We're here in Amsterdam. Today, a ship can travel at sea with relative safety. But in the 17th century, many of the ships that left this harbor never came back. The reason so many sailors' lives were lost is because the crew answer a simple question--where are we. In this unit, we'll learn about simple harmonic motion and in doing so, we'll learn one method for determining the position xe.
Answering the question of where am I when you're at sea is not so easy, but luckily you'll only need to know two pieces of information. The first is latitude and may be you remember from Unit 1 that lines and latitude run parallel and this 0° line of latitude is known as the equator. The second piece of information we need is longitude. The lines of longitude run from north to south and since they run this way, they actually tells us our position east to west. Now it turns out that one of these is very difficult to find and the one is very simple. And let me give you a hint, you have used may be in one of the previous units an effective method of calculating one of these--are the latitude or longitude. Can you tell me which one?
In Unit 1, we talked about the Tropic of Cancer, but I'm not sure if we actually used that word. The city of Siena where Eratosthenes saw the sun shining vertically down the row is located on the Tropic of Cancer. The sun isn't always directly above the Tropic of Cancer as it happened to be on the summer solstice. In fact, due to the tilt of the earth, the sun can be directly overhead anywhere between the Tropic of Cancer, the equator, and another imaginary line 23.5° south of the equator known as the Tropic of Capricorn. The sun is directly over the Tropic of Cancer on the summer solstice and directly over the equator and what's known as the equinoxes--it's directly over the other tropic on the winter solstice. I'm using a northern hemisphere bias here. If you're in the southern hemisphere, you'd probably refer to this as the winter solstice. Now thinking that's what we did in Unit 1, let's say you're standing at some unknown latitude. Where one where your latitude would just be this angle here which I've labeled α, and let's say it happens to be one of the equinoxes said the sun is directly above the equator and we do an experiment. You look at a vertical object, measure its shadow, and do some trigonometry to find that the angle here I've labeled α is equal to 40°. So I want you to tell me for this measured angle of 40° and remember this is when the sun is highest in the sky and the sun is directly overhead on the equator so it's one of the equinoxes, what is your latitude. Enter your answer in degrees here.
Well, this one is not so bad. The answer is 40° and let's look at why. Well, here comes the sun's rays, arriving in parallel all over the surface of the earth. These lines are supposed to be parallel. Well, let's replace our little guy with just a vertical object. Well, check it out-as our rays drawing in some parallel lines has really helped us solve this problem because look at what we have here--our angle α which by definition--that's the longitude-- that's how we define longitude and that's exactly equal to this angle here, which we also happen to call α there, but hey, good coincidence they're the exact same angle. And so if we measure this angle to be 40°, this must also be 40°. We'll see things change a little bit. If it's not on the equinox, it gets a bit more tricky but I think you can handle that and you'll do that in the premise that at the end of this unit.
So we're halfway then--we've got one of the two numbers we need to know our position on the earth--latitude. So, this unit is about covered up but not quite. Longitude is much more difficult. Do you have some intuition for why longitude is so much more difficult. Let's take a look at these lines. Here we've drawn five lines of the longitude and in the middle we have, of course, the equator. Now, the equator wasn't chosen randomly. The earth rotates around an imaginary axis that goes through the center of the earth. Now the equator define this sort of flat plane through the middle of the earth-- that makes a right angle with the axis of rotation. The equator isn't some sense real. It's based on actual physical property of the earth. Now let's see. Which of these lines is the 0 line of longitude. Is it line 1, 2, 3, 4 or 5, and I've labeled them here. If you can't tell these to my drawing, you can also select this option.
And the answer is--you just cannot tell. It could be any of these. It could be a line that I haven't even drawn. The truth of the matter is longitude is in many ways completely arbitrary. Right now the prime meridian or 0° longitude line passes through Greenwich, England. In the past, it's passed through Rome; Florence, Italy; Paris, France; Stockholm Sweden and many, many other locations. The fact that this line is so arbitrary, actually hints that how difficult longitude in general is to calculate.especially when you're at sea.
One method to get from A to B across a vast body of water is known as dead reckoning. So what's dead reckoning. In dead reckoning, we start with a known location--city A. Then we're going to keep track of our ship's heading and its speed to determine where in this route from A to B we might actually B located. So for example, let's say our ship's started off moving for 1 hour east at a speed of 10 km/h. Well, east is this way and 10 km/h for 1 hour means I've moved 10 km, and I can follow this sort of thinking as long as I keep good track of duration, heading and speed. So 2 hours at southeast at 8 km/h but by 1 hour south at 5--so you get a feel for where you are just by using this sort of dead reckoning method. Of course, this method has some problems. For example, how do you know the speed of your boat.
So how do you calculate speed? Hopefully, your boat is going forwards to the water with some velocity we'll call v. Well, actually anything use to calculate your speed at sea is called a log and why is it called that. Well, the reason it's called a log is because what mariners used to do was throw an actual log over the back of the boat. They would let the log float as the boat continue forward and is it did this, they plant some rope. Tied in this rope were some knots and by counting the number of knots, they can refer the boat speed. This is where the term knots comes from in reference to a ship's speed. Now you maybe saying this is a ridiculous way to measure speed. How accurate could this whole procedure be anyways--well, you're right. This is now the best way of measuring speed, but it was somewhat functional at the time. So what do you think are the problems with this method. Is it a problem that the line could stretch, that it's hard to accurately measure time or what about currents, may be there's some lateral currents in the water behind the boat. Check all that you think apply.
Well, the line could certainly stretch--that would throw off your measurement. It actually was hard to accurately measure time. They often use things like hourglasses and some mechanisms but it is still difficult to be accurate and currents is the big one. If all this water is also moving to the right with some speed, well I would expect this log to more or less follow behind your boat-- that doesn't mean your boat isn't moving--it just means that it is not moving relative to the water. Clearly we knew the better method, what can we use.
