https://youtu.be/-uvcPUv8Hk4
PHYS 1101: Lecture Twenty-Three, Part Four
Okay. To answer these next questions, I want us to watch the motion of this object more carefully. So here, I am stretching the spring, and now I let it go, meaning the only force on the object now is due to the spring and you’re seeing the size of the force there change in real time. Watch this motion. We’re going to let it go through several cycles, because I’m going to ask you several questions about the nature of that motion in the following quiz questions. Be thinking about the motion diagram. Where’s the object going the fastest? Where’s it the slowest? Does it ever stop?
Think about the acceleration. Acceleration always compares two velocities. What’s the acceleration look like? Think of the motion diagram and how you would generate the acceleration vectors by looking at the difference between two nearby velocities. Remember, at a turning point, the acceleration is never zero. The velocity might be zero at a turning point like there, but the acceleration is not. It can’t be if that object turns around.
Okay. Here’s a snapshot from that animation and go back, play the movie again. Rewind the movie to answer these questions, but here’s in essence what I’ve drawn. Here’s the equilibrium position, and you’re watching this mass oscillate about this equilibrium position, meaning its x coordinate sweeps to positive values, back to zero, and then to negative values. We use the variable “a” to denote what’s called the amplitude. It’s the maximum positive x value. -a is the extreme negative x value, so if this is 5 meters, the x coordinate goes from +5 meters back to -5 meters, so you’re picturing then the extent of this motion.
Question 13: what’s the right motion diagram for this oscillatory motion? Here’s a statement I put in red to emphasize what you see in that animation, which is that the force is not constant during the motion. This makes it fundamentally different from our chapter two previous work where f stayed the same. It was the same, say, 5 Newtons the whole duration of this motion and that’s not the case. The force depends on x. If x is changing, so is the size of the force.
Okay. Question 14: as you watch this motion oscillate back and forth, where is the speed and minimum? Choose a if you mean the snapshot when the object’s right in center. It’s sweeping just through equilibrium or the instant of the snapshot when it’s about halfway in between, or is it this instant right at the end where it reaches this maximum position?
15: where’s the speed a maximum? At A, B, or instant C? Is the speed ever zero? At position C, at this instant, I want you to think about the acceleration vector. What direction does the acceleration vector point? Is it zero at this turning point? Does it point to the left or to the right? Think about your motion diagram and work that out. Actually do the vector operation and figure out in what direction A would be. At instant B, think about your motion diagram and compare a velocity just before and just after as I sweep by C and I head back to the equilibrium. Think about A as I’m moving away from equilibrium and I pass this spot B. What’s the acceleration then? That’s question 18.
Question 19: is the force ever zero? In that animation, you’re seeing the force vector there. That’s the blue arrow. Does it ever go to zero and if so, at what position? Okay. Question 20 is a little tricky. I’m going to give you six points for this, because I want you to think about that. Is the acceleration at position A positive, zero, or negative? Position A, again, is equilibrium. Just as this object sweeps through equilibrium, whether it’s going to the left or to the right, at this instant, what’s the acceleration?
Think about this from two perspectives, and both must be true. We know that acceleration is a vector that’s in the same direction as Delta v or the difference between two velocities that are right next to each other. So if you compare a velocity vector just before you get to equilibrium to one just after you’ve gone through equilibrium, what does the difference look like? Think about the motion diagram. What also has to be true is that whatever acceleration that object has, it has to result from the force on it from the spring divided by m. At equilibrium, how big is this force? That would tell you how big the acceleration would have to be.
I want us to look now at the graph of position versus time as time goes on. We saw, as we watched the movie if you want to go back, that this object oscillates to the right and to the left, meaning it’s coordinates go from zero to a positive value, back, and to a negative value. I’ll show you in a minute another way to look at it, but if you were to plot the x coordinate, here’s equilibrium right here in the middle, versus time, so this is in seconds. So as time increases, my position, I start out with positive position values then it goes back to zero, then I start having negative position values. My position oscillates up and down.
This is a figure from your book, and it’s a great combination way of picturing how these plots are generated position versus time. Imagine, to this object attached to this spring, going back and forth, that I attach a pen to it, and I have this pen tracing out, against a sheet of paper that’s rolling up as time goes on. Can you picture as this mass swings back and forth, it’s going to map out this oscillatory trace on the paper. The pen is marking out the time history of what that position did. All we’re doing when looking at a plot of x versus t is turning that on its side, so we look at x horizontally here.
