40 min.
https://youtu.be/4C3f4ekhttY
PHYS 1101: Lecture Twenty-Four, Part One
Welcome to Lecture 24. This is the last topic we need to cover with respect to simple harmonic motion. All we need to do in this lecture is connect previous ideas of energy and see how they apply to simple harmonic motion.
We’re going to have two types of energy to consider, kinetic energy and potential. As before, kinetic is the same. We calculate it as 1/2 M times the speed squared.
Remember that when something’s undergoing simple harmonic motion, its speed varies. As it changes its position, it slows down, comes to a stop at the end, for example, then turns around, speeds up, so then does its kinetic energy. Where the speed goes to zero, so does the kinetic energy. Where the speed is a maximum in the center, so then is the kinetic energy.
For springs, we can define a potential energy just like we did for gravitational potential energy. For a spring, the potential energy is 1/2 times the spring constant, K, measure of the stiffness of the spring, multiplied then by the amount that the spring is stretched or compressed, squared.
My potential energy also changes with position as this object rotates or oscillates back and forth. Wherever the spring is stretched the most, X will be the largest, and I’ll have a large amount of potential energy. Wherever the spring is compressed the most, I’m also going to have a large amount of potential energy.
Right in the middle, when that mass, that object sweeps through the equilibrium position, at that instant, the spring is not stretched or compressed. I won’t have any potential energy there.
Like I did before, I can describe or talk about the total mechanical energy of this object as being a sum of the kinetic plus the potential. For a spring, what we’ll see in this lecture is that there are no other forces on this object if I just have a mass on a spring oscillating back and forth. Therefore there’s no non-conservative work being done.
All I have is mechanical energy that’s always the sum of kinetic and potential, and it has to stay the same the entire problem. At any instant, it’s just a question of how much kinetic I have then compared to how much potential.
In real time, as this mass swings and moves back forth, I have this conversion that’s going on, that kinetic energy is turning into potential, and potential into kinetic as I go back and forth.
Here are some key points. I have all potential energy at the turning points. All of my mechanical energy is potential energy. At the turning points, I have no speed, so no kinetic.
Complementary to that is that at the instant in the center, at the equilibrium point, I have no potential. Now my total mechanical is all made up of kinetic. That connection, the fact that the kinetic in the beginning, in the middle, is the same amount of potential I have an instant later.
When all of that kinetic has been converted into potential, that potential amount is the same value. One-half K, A squared, as my total kinetic, my maximum kinetic, when the object’s in the middle. This is an important relationship that allows you to connect the maximum speed of an object undergoing simple harmonic motion to its amplitude.
At any snapshot of time, that total mechanical is always a combination of the kinetic at that instant plus the potential. What would you substitute in for the left-hand side? Well, this is the total mechanical. If you happen to know what the amplitude was in the spring constant, for example, you could calculate the maximum potential.
That is the same value as the total mechanical, because at the end points, my mechanical is all just potential energy.
Likewise, the left side, you could get this number for the left side by calculating 1/2 M, Vmax squared. At the middle instant, all of my mechanical is only kinetic. I get the proper number for mechanical by just simple calculating 1/2 MVmax, squared.
What I’ve highlighted here in yellow are the key equations for this lecture.
Let’s warm up here, before we begin our new material, with a couple of quiz questions going over previous content. You’re going to compare two scenarios here. In an instant, you fire this red object, the cannon does, with a velocity that’s horizontal. At the same instant, this orange object is simply dropped. No air resistance for either case.
Question one, which of those objects hits the ground first?
Question two, as they’re making their way to the ground, which of those objects is in equilibrium?
Question three, when a disk rotates counterclockwise at a constant rate, the tangential acceleration of a point on the rim is what?
I highlight here the words “constant rate,” because those always jump out at me. Constant rate right away flags to me, and makes this distinction between whether something is speeding up and slowing down, or if it’s in equilibrium. It’s going around at a constant rate. That makes all the difference in the world. If it’s in equilibrium, I know that I have no acceleration. I have no angular acceleration.
Question four, you’ve got to rank the torques for this object. The door here, this bar is your object. Consider that in red. My rotation axis is shown here, and I’ve got four forces. The location that they’re applied and their relative lengths are shown.
I want to emphasize with this pink dashed line an important point. This line is parallel to this object. You see that both of these forces reach or meet this line.
Question five, complete the following statement about a collision. Momentum will be conserved in two body collision only if… Which of those statements is true?
Here’s our summary of the last lecture, pulling out the key equations. Last lecture is where we began our discussion about simple harmonic motion. The first, most important concept is the equation that tells us what the force is on an object that’s undergoing simple harmonic motion.
The simple case we start with is an object attached to a spring. If we take that object and we pull it to the right and then stretch the spring and then let it go, the only force then on the object is from the spring, and it’s going to cause this object to oscillate back and forth and undergo this motion.
What is that force at any instant? How big is it and what direction? This equation allows us to calculate that. The force, both magnitude and direction based on the sign, is equal to the negative of the spring constant times X.