It turns out what you need is a good timekeeping device or a chronometer and if it's not clear why, why don't we go to Leiden in the Netherlands and talk to Tjeerd Bucker. You need a timekeeping device to find longitude because they the world turns on it's axis in 24 hours, and we've probably seen the sun at its highest point at 12 o'clock. When you're, for instance at the ocean, and you look at the sun and you see it reached its highest point, you can say it's 12 o'clock and you look at the clock which you take with you it shows the time of your home harbor. For each hour in difference there is a 15° difference in longitude.
So the question is how do we find the time. Actually, sorry that's not the question. Finding the time is easy--we'll just look at the sun. The real question is how I can keep the time from my local harbor when I go to sail at sea. What I need is periodicity. I need something that happens at regular intervals. For example, we're all familiar with the ticking of a clock and that seems like an obvious solution but clocks are difficult to make--what else could we use that has a regular periodic nature to it. One attempted solution to this problem of finding something periodic that we could use as a timekeeper came from our good friend Galileo-- standing as he always does next to the Leaning Tower of Pisa. Now Galileo had invented some incredible telescopes and with these telescopes, he was able to observe the ring around the planet Saturn, mountains on the moon, dark spots on the sun and the planet Jupiter and its four moons, each of which orbited around in a predictable fashion. Now of the four discoveries Galileo made with his telescope, which do you think could somehow be used for keeping time at sea.
And this is a tricky one, but the keywords are used when describing the moon's Jupiter where predictable and regular. Predictable means Galileo could make a chart or some catalogue of the times when each of these moons were supposed to let's say go behind the planet Jupiter. If we're looking at the zoomed in version of Jupiter, we could tell when this transit occurred because the moon will go from being visible to no longer visible when it passed behind the planet, and it didn't matter if you're observing over here in Italy or may be over here somewhere in Africa let's say--this would happen at the exact same time. It doesn't matter if you're Galileo here in Italy or some sailor out in the ocean. You would've observed this transit to occur at the exact same time. So if you knew for instance that this occurred at 8 o'clock in Italian time, this sailor could immediately get a reference for what time it was back at home. By comparing that with their own clock, they could determine their longitude. Seems pretty brainy right?
I want you to tell me--what do you think were some of the problems we're trying to use this telescope to look at Jupiter to ascertain your time? Which of the following do you think our problems you might run into when trying to use Galileo's method of observing the transits of Jupiter's moons to keep time? Do you think it's a problem when you can only see Jupiter at nighttime or that telescopes are hard to use on a ship that's bouncing at sea or may be clouds could get in the way and obscure your vision--check all that apply.
Well, this was an easy one. No problems. I don't want to only be able to tell time at night under special instances when the moons are passing behind the planet. Furthermore, telescopes are going to be almost impossible to use on the ship as it bounces around on our waist. This is going to wind up being the biggest problem, that telescopes are so hard to use on a ship. So Galileo actually invented something that he thought would help adjust this problem. What he invented was this ridiculous contraption called the celatone which you can see here in the black and red and it's basically just a helmet where one eye had the telescope over it and the other eye was just an open hole, and the idea was you could sight and find the idea for where Jupiter was through this hole and then what he's doing then in focus and watch the transit of the moons through the telescope. As you can maybe imagine this didn't work so well.
All right. Enough about these crazy contraptions. Let's go to clock. Actually, let's go to something even better than a clock--it's called a chronometer. And let's think, what are the necessary properties that our chronometer has to have if it's going to be used for timekeeping device at sea. Does our chronometer have to run forever without any external source of power. Does it have to be periodic--what periodic means, it has some cycle that repeats itself. If it must be periodic, does it have to isochronous as in does that cycle have to take the exact same amount of time every single iteration. Finally, does it have to be predictable and by predictable I mean even if we know it's isochronous, do we know exactly how long each cycle takes. Check whatever answers you think apply.
Now, I'm guessing is there could be 3 options. And I'm going to walk through backwards to talk about why they are all so vital to our chronometer that we want to make. Even if I device the pit itself and does so in a consistent manner, if we don't know how long those repetitions take, well what good is it. If 10 ticks of our clock corresponds to 10 seconds or 15 or 20, but we just don't know, well it's not a very useful clock. Obviously, it has to be isochronous if one tick of the clock is one second and the next is seven seconds, well again I use this clock. And all of these required that something be happening over and over at regular intervals, which means periodic. Well, it would be nice if our clock could run forever without power. If you know how to make a clock that does this, you just invented a perpetual motion machine and send me the diagram please. Now, what's really cool is there is a whole field of physics dedicated to phenomena that meet this criteria, simple harmonic motion. Underlying this field of study, simple harmonic motion is the concept of equilibrium. So let's talk about that.
What is equilibrium? Well, let's think about it as a state of balance. And that balance I mean well, balance forces. Before we get too deep into a definition, let's talk about some examples. For example, let's consider a ball in a well. If I release this ball from right here, well we all know what's going to happen. It's going to roll back and forth and eventually it's going to settle in one spot. A place where it settles somewhere near here is its equilibrium. If you disturb the ball from its equilibrium, say by shifting it a bit to the right, well now it's on a hill and we could do some force decomposition, and we would see that there's actually a force pointing down the hill. It's going to tend to restore this object back to its equilibrium. And this is such an important concept given its own sentence here. If I'm being specific, I should change the title to stable equilibrium. This idea that when disturbed the object would try to return to equilibrium that's the characteristic of what's known as stable. There is also semi-stable and unstable equilibrium, but we're not going to talk about those today because this is the sort of equilibrium that gives you the simple harmonic motion we need for our chronometer. There's lots of other systems that have this exact same property. For example, if you have a mass hanging from a spring Its default state just hanging in equilibrium, but if I pull the mass down, the spring will exert a force back upwards which tends to restore it to equilibrium. Likewise with a pendulum or a mass swinging from a rod, its default state, its equilibrium is vertical. If we displace it from vertical, there will be some component of force in this case gravity, which tends to restore it. There is one more key point for simple harmonic motion. The criteria that must be fulfilled if we're going to get isochronous periodicity that we want, and this one I want you to complete for me. So, you could look at this for example as I move this ball further and further up the hill further from equilibrium, does the force tending to push it back increase, decrease, or stay the same? Enter your answer here.
