blur, so surely it’s travelling at a terrific speed. 50 miles per hour? 100 miles per hour? What do you think?
Walking pace. A typical speed for the middle of a guitar string given a good twang is walking pace. And that’s the middle of the string. Near its ends, it’ll be moving much more slowly.
How can that be so? Well, although the string is going back and forth hundreds of times a second, it’s only travelling a few millimetres on each trip. So the distance that it travels in each second isn’t as much as you might expect. It certainly isn’t as far as I expected.
If you think that the string moves slowly, what about the body of the guitar? The string itself radiates very little sound into the air; its job is to set the body of the guitar vibrating. The body of the guitar, with its much larger area, is much more effective than the string at setting air into motion. Yet we can’t even see the body vibrating. At what snail’s pace must it be moving?
Remember also that the air molecules on which the guitar body acts are already travelling at something like 500 metres per second. Isn’t it astonishing that the sub-pedestrian movements of the guitar affect the movement of the air molecules enough to produce a sound that we can easily hear?
Suppose that we look near the centre of the string, where its movement is the greatest. Shortly after being plucked, the width of the blur that we see is going to be something like 5 mm. So for every complete oscillation, the string does a round trip of about 10 mm.
The frequencies of the strings on a standard 6-string acoustic guitar are (to the nearest whole number) 82, 110, 147, 196, 247 and 330 hertz (one hertz is one oscillation per second). If we multiply these frequencies by the 10 mm round trip, it tells us how far the centre of each string travels in one second, that is, its average speed. I’ve converted these speeds into metres per second. For comparison, a brisk walk at 4 mph is about 1.8 ms-1.
Mean speed of middle of string in metres per second (to 2 sig. fig.)
Our brisk walk is right in the middle of this range. And remember, we’ve done the calculation for the part of each string that’s moving the most. Near its ends, each string will be moving much more slowly than this.
Complication 1 – how long is the round trip really?
So far, we’ve assumed that each part of the string does a simple back-and-forth movement along a straight line, but if you carefully watch a vibrating guitar string you’ll see that the string often moves in an irregular but roughly elliptical orbit. The wire-wound lower strings show this most clearly; you can see a hint of it in the image at the top. This makes the round-trip distance a bit longer than the 10 mm that we used in the calculation earlier. Does this affect the string’s average speed much?
We’ll take the extreme case where each part of the string moves at constant speed in a circle of diameter 5 mm rather than along a straight line 5 mm long. The circumference of this circle will be mm, or about 15 mm. So the speeds of the strings (in this rather unlikely extreme case) will be about 50% greater than the ones listed above. They are still hardly impressive.
Complication 2 – the peak speed
So far, we’ve calculated the mean (average) speed of the string over its round trip. However, unless it’s moving in a perfect circle, its speed changes constantly, and its peak speed will be higher than its mean speed. How much higher?
Imagine that part of the string is vibrating back and forth along a straight line in the simplest possible way. At one end of the movement the string is stationary as it changes direction. It then speeds up, reaching its peak speed at the centre of its range of movement. Then it slows down until it reaches a halt again at the other end of the movement and changes direction again. How do we calculate the peak speed if we know the time taken for the round trip?
You’ll need to know a bit of maths for the next bit. The simplest vibration of the string is where each part undergoes simple harmonic motion, that is, where its position varies sinusoidally with time. This means in turn that the velocity of the string also varies sinusoidally in time. So we need to ask: how does the mean value of a sinusoid compare to its peak value?
Consider the function , for half a cycle, that is, for theta from to radians. We construct a rectangle between these limits, such the rectangle’s area equals the area under the sine curve. The mean value of the sine function is the height of the rectangle, which its area divided by its width (which is ). So we need to work out the area under the sine curve between and , which we can do by integration:
We constructed the rectangle to have area . Its width is , so its height, and therefore the mean value of the sine function, is . The height to the peak of the sine curve is 1, so the peak value of the sine function is times its mean value. This means that the peak speed of our guitar string is times, or 50% more than, its average speed. Again, nothing to write home about.
(Rather pleasingly, this ratio of is the same ratio that we got earlier in Complication 1. It means that the peak speed of a particle doing simple harmonic motion with a given amplitude and period is exactly the same as the (constant) speed of a particle moving in a circular orbit of radius with period . This isn’t a coincidence. It arises because the circular motion can be considered as two linear simple harmonic motions at right angles to each other.)
It’s sometimes said that if you sit an immortal monkey in front of an equally durable typewriter and leave it to tap randomly away at the keys, then eventually it will produce the entire text of Richard III (or any other Shakespeare play of your choice), completely by chance. All you have to do is wait long enough.
I was thinking about this one day, and also thinking about air molecules. In the room I’m sitting in at the moment, there are at least 1,000,000,000,000,000,000,000,000,000 air molecules, all frantically dashing around bumping into each other. How often, I wondered, do little clusters of these molecules fleetingly arrange themselves, by chance, in arrangements that we would regard as being somehow regular or remarkable?
It’s not possible for me to directly observe air molecules, so instead I used my computer to make a 2-dimensional simulation of some atoms of gas doing what atoms of gas do. I set it going, and waited…and waited…and waited…
I promise you that my simulation didn’t involve any secret forces drawing atoms towards certain positions. The movements and collisions of the atoms all occurred in accordance with the laws of mechanics. But I did cheat a little. Can you figure out how?
Molecules and atoms
In case you’re fidgeting on your chair wondering why I started talking about air molecules but finished talking about gas atoms, let me explain.
Nearly all of the air is nitrogen and oxygen. Nitrogen and oxygen atoms are essentially spherical. But in the air, nitrogen atoms are bonded together in pairs to form nitrogen molecules that are, roughly speaking, a stubby rod shape. The same goes for oxygen. Now a collision between two moving rods is much more complicated than a collision between two spheres, because the rods can spin end-over-end in a way that spheres can’t. In fact, I’m not sure that I know how to do the calculations. As the point of the video could be made just as well using atoms rather than molecules, I did the simulation using atoms. If you like, you can think of them as atoms of helium or argon, which do go around on their own.