X, remember, is the coordinate of the position of that instant. My origin is always at the equilibrium position, where the object would be sitting if the spring were not stretched. Coordinates to the right would be positive, and to the left would be negative.
This negative sign is very important. It flips the sign and therefore the direction to give us the proper restoring force. In other words, if a spring is stretched, the spring is always pulling the object back. The force would be in the negative direction. If the object is to the left and the spring is compressed, the spring is pushing. It would be a force in the positive direction to try to take the object back to equilibrium.
The three equations shown here, you can think of as the kinematic equations for this motion. I have an equation that tells me how the X coordinate changes as time goes on. The next one tells me how the velocity changes. The last, the acceleration.
For this oscillatory motion, all three of these quantities oscillate between positive and negative values as time goes on. All of them oscillate. None of them are constant for all time. The acceleration oscillates between maximum values of plus-A times omega squared and minus-A times omega squared.
The velocity oscillates between plus and minus A times omega. My position coordinate oscillates back and forth between plus and minus A.
The three key new variables we also emphasized last lecture were omega, the angular frequency, radians per second. The plain frequency would have units of Hertz, or cycles per second. Then the period, the time to complete one cycle. That’s always in units of seconds.
Here were the important relationships that related those three variables. The last one I’ve highlighted in yellow for you tells you the angular frequency, the characteristic rate that this is oscillating back and forth with, depends only on how stiff the spring is and how much inertia is attached to the end of the spring. The mass of this object. This equation allows you to calculate that.
Here’s our new material. We’re going to be talking about the energy associated with this kind of motion. Let’s first think about, focus on the kinetic energy. Let me pull my animation back here for us so we can watch this. I’m going to grab this object and pull it off to the right and let it go.
As it oscillates back and forth, remember you’re seeing the blue arrow there representing the force of the spring on the object. For the kinetic energy, you want to train your eye on the speed of this object. Where’s the speed maximum? Where is it zero?
See how the object comes to a brief stop at these turning points? Its speed at that instant is zero, so it has no kinetic right here. It then starts accelerating, speeding up. It reaches a maximum value at the equilibrium point in the middle. Then as it goes past that, it starts slowing down again and losing kinetic.
Kinetic is a maximum here at the middle, and it goes to zero at the two ends.
You always calculate the specific value as 1/2 M times the speed squared at that instant. V, remember, the speed, is just magnitude of the velocity at that instant.
Question six, where’s the kinetic energy in maximum?
Question seven, is the kinetic ever zero, and if it is, where is it? In the middle, as the object passes the equilibrium point, or at the ends?
Potential energy. We can assign a potential energy associated with a mass on a spring as being 1/2 times the spring constant times X squared. Where, again, that X is the compression or the stretched amount.
The origin is always at the equilibrium position, so that’s zero at equilibrium.
We can do this, assign this potential energy, because the force that a spring puts on an object is a conservative force. If you remember back to what that definition is, it just means that if we consider the work that this force does on the object as we go to position X and, say, back to zero, the amount of work we do going in one direction, we get back going in the other direction.
In other words, test that out in questions nine and 10. I’m going to show you the animation again, and I want you to watch that as the object moves to the right, away from equilibrium, what’s the right sign for the work that this force that the spring provides? What’s the sign of that work?
Let’s start the thing oscillating. My question is about, as I pass equilibrium here and I start moving to the right, past equilibrium and during that motion, as I move to the right, what’s the sign of the work that the force of the spring here, this blue force, does on the motion?
Remember this trend, that a force will do positive work if it’s contributing to speeding up the object. That’s if the force has any component in the direction of motion. If a force opposes the motion, it’s doing negative work, and it’s working to slow the object down.
Which of those scenarios is true during that snapshot, that little segment there, after I’ve moved past equilibrium and I’m headed to the right?
Question 10 then asks, on the return trip, meaning now I’m headed back toward equilibrium, my motion is to the left, my force is to the left. In this case, is the work that this spring force doing positive or negative? Is it adding energy and speeding this object up, or is it taking energy away and slowing it down?
Remember, the sign of work is a reflection of how the direction of the force compares to the direction of the motion. If they’re in the same direction, the work is positive. Opposite direction, the work is negative.
If you’ve thought that through, then what I want to emphasize or impress upon you is that as I swing to the right, past equilibrium, the spring does negative work. When that object turns around and then starts swinging back to the left, now I start doing positive work, and I do the same amount. Whatever energy the spring took out, it gives back as I repeat that motion in the opposite direction.
This trend or characteristic is what allows us to say that the spring force is a conservative force, like the force of gravity. Therefore also like the force of gravity, I can assign a potential energy that’s equal to the negative of the work that that spring does.
In my energy conservation equation, I can consider that potential energy from the spring over on the right side of my equation. You can check the book if you’d like to see the derivation, but I’ll just show it to you here.