The restoring force must grow. In fact the restoring force has to be proportional to this displacement from equilibrium. Only then will we have our ball making a back and forth motions, our spring making up and down motions, and our pendulums swinging with equal time intervals between well swings or bounces up and down or back and forth rolling. Let's take a deeper look at one of these examples--specifically, the mass on the spring.
When we think of a mass on a spring, there's only two numbers initially that we really care about. One is the mass of the block and the other is K, which is a constant associated with the spring that measures its stiffness. For example, a rubber band sort of spring. And it would have a very low K since it's not very stiff. The springs in your car on the other hand would have a very high K because they're very stiff and hard to compress. Now right now this mass spring totally in equilibrium. This is exactly where the mass wants to be. Let's imagine displacing this mass from equilibrium. So let's move the center of mass over to here. So we've displaced our spring from equilibrium by a distance that I'm going to call Xmax, and you'll see why I called it that in a second. Now if I pull this mass back to Xmax and release what's going to happen is it's going to undergo a series of oscillations. In fact, if we assume that this surface is frictionless it's going to oscillate forever. Of course that would never happen in real life but let's talk ideal-wise here. In fact, this mass will swing through the equilibrium position here and all the way back to distance of -Xmax from the equilibrium. So now you can see the reason why called this max is because as these oscillations continue this is the farthest from equilibrium that the mass will ever reach. Now what I want to do is analyze this spring in terms of energy, which we talked about a ton in the last unit. Well it turns out that the potential energy stored in the spring is equal to half the spring constant K times x². X in this case would be whatever the current displacement from the equilibrium is. So when the mass is at the equilibrium, x will be 0. When the mass was over here, it will be Xmax. Maybe here it would be 1/2 of Xmax and so on. And so if we looked at let's say three possible positions for where the mass could be in its back and forth oscillations. So the position 1, equilibrium. Position 2 and this is sort of midpoint. And position 3, Xmax. So you need to tell me for each of these three positions as the spring is oscillating back and forth what kind of energy does it have at each of these positions? Check all boxes that you think apply. And remember we're talking about kinetic energy, the energy of motion and potential energy, which in this case is spring potential energy and not gravitational potential.
Well let's start by thinking about position 3. When the mass is way back here as far away from equilibrium as it's going to go it has only potential energy. All the energy is stored in the spring for an instant the block stops moving before it changes direction making the kinetic energy 0. Let's turn to position 1 next where here the spring isn't stretched out. It's at equilibrium. X is 0, so there can be any potential energy. So what happened to all the potential energy that was storing the spring back here? Well, it must've been converted to kinetic energy. At position 2, midway between positions 1 and 3, I know the spring is stretched somewhat so it has potential energy but it's not as stretched as much as it was at position 3. So some of that potential energy must've also been converted to the energy of motion, kinetic energy. So these are the answers.
Then we have a qualitative understanding of the energy involved with this spring moving back and forth. Let's do some actual calculations. So let's say we told you the mass, the spring constant, and Xmax the amplitude of oscillation. Could you tell me the maximum velocity this object will reached and remember think in terms of energy. We know the potential energy it has when it's back here. We know something about the kinetic energy, at least the equation which describes kinetic energy over here. Enter your answer here in m/s.
Well, let's see. How do we solve this. Well, I know that the potential energy over here is going to be equal to the kinetic energy it has when it passes through the equilibrium. Remember at the the equilibrium point, there's no potential energy. All of that energy must have been converted into kinetic energy. Okay, so let's do the math. Potential energy gets converted to kinetic energy. Use the appropriate equations and when I solve for V--that should be a max and I'm solving for the Vmax as well. I get this for an answer and plugging in the appropriate values--I get an answer of 4.5 m/s. Now something that's interesting here is that this could've been positive number or a negative number--a positive number would have corresponded to when the mass is moving this way and negative would have corresponded to when the mass is moving that way. In fact, there's a lot we can learn about simple harmonic motion by just calculating velocity, keeping track of sign and plotting it on a diagram versus position. What do I mean by that? Well, I'll show you.
Okay. So this next part what we're going to talk about I have to admit is a little bit tricky. It will require some serious focus, and it's something I was never shown in school but after having seen this what I'm about to show you I had to say my mind was fairly blown. So just for some anticipation, I want to show you what we have. We have a one-dimensional motion. We have a mass moving back and forth in a straight line on a spring. I don't see any right triangles here, do you? I don't think so. Yet, what we're going to get pretty soon is an entirely new way of looking at trigonometry— the science of right triangles. This graph is a graph of position versus velocity so on the x axis here we have position and I've divided by Xmax, the maximum possible position. So if I were to put a dot here that would mean that at some point in time the mass was all the way at the maximum possible position. That's why this number is 1 and notice that I put it here. I didn't put the dot here or here. And that's because on the vertical axis, I have velocity. Here velocity can be as much as positive Vmax, which we calculated earlier. Or negative Vmax. Positive this way. Negative that way. But we know that for the instant when the block is over here its velocity is equal to 0. It stops for that instant before it gets pulled back to its equilibrium. Okay. So this is a point on our graph of position versus velocity. Let's see if we can fill out some more points. I promise you we are going to see something very interesting. I've given you four possible suggestions for points that should exist on this graph. Can you select the appropriate one?
And you put the point to say that is this last one, all the way at the end. This corresponds to the mass having fully compressed the spring and again it's stopping with 0 velocity for just that instant. These other three points don't really make sense because they correspond to since they're on this horizontal line here--this is the line of 0 velocity. Anytime the mass is in between these extreme points of Xmax and -Xmax, well it's going to be moving--so this doesn't make sense. I will also tell you that when the object is swinging through so let's say it's coming from this end and swinging through equilibrium. When it goes through the equilibrium point, its velocity is maximized. It's going very fast, as fast as it'll ever will go but will in this case it's going to the left. Leftwards means negative, so that will correspond to this point. We've already talked about what happens--it's going to compress the spring, stop for a second, and then swing back through. When it get through equilibrium on the swing through, well then it's going to the right with maximum speed again corresponds to this point.