Here are some important variables that we use to characterize this simple harmonic motion. This oscillatory motion as time goes on. The first variable to highlight is a. I’ve alluded to it before. a is called the amplitude. So remember, I emphasize that here is my equilibrium position in the middle and then this mass would move to an x-coordinate here of plus a and then it would sweep back to minus a. So when you see how that relates to this piece of paper graph that I’m generating, you appreciate that a is half of the distance between the maximum and the minimum, the total excursion. Okay. That’s worth noting. If you want to read off the amplitude from a graph, it’s not this number minus this number, but you’ve got to measure from the center. That’s a.
The next variable is called the period. This has the analogous interpretation to what we did in chapter five, where we talked about the period to complete one cycle. Here is a time trace. If I follow through one cycle, the time that has evolved as I go up and then back down to the same position through one cycle, that time, the number of seconds between these two tick marks would be the period. The time to complete one cycle.
The frequency is another important variable and it’s related to the period. The frequency is measured in units of Hertz and the interpretation of Hertz is a cycle per second. So the frequency is a measure of the number of cycles per second that this oscillatory motion undergoes.
Here’s how they’re related. The period is the seconds per cycle. It’s the time per cycle to repeat one cycle. This is really how those units would read. It’s the seconds per cycle. To get to frequency, I want to invert that fraction. I want it to read cycles per second. So here’s how the two variables are related. The frequency, you can always calculate if you take one and divide it by the period, the number of seconds per cycle. The inverse of that gives you cycles per second. So here I’ve highlighted definitions of variables and this key equation that relates these two variables.
Okay. Here’s an animation I want you to watch because there’s a very useful analogy between uniform circular motion and this simple harmonic motion. Very useful. So let’s think about what we know in this context and apply it to here. If you were to carry out the exercise of hanging a mass from a spring, and then plucking it, letting it go, it’s going to oscillate up and down. If you turned on your record player, put a ball on it that has the same diameter as this distance from +a to -a, and then you slowed your record player down so that the time it takes to repeat one cycle is the same as this time to repeat one cycle, you’d see that the motion, if you look at the up down motion of these two objects, it’s exactly the same. They both are at their highest point at the same instant, they both are here in the middle at the same instant, they reach the lowest point at the same instant, and they come back to this middle position at the same instant.
In the middle, corresponds to the equilibrium position for this spring. At the high end and the low end corresponds to the two extreme amplitude coordinates of a at this end and the a at that end. The graph here shows you the y coordinate. If I were to draw a y-axis here with this being the equilibrium, plotting the y coordinate as time goes on, I see that same oscillatory curve that I saw before.
The other thing I’ll point out is the velocity analogy between these two. The arrow that they’re showing you there is the vertical component. The v-y part of the velocity vector for this object going in a circle. Notice what it does as we play it. It looks like it’s largest here as I sweep through the middle. It goes to zero at the extremes and now it’s flipping direction. It goes back to the maximum here through the center, and it goes back to zero. That velocity vector, the v y part for the oscillatory motion is exactly the same as the v-y for this ball going up and down. I stop briefly at the ends, I go the fastest through equilibrium, and I stop briefly at the other end.
Okay. I’ve got a snapshot here from that animation and I’ve just added a few things to emphasize this connection. Let me draw on here y=0. This is our equilibrium and this then here is our y-axis. This would have the y coordinate of +a. At this point would have the y coordinate of –a as this object oscillates between +a and -a. The one important variable that I want you to connect in particular between these two is the notion of the angular velocity. The Omega, in radians per second, we used to describe uniform circular motion, is the same Omega we want to assign to this oscillatory motion. Omega is the radians per second. One cycle is analogous to 2π radians. Because, for oscillatory motion, Omega is related to the frequency of this motion, Omega is also called angular frequency in addition to angular velocity. It’s interchangeable, these two terminologies.
Here’s an important connection. Omega, with units of radians per second, when I multiply that by the time to complete one cycle, the product should be 2π. That’s the radians of one cycle, one complete circle, 360 degrees is 2π radians. And angular velocity, angle per second times a certain number of seconds, give you angle. Like velocity times time gets you distance. So I’m going to highlight in yellow, this important connection.