Okay, so we got four points on a graph, let's fill in a few more and see if a shape starts to emerge. I want to know when X=1/2 Xmax, so that would correspond to this position right here or this position here on the graph and the mass is moving to the right, what is V/Vmax. Now you maybe thinking but Andy, you haven't given me k, you haven't given me Xmax. How can I calculate anything? Well, you have two options-- You could make up those values and I promise you when you calculate this out it won't matter what values you chose or you can just manipulate these algebraic symbols until you solve for V/Vmax. That could be a little trickier--you've either option to shot. I know they are both tough. If you need help, don't be discouraged. These are tricky topics we're dealing with. Feel free to go to the forums.
Well, normally I would solve this algebraically and I think that's the best way to do it and I encourage you to do it in the forums. But since we already calculated Vmax numerically using the numbers I had made up last time, k = 500, m = 1, and Xmax = 0.2. Let's go ahead and get ourselves an answer. So how do we do this? Well, I know that the system has a total amount of energy. When it is at the extreme points, the energy is all potential. In the equilibrium, it's all kinetic and this middle point here is both kinetic and potential. I can phrase this mathematically as the total energy is going to be the kinetic plus the potential. Well the total energy is just whatever energy I had when I pulled it all the way back to Xmax. The kinetic energy? Well, I'm not sure what that is. But I know that this is the equation that describes it. And actually, I know all of these numbers except for v. To solve these for v is just a matter of doing a little bit of massaging, which I will leave to you. When I carry out all these algebra and arithmetic, I find that when I'm at this point when the mass is moving through this position here, the velocity is 0.87. It's maximum velocity. 87% and it's moving pretty fast. My plot will go up to here, up here, and probably somewhere around here. Now, I could do very similar algebra to find that there's also a negative solution and this solution corresponds to the exact same thing but when the mass is moving this way. I could also find that there are solutions over here and if I solve for the position when the velocity was 1/2 of its maximum I would get 0.87 Xmax. I could put points here and here. Can you see what shape is starting to emerge on the points we're plotting? It's a circle or at least it's supposed to be. Sorry some of my points are a little bit off. This is something in Physics that we call a phase diagram. And it's a very powerful tool once you understand how to use it. The way to interpret this diagram is as follows. We start with a mass put all the way back to Xmax. We let it go. Well in this case, it starts with negative velocity. It moves that way. That would correspond to going into the negative realms of velocity down here. Eventually, it maxes out in speed as it passes through equilibrium but keeps moving to the left as it can be seen by the negative velocity. Eventually, it reaches this extreme over here and finally it turns around. Positive velocity moving to the right passes through the equilibrium at maximum speed and keeps moving through the right and returns to where it started. The Wikipedia page in simple harmonic motion has a really great illustration. Now, a couple of things that I notice that are different about this illustration is position is vertical, velocity is horizontal, the spring itself is moving up and down but you get the same idea. Here, the vertical position is tracking the object's position and as the object moves downwards we see that the ball is on the left side of this line corresponding to a negative velocity as it moves upwards. This ball on the right side is corresponding to a positive velocity. Okay, so why did I spend all this time to show you this?
Now you're maybe asking yourself, so what? Who cares about this phase diagram? You're in for a treat. Notice that when I pick out a point on this circle. And remember this circle tells us all the locations in what we can call phase space which is a combination of position in velocity. All the locations in phase space where we might find this object. Let's look at this point for example. Well, this point corresponds to a position down here and to a velocity over here. Well, I'm seeing something very fascinating right now. Let's just draw a line from the center over to here. Let's draw a line from the center out to the point we're talking about. Oh, I think you're seeing it too now. Well, check it out. This point has an angle associated with it. I'm going to call that angle θ. It has an opposite side and the length of this opposite side well that corresponds to this length over here same as the velocity, V/Vmax. And this triangle has an adjacent side and this adjacent side corresponds to the position of the particle or the block at this instant in time. Another great thing is that since we've chosen to write the speeds and positions this ratios of X/Xmax and V/Vmax this circle has radius 1. The hypotenuse is just 1. That's such an easy number to deal with. Now, can you figure out a very important equation namely what trigonometric function goes here and which goes here? I'll give you a hint. It's either sine, cosine, or tangent. And it's either sine, cosine, or tangent. So is X/Xmax the sin θ? The tan θ? Cos? Fill in your answers here and here.
Well what is the cos θ for example? Well, it's adjacent over hypotenuse. But the hypotenuse is just 1. So X/Xmax is just equal to the cos θ. Similar thinking tells us that V/Vmax is equal to the sin θ. Now these two equations can be slightly rearranged to be written like this. Okay, so we've learned that we can express position in velocity in terms of the maximum position in velocity and some trigonometric function of this weird angle θ. Theta only shows up on this strange diagram I've drawn. It seems like it might somehow correspond to time since you know that as time progressed I moved along this diagram. But before we can move any further, let's talk about radians.
Let's get this tool under our tool belt. Let's figure out what radians are. Well, I'll tell you what they are. They're unit of angle--just like the degree Whereas the trace of degree was somewhat arbitrary, who thought there should be 360° in a circle--where that number come from. The choice for the radian is a little less arbitrary, and so let's talk about that. Here we have a circle of radius 1. Can you tell me what's the circumference of this circle. You can put your answer here and round to two decimal place.
What we know from unit 1 that the circumference of a circle is 2πr. If the radius is 1, well that's just 2π--so that's 6.28, but let's not think of that. Let's think of this as 2π. Now, if the units of this circumference of the radius were meters for example. And now I was a little man walking along this circle--that means I would have to walk In fact, this is how we define the radian. This angle which previously we may have called 360°--we now call 2π radians. If this is all you remember about radians, you can do everything else. I'm not saying this is all you should remember but this is enough.
So if 2π radians is one full circle 360°, what do you think will be 1π radian or just π radians. Can you tell me where in the circle you would wind up if you walk 1π radians? Select the appropriate check box.
Well, let's see.2π radians get 0 here and the full circle, 1π radian just get 0 and half the circle. If we want to go a quarter of a circle, well that would be π/2 radians. If we want to go one-eighth of a circle, that'll be π/4.
How about 5π/2 radians. Now, this is a slightly trickier one. Where in the circle would you wind up if you move through an angle of 5π/2 radians--choose the best answer.
Well, let's see 5π/2--5/2 is 2.5π. Okay. Well, I know 2π is a full circle. So that's the sign we go back to where I am, but I still have half of π radian left to go--0.5π. Well, I know 1π/2 does--0.5π--this brings me right here. So actually 5π/2 is exactly the same as π/2.
Now, let's go back to our mass bouncing up and down in the spring. You noticed I've changed it from being a horizontally sliding mass to a vertical one. The thing I like about this picture is that the mass goes this way counterclockwise though this phase space. That's the same direction that we measure angles from, so we want to stick with that convention for now. Other than that, all the physics are the same, so let's say our particle is here, our mass is here-- that's almost all the way at x max and it's moving upwards so that it'd be may be somewhere here and going up--this should be negative x max--that's positive. Okay. So let's say this is an angle π/4 radians or we can draw a right triangle. And in this case, since we've flipped the direction--we're not going vertically, and now the cosine goes with the velocity component and the sine goes to that position but you could've figured that out from the trigonometry. Now it's convenient because if we want to calculate let's say the cosine of π/4 radians-- remember when we're using Google before, we were typing something like the cosine of 45° and we're actually had to spell out degrees--yeah, that wasn't very fun. Now if you type this into Google, it'll just give you your answer--you don't have to specify radians. It assumes radians because radians really are more fundamental than degrees. So if you type this in, you'll find that the velocity at this point is 0.71 Vmax and the position is 0.71Xmax. I've just plugged in the cosine of π/4 and sine of π/4, and for this angle, cosine and sine happen to be completely equal.
All right. Are you with me? It's time to push forward into uncharted territory. We're going to go out of the first quadrant. When we look at our axis like this. Remember we called this the first quadrant, this the second, the third and the fourth. So far everything has been between 0 and 90° and we found that all of our trig functions have been positive. I could take the sine of an angle, the cosine of an angle, the tangent of an angle. It didn't matter--all of these trig functions are positive. Let's see what happens if we go into new quadrants. For example, let's look at this angle in the second quadrant. So now we're looking at this entire angle. Now I want to know--what is the sine of that angle or the cosine of that angle, but first let me tell you what the angle is. This angle has 3π/4 radians--now, what I do in these situations is I still make my right triangle relative to this x axis baseline--we can think of it as. Now you go to Google and type these in. I know this is sort of a menial task, but there's a good reason to type these in, get an answer and tell me what you get.
When you do this, when you type in 3π/4 but for the sine you get a positive answer--0.71. For the cosine, you should got a negative answer of -.071, and let's think why is that--well, let's see. Trigonometric ratios have to do with the sides of triangles. In this case, the sine was comparing opposite to hypotenuse. Notice that the opposite side well it's pointing above the x axis--it's a positive number and the hypotenuse, well we always call that a positive number. So opposite over hypotenuse, positive over positive gave me a positive number. Cosine, well let's see. The adjacent it goes to the left of the y axis--that's a negative value. Adjacent of hypotenuse negative over positive gives me this -0.71.
We see in the first quadrant all trigonometric functions are positive and that's because well the opposite side, the adjacent side and the hypotenuse of any triangle you're trying to draw will also be positive. We saw that quadrant II only designs positive, let's think quadrant III. Let's imagine this giant angle. Well, to talk about this angle, we're going to think about the smaller version of the angle and we're going to make a right triangle. Now, if this right triangle the opposite side goes below this zero line, so it's negative, and the adjacent side goes to the left of this zero line so it's also negative. I want to know in the third quadrant, which trig function is positive. Is it the sine, the cosine, the tangent, all of them or none of them? Choose your best answer, and this is tricky, but remember think of what happens when you perform the divisions required of each of these trigonometric ratios. Also to remind that the hypotenuse is always positive.
Let's see sine and cosine, we're comparing either opposite or adjacent, which are both negative to the hypotenuse, which is positive. Negative divided by positive well that's negative. Those can't be the answer. Tangent compares opposite to adjacent, negative over negative. That does give us a positive. All functions are positive in the first quadrant. Only sine in the second, only tangent in the third, and only cosine in the fourth, which you can show using similar thinking that we just used. Now, I hope you believe that trigonometric functions can actually be negative. This is very important because we're going to use all of these everything we've learned, radiance, this idea of this circle and the trigonometric relationship to the circle, and the concept of a negative trigonometric ratio. All of these is going to come together when we fully described simple harmonic motion.
I think we have enough knowledge to begin tackling the mystery of the angle and what do I mean by this. We've shown that if we know, for example, what angle describes our position and what I've called face space, which is somehow a relationship between velocity and position, we can determine exactly what we are in the cycle with this mass moving up and down, but how we do get this angle? This angle seems so important. I don't know how to find it. It seems to be related to position somehow, to velocity somehow, and also somehow to time, because if I knew, for example, that I started here and released the mass from Xmax as time went on, it would just trace out the circle. And we've discussed a little bit what tracing the circle means. So, I guess my question is, which variable can we use to determine this angle θ? Is it velocity? position? or is it time? This is a very tricky one but think it through and see if you can get the answer.
Let's see, if someone tell me the position, could I figure out where the object was. Let's say they told me they were here. Well, to find out this position, I know they're either on this part of the diagram or over here in this part of the diagram. But that doesn't tell me what the angle is--it might be this angle or it might be that angle. I can do the same sort of thinking with velocity--it's not enough information to determine the angle. The only thing that would tell me the angle, assuming I know the starting position, is the time, and don't worry if you didn't get this one--this is a very, very tricky one. Now, how does the time relate to angle.