Now that we’ve looked at that carefully, I think we’re ready to define our kinematic equations. The first one, we’re going to use the variable x and just generically think about horizontal oscillation, oscillatory motion. I have the position to think about for this motion, the position coordinate, the velocity at any instant in time, and the acceleration at any instant in time.
Okay. The right way to describe that oscillatory graph that you saw, this graph that I’ve shown here, I have pasted for us on the left. This oscillating nature of that motion as time goes on is with this trigonometric function, but for the argument of cosine, instead of just putting in an angle, which is what we’re most accustomed to, the argument is our angular frequency Omega multiplying by time.
As time goes on then, the value here for my total angle here that I put in for the cosine is changing. Remember, Omega is equal to 2π radians over the period. When I multiply by so many seconds, it’s that fraction of the time compared to the period that will tell me how many radians have gone by. The proper angle that I need for this cosine function.
Okay. This is what properly describes this function. a then, is as I’ve outline for you. Zero is in the middle. This is the distance that’s a. So my position coordinate oscillates between +a and -a. What does the velocity do? Well, as I had you watching that motion carefully, I want you to appreciate that the velocity is really changing as this thing oscillates back and forth. It briefly comes to a stop out at the two ends and it’s going the fastest as it sweeps through equilibrium.
Okay. The right function that describes that velocity is this. I have again another sine function. As time goes on, the angle that I plug in to evaluate the sine function, it’s changing, so at each time I have a unique t times Omega, to give me the right radians for the function, but then whatever that results in, I multiply it by a times Omega. a times Omega is the maximum speed. I have a -a times Omega for my velocity vector when the mass is sweeping to the left through equilibrium, and I have +a Omega when it’s sweeping to the right.
v oscillates between these two extremes, so the maximum speed, the maximum value, is a times Omega. You may need this. Let me highlight in blue here our x. You may need this. It comes from this factor out front. The acceleration also oscillates between zero and two extreme values. It also has a trigonometric function here. It has the same trig function, notice, that the position does. Multiplied out front here though, is a different factor.
The acceleration oscillates between extreme values of +a times Omega2, meaning the acceleration vector points to the right and -a times Omega2, the acceleration vector points to the left. The maximum acceleration is a times Omega2 and I achieve that at the same positions where the x position is its smallest. I have the largest acceleration at the two turning points at the ends. The acceleration vector, as is always the case, always points to the inside of a turn and if you turn around, it’s large. For oscillatory motion, it’s the largest value at the turning points.
There’s one last important equation that you may need that I want to highlight for you. If you go through the analysis of this motion, you can derive, and the book does this for you, the fact that the angular velocity, the radians per second that describes the characteristics cycle time of this motion, it’s not arbitrary; it depends on two important characteristics. How stiff the spring is and how big the mass is that’s attached to this spring, that’s oscillating, undergoing this oscillatory behavior, and here’s mathematically the exact way that it’s related. Omega is equal to k over m, that ratio, but the square root.
Okay. Let’s do some examples now. For quiz question 21, I want you to think about the velocity and the force when you’re looking at the x versus time plot, so this is asking you to be able to look at this graph and picture what the analog is for the mass on a spring going back and forth. It might help you to think about this graph. Let me compress it a little bit and suggest you do this. Think about drawing your mass on a spring up here or on the left, so I’m lining up here my equilibrium position.
Here are my two extremes and here’s my mass attached to a wall. This is undergoing this oscillatory motion back and forth and it’s leaving this trace as time goes on, so at this instant, when the object is at this spot, what can you say about the velocity? Remember, it’s a vector. The sign would indicate direction, that the object is moving at that instant, and the same with force. It also has a sign indicating direction of that vector.
For the next question, you need to think about the maximum velocity, but what you’re given here are the graphs of position versus time. Object one has a small amplitude and it’s oscillating really slow. The time it takes to repeat one cycle is actually three times longer than the time it takes this object to repeat one cycle. This is the same time interval as this. This object, it’s going faster and it has a larger amplitude. Three times larger. It undergoes one, two, three cycles in the time that this object undergoes one cycle. Which of these two, a or b, has the greatest maximum velocity? Remember that v max is set by the product of the amplitude and Omega. I can equate that to a time, let me substitute in for Omega, 2π over the period. That may help you. I’m just doing this substitution here for Omega.