The connection between angle and time is what's known as angular velocity, which we give this symbol ω--it looks sort of like a funny w. In this quantity describes how quickly in radians we progress around this circle and so it's units are in radians/second and the equation that describes this is that the angle is equal to the ωt--the angular velocity times the elapsed time. I should put a Δ here. This is how much you've changed your angle. So let's say we started an angle of 0 degrees, 0 radians were down here, and I have an angular velocity--let's keep it simple of 1 radian/sec. So what I want to know is how long will it take to complete one cycle and you'll have to remember what one cycle means in radians when four cycle given an angular velocity of 1 radian/second--enter your answer here.
Well, four cycles is 2π radians and it has to equal well our angular velocity which is 1 times our time. So the answer is 2π, 6.28 seconds. We call this number the time it takes to complete one full cycle--the period. So if this mass had a period of 6.28 seconds, that would mean if I lifted it up all the way to Xmax and release it, it would take 6.28 seconds for it go all the way down and then all the way back up--one full cycle.
We know about the angular velocity which we call ω--that's in radians/second and you know about the period, which we used a big T to describe-- to remind us that it's units of time or seconds. It turns out there's actually a really wonderful relationship between these two and that the period is always just 2π/ω. For this quiz, I want to start up here--so that means we've lifted our mass up our mass up and we're about to release it. And let's say it'll tell you that this particular combination of spring and mass has a period equal to 2 seconds. I want to know after 7.5 seconds, where will the mass be. Select the best answer.
Well, there is a couple of ways to answer this question, at least a couple, and I'm going to show you too. The period is 2 seconds and we're talking about 7.5 seconds. If a mass starts off up here, what are 2 seconds is complete one period; two more, it's gone up here; two more, another period. Okay, so far we've had 6 seconds elapsed. What are 7 seconds that's half a period, and then another half a second, well that's another half of what's left. We'd wind up here. How else could we have solve this? Remember that the amount of angle we go through in a certain amount of time is equal to ω t. Okay, I know that t is 7.5 seconds. How can I find ω? Well, we have this relationship, T = 2π/ω, or rearrange, ω = 2π/T. In this case, with a T of 2, ω is equal to just π radians/sec. That means in each second, it's going to go π radians. Well, 7π radians would be seven-halves of the circle. Starting from here, that would be this angle, one-half, two, three, four, five, six, and seven, but it was 7.5 seconds, so the extra π over 2 radians, and again that puts us right here.
All right. So we've got our spring with our mass and it's bouncing up and down. And the spring starts from up here—the mass starts from up here. This would correspond to the mass fully compressed with no velocity because it's right at the top of its trajectory. Well after 7.5 seconds, we've made some common sense arguments to show that it has completed a few full cycles and then wound up here. Going through the equilibrium position, going upwards with maximum speed. But sometimes common sense isn't enough. So my question is, what is the mass' position after 7.2 seconds? The only information I am giving you is that the period is 2 seconds. The mass starts from here at the top of its trajectory when we're just about to release it and then it bounces up and down. So what's the mass' position after 7.2 seconds? And I want you to write this in terms of Xmax. So enter you answer here. Be really careful with this π/2. Be careful with which trig function you use. Good luck.
Well let's see. Position is vertical. Here. Vertical. That's opposite the angle. Opposite the hypotenuse. Okay. We're using the sine. Sine is opposite. So we know that x is equal to Xmax times the sine of whatever this angle is. But remember, we don't know what this angle is until we do some math. The angle is omega t plus this phase. I didn't show you this letter before but we call this phi. Phi is how far along in this cycle we started off, and we started off π/2 1/4 of a cycle ahead. So I evaluate this for omega. I plug in π. For t, I plug in 7.2. And I get -0.81Xmax. If you got this, I'm very, very impressed. This was not easy. And if you didn't, don't worry. Go back and do it again. The beauty of online education.
Okay. So we've already talked about one way to visualize a mass bobbing on a spring. Let me draw that picture up here. Okay. So we're thinking about this mass going up from Xmax to -Xmax. And we've talked about this interpretation over here. But this was a graph of position versus velocity. What if I want to know position versus time? Well, time = 0. Our mass started up here at Xmax. Remember the period was 2 seconds. After 2 seconds, well, one cycle one period means that the mass has returned to its original starting point. What happened after 1 second halfway between these two? Well after 1 second, our object was way down here all the way to -Xmax. After 0.5 seconds, the object passes through equilibrium so we're going to draw out the position on this 0 line and same with after 1.5 seconds. It turns out that if you fill in more and more parts of this graph, more and more dots, you'll see a shape start to emerge. And you can fill in those dots using the math we learned just in the previous video. This is a graph of position versus time, which we call x(t) and this is a wonderful equation. It's just that we have some maximum, let's call this amplitude, that's modified by this function sin(ωt+π/2). And this is just some function that whenever I plug in the t, it gives me a number between -1 and 1 and it repeats forever. Now, I'm going to draw some potential graphs for the velocity as a function of time. And I want you to tell me which one is the best. So I've drawn you some graphs. No, it's not just a bunch of pretty pictures. Though I agree they are very beautiful. This may look confusing but the way to read this graph is to look one color at a time. We know that the blue graph depicts a position versus time. And what we want now is velocity versus time and there's four candidates. Well there's the blue graph. Maybe you think it has the exact same graph as positioned. There's the green and the pink and the red. Now, I want you to look through these and think which makes sense. And the best thing to do will be to look at this diagram and think. Okay. If our motion is starting up here and progressing around this phase diagram, which of these graphs best makes sense for the velocity as function of time? And think, does the particle start with velocity? Or does it start with positive, negative, zero velocity? When is the velocity negative? When is it positive? Everything you can learn from this, you should be able to see reflected in this graph somehow. Go ahead and take a stab at it. This is a tricky question.
And so I sort of did process of elimination on this one. Let's see. If it starts up here, that means the velocity is 0, right? Well it can't be the blue graph because the blue graph starts with some positive velocity likewise the green starts with some negative. That just doesn't make sense. We know it starts at rest as both the pink and the red do. Look at the difference here. The red immediately goes towards negative velocity. The pink goes towards positive. What actually happens? Well it goes downwards. Of course it goes downwards. We're dropping it. So it's actually the red graph is the best answer. There's a lot more to say about these graphs. Ways to manipulate them and change the amplitude period shift them and a lot of interesting stuff to learn. We're not going to cover it in this class. Maybe in a later one. But you should definitely read up on it on your own if it's something you're interested in. But these graphs sort of highlight the periodic nature of simple harmonic motion. They just repeat themselves over and over and over, and that's exactly the property we want from a chronometer. We want it to be able to repeat itself. We want to know that every period is the same length of time. One tick of the clock. Now, the real question is, how do we control what this period is?
Okay. So here's our mass hanging from the ceiling on a spring with the spring constant K. If I started shaking up and down and oscillating, we know it will move with a certain period. How do we adjust that period? Well let me first tell you the force law of a spring. And to do this, it actually may be better to think of the spring being horizontal. Here's our spring at its normal happy equilibrium position. We're going to call the equilibrium position x = 0. Now let's see, when we stretched the spring a bit and when we moved the mass back to let's say here to some position x. Well the spring exerts a force backwards back towards the equilibrium. The force of a spring is somehow negative. If I pulled to the right, it pulls to the left. And in fact, if I pull further to the right it pulls harder to the left. If you ever pulled on a spring you can tell pulling it a little bit is easy. Pulling it more is more difficult so actually the force is proportional to how much you stretch it. And what's the constant of proportionality? Well, that's why we have this letter K. This is the force on a spring and we know from our good buddy Newton that forces cause accelerations. In this case, the spring will accelerate this way due to the spring force. Why don't we replace this F with this -Kx? Well in the case we get ma = -Kx and so a = -K/mx. Now I hoped you followed along so far. The acceleration is equal to -K/mx. The reason why I said I hoped you followed along is because unfortunately now I have to do something that I hate to do. I have to just tell you something and that something is that whenever you can write a force law like this that the acceleration is equal to minus something times x times its displacement from its equilibrium, well it turns out that this something is always equal to ω². And remember omega is our radians per second, our angular frequency. Now let's say you have a spring with a spring constant of 50 N/m. And let's say you want your period to be equal to 1 sec, and remember T = 2π/ω. And why do we want 1 second? Well we're trying to design a clock here. Of course we want a 1-second period. We want it tick tock, tick tock. The question is with this spring and this desired period, what mass do we need? Enter your answer here in kg.
Let's see. I said whatever this something is could be replaced with ω². Check it out. The something in this case--the something that multiples our x is k/m. If I know k and I know ω, I can get m, but I don't know ω. I know period. Why don't I get ω from that period. Well, in this case the period is 1 second so ω is just 2π. Okay, well, now I know that ω², 2π² is equal to k/m--50/m. Well, all I have to do now is solve this equation for m. When I do, I find that I need a mass of 1.27 kg. Okay, this was a little tricky. The take away I want you to remember is that any time you can write your force like this, acceleration equals negative something times x, you're going to get simple harmonic motion. That's really incredible. Not only that--whatever this something is will be the square of your angular frequency. Now, that's really cool, but unfortunately all this stuff we've learned about springs-- well, springs weren't how they made the first clocks or first chronometers. There was one day when our good friend Galileo while sitting in a church saw something that blew his mind and would forever change navigation. Le'ts talk to Tjeerd Bucker again to see what Galileo found. [Tjeerd Bucker, Museum Booerhaave] Well, it's a rumor or an anecdote--however you want to call it. Galileo was sitting in this church, and see the chandelier swinging from left to right, and he checked with his pulse to see if it kept an accurate rate. Due to counting, he found out that there was an equal line in oscillation, an isochronism, in the pendulum in the swinging of the chandelier. That's how the pendulum started life, really.
So, here's Galileo sitting in his church staring at this pendulum-- I mean, chandelier--as it rocks band and forth. This is a chandelier. These are candles. If you ever want to teach intro to drawing, I would be happy--in fact, thrilled--to take that class. Here he is measuring his pulse with his fingers up against his neck, and he finds that--what do you know?--the chandelier is swinging periodically. It takes the same amount of time every single time it swings back and forth. This gave Galileo an idea. Galileo realized the potential here. This could possibly used as a time-keeping device and solve once and for all the problem of longitude. But it wasn't Galileo's bright idea that caused the birth of the pendulum. It was the equally bright, but apparently more energy efficient, idea of Christian Huygens that actually lead to the birth of the pendulum. Why don't we talk to Ad Maas, the curator of the Museum Booerhaave in Leiden, to get a bit more perspective on this. [Ad Maas, Museum Booerhaave] Christian Huygens was mainly known for his role in what is called the Scientific Revolution. For example, his cutting edge work on the pendulum motion was combined with thinking about making precise, steady clockworks. Making steady clocks had to do with being able to navigate at sea. The only way to be able to determine your longitude is if you would be able to compare your time on your ship with the time of your starting point. In order to keep the time of your starting point, you need a very steady clockwork. Well, that was a big challenge for inventors throughout the 17th and almost all of the 18th century.
Okay, so Huygens wanted to look at a pendulum to measure time. Let's analyze what's going on exactly with a pendulum Well, a pendulum is just a mass on a rod. When we pull it back, like we have here, away from the equilibrium position, which I've drawn in dotted lines, it's going to swing back and forth. If there's no friction or air resistance, it'll just do that forever. We expect this pendulum to give us simple harmonic motion, which means we expect that when we work out the forces in the acceleration we want to get acceleration that looks like minus something times x. Let's look at the forces and see if we can get that, because we want to understand what's going on here. Let's imagine pulling our pendulum by some angle θ. Well, of course, our pendulum has some mass. I said simple pendulum so that what means--I should explain-- is that all of the mass is in the bob at the end, which is just some really tiny little point mass at the end, and there's no mass in the rod, and everything is going to swing without friction. The pendulum has mass or weight pulling down, and there's only one other force. That's the force of tension supplied by the string or rod. I know that when I release this pendulum, if I've pulled it back here and I release it, it's going to swing this way--down back towards the equilibrium. Actually, what we want to do is break this mg into a component this way that will balance the tension and a component in this direction and a component in this direction. This picture is starting to get a little cluttered, so let me pull it off to the side and make it a little bigger. I can see that this angle is the same as this angle over here. If that's the case, well, I can do my trig. This is a right triangle with a hypotenuse, and there is this side, which is mgcosθ. This side is pretty uninteresting, because I know it's going to exactly balance tension. How do I know that? Well, because the mass never accelerates towards the pivot point or away. It only accelerates perpendicular directly back towards the equilibrium point. That force that's doing this acceleration is mg times sin θ. Okay, this is the force. Actually, I should say that it's negative, because if I pull it back to the left I would normally call that a negative angle. The force is to the right, so this negative sign is actually important. So, we have this force doing our acceleration minus mgsinθ. Let's use our good old friend F = ma. In this case -mgsinθ = ma. When we check it out, the m's cancel, and we're left with this. The acceleration is equal to -g times sinθ. We're in a big of a pickle here because--uh, oh--this does not look like a = 1 something x. Have no fear. We're going to do some sly massaging of this equation to figure out what's going on.
The first thing I want to do is show you something very fascinating about sin θ when θ is very small. If θ is 0.01 what is the sine of that angle? Remember, we're measuring in radians, not degrees. If I type this into Google, the sine of 0.01, I get 0.0099983333, but if I round that up to any reasonable number it's 0.01. Can you do the same thing for 0.02 and 0.04? Just enter your answers in this box.
Look at this. This is amazing. Type it into Google for yourself. You can see for any reasonable amount of rounding for small angles, sin θ is equal to θ, and it's so close. It's such a good approximation that we can actually do physics with it. Instead of writing a = -gsinθ, we can now write a a -gθ. Okay, we're making some progress. We're not quite there yet. This doesn't look like an x or anything. This is an angle, not a distance. What can we do?
Well, look what we have here. If this is our pendulum and this is our pivot point, when I pull back the pendulum it's going to trace out this circle. Now, of course, I'm constraining myself to small angles, so let's get rid of most of that circle. But when I pull back my pendulum by angle of θ, let's say this pendulum has some length L that's the distance from here to here. We can do the same sort of ratio that we did way back in Unit 1 when we were measure the circumference of the earth. This angle θ, when compared to the entire angle of the circle, 2π radians, is in the same ratio as this arc length, which I'm going to call x, to the entire circumference of the circle, which is 2π times the radius or L. If this isn't clear, let me just explain this one more time. What I'm saying is the ratio of this angle to the entire 360° or 2π radians of a circle. That's identical to the ratio of this distance to the total circumference. This angle to the total angle is the same as this distance to the total distance. What I'm trying to do is replace this θ, so I'm going to solve for θ. When I do, I get θ = x/L. Now we're making progress. Let's see what happens. Now I have that a is equal to -g(x/L), having made that replacement. Or put another way, a - -g/L(x)--minus something times x. We have simple harmonic motion in our pendulum. Wonderful. Now, I'm going to ask you a very hard question, because it's going to have you put together a lot of the things we've been talking about. We're trying to design a clock. We want a period of 1 second. I want to know how long should our pendulum be if we want a period of 1 second? Don't forget that this something here becomes ω². Enter your answer here in meters.
Okay, let's see. We want a period of 1 second. Let's convert this to an angular frequency ω using this equation. So, we want an angular frequency of 2π, an I know that ω² is equal to whatever this thing is, the minus something times x. In this case, ω² is equal to g/L. Let's solve this for L. This gives me a length of 0.25 meters. Now, it turns out that for pendulum clocks, actually, we might want a period of 2 seconds. The reason is because let's say we had a pendulum with a period of 1 second. That means that it takes 1 second for when I pull it back to go over here and then all the way back. That's actually quite a bit of motion and a period of 2 seconds is actually a little bit better, because then every time the pendulum reaches one if its extrema, either all the way over here or all the way over there, we know 1 second has passed--half of the period. I wonder what would be the period for a 2-second pendulum. Why don't ask Tjeerd the clockmaker what he thinks. [Tjeerd ] When you design a pendulum or pendulum clock, the thing that mostly influences the period of oscillation is the length of the your pendulum. If you have an ideal pendulum, a fictional pendulum, made out of no materials as a mathematical piece, the pendulum of a second, oscillating in a second, will have a length of about a meter. A pendulum is made out of materials, and the rod, for instance, is made out of steel or brass, which brings the center of oscillation up. That's why you need a heavy pendulum bulb underneath the pendulum-- to bring the center of gravity in the pendulum down to have your proper theoretical length.
Now that you know the basics of simple harmonic motion, how to construct springs and masses or a pendulum with a certain period, and how to describe that motion mathematically, you could more or less design a good clock all by yourself. Now, in a real pendulum clock we need to answer the question of how do we convert this periodic swinging into the ticking of a clock? What we do is we take this gear mechanism here, and we allow this to rotate somehow. Then we attach our pendulum to this thing, which is called an escapement. Now, this gear mechanism would love to just unwind. In fact, if the escapement wasn't here it would just unravel immediately. As you can see, if the pendulum swings to the left this moves out of the way, which allows it to move forward by 1 tick. As it swings to the right, this one moves out of the way--tick, tock. Basically, that's how a pendulum clock works. Now, there are some problems. We have to deal with friction. This isn't going to run forever on its own, but we'll do a problem set problem about that.
Congratulations. You've answered a question that was vitally important for centuries. How can we locate our ships at sea? You've seen that even though finding latitude is relatively easy, finding your longitude can be incredibly difficult. With a reliable time-keeping mechanism, which we can create by taking advantage of simple harmonic motion, we can find our longitude by comparing the time on our ship to the time at some known location. In the next unit, we're going to visit the 18th century when buildings were frequently struck by lightening. In the next unit, we're going to answer the question how can we prevent buildings from being struck by lightening. See you then.