How do you calculate that work and then the negative of it? At the end of the day, this is what you use for the potential energy for a spring. At any snapshot, the amount that the string is stretched or compressed, I have to square that amount times the spring constant, times 1/2. That gets me the number in joules that represents my potential energy.
Question 11, move back in the movie lecture here, replay the motion, watch it a bit more. Now ask yourself where’s the potential energy a maximum? You’re alternately asking where is X a maximum.
That’s what you’re asking because the potential energy from a spring reflects that. It depends on this amount I am stretched or compressed.
Question 12, is potential energy every zero?
Question 13, if it is zero, where does that happen? Is it the instant that the object zooms past equilibrium, or is it the instant that it comes to a brief stop out at the two ends?
When that potential energy is a maximum, think back to where you chose the kinetic energy as being a maximum or a minimum to answer question 14. I want you to take your time and appreciate the complementary nature of the kinetic and the potential for this kind of motion.
Here’s how we can use this in the end. We’re still going to talk about the mechanical energy as being the net result of all of the kinetic at that instant plus all of the potential. Mass on a spring, the only potential I have is from this spring. I’ve got my kinetic plus potential.
My basic energy conservation equation still holds. If the only force I have is from this spring, then the left side of my equation is zero. Remember, when that happened, it meant that the right side, I could think of as saying that the mechanical energy at the end, minus the mechanical energy at the beginning, also has to add to zero.
That’s what these terms represent if they’re just rearranged. That means if this is plus five, this has to be minus five. In any two snapshots, if I compare, the kinetic plus the potential has to add up to five or two, whatever it happens to be for that problem.
Let me walk you through a little diagram here to try to emphasize this to you. Here’s a sketch emphasizing I’m going to be thinking about this oscillatory motion as this object moves back and forth. It’s oscillating between an X position of plus-A and minus-A. Here’s equilibrium. This is X equals zero.
What I’ve drawn here in these bars would represent the total mechanical energy. I’m going to fill up these bars with the amount of potential and kinetic that I have at each instant. When this bar is positioned here, for example, I mean when this object is at this position. At that snapshot in time, what’s the make-up of my mechanical energy? How much of it is kinetic? How much is potential?
Let’s do our potential first. In the center, my spring is not stretched. I have no potential. Out here at the turning points, I stop briefly, I have no kinetic, and I have all potential. I have the maximum potential value. My X coordinate now is plus or minus A.
My bars here on the end should represent that all of my mechanical is potential at those two extremes. I stop here briefly, so I can’t have any kinetic.
Conversely, in the middle here, my spring is not stretched at that instant, so my potential is zero. Boy, I’m going the fastest at that instant. Now all of my mechanical has to be filled, so to speak, with kinetic. All of my energy is in the form of energy of motion, kinetic energy.
What about these two spots in between? Well, somewhere in between, the spring is compressed some amount. I do have some potential. I will have some speed here as I cross this point. I have a combination. I have some kinetic and some potential.
The same thing will happen here on the other side. I now have a make-up of energy that’s partly kinetic and partly potential. Maybe you can picture these bars more continuously as this object swings back and forth.
If I were to put in a bar here, for example, I would have mostly potential, but a little bit of kinetic. I’d have a little bit of speed near the turning point, but not much.
Same thing here. As I get near the center, for example, my bar… Let’s put in a couple more bars here just so we see that. Of course, the total length of the bar still has to be the same. I don’t lose any energy. Near the ends, I only have a little bit of kinetic. Here is a bar that’s near the center. Now I’ve got mostly kinetic, but still a little bit of potential.
I just changed that one box and filled it in the proper orders just so we have more… It doesn’t matter if I put green on top or pink on the bottom, but I just want it to be consistent.
I went ahead and sketched in more bars here so you have more of a complete picture. You can see how this object, as it moves away from the turning point, speeds up, and then slows down, I go from all of my mechanical being potential energy to it all being kinetic. Then it’s all converted back to potential, and this keeps happening as this object oscillates back and forth.
If you think about that, where I have all potential at the ends, and I have all kinetic in the middle, that gives me an important relationship. When I write that mathematically, I end up with an equation stating the truth of that. All of the kinetic at one instant all is the same value as the total potential at a different instant.
At any other instant in between, it’s not all just kinetic or just potential. I’ve got some combination. My bars are some combination of pink and green when I’m anywhere except these key points, the turning points in the middle. I would have to find that my kinetic in that instant plus my potential has to add up to the same total bar, the same total amount of mechanical energy.
As an aside, this bar analysis, I probably should’ve introduced it to you back when we were talking about potential energy due to gravity and stuff too. You can think about the same kind of bar picture of my total mechanical energy being converted back and forth between potential and kinetic as a rollercoaster rolls up and down hills, for example.
Let’s do just a couple of examples and end this lecture. I have here a horizontal spring on a frictionless table. One end is attached to a wall, and the other end is attached to a block, a mass of 0.2 kilograms.
It says here that the block is offset to the right from equilibrium by this amount, 0.1 meters, and then it’s released. I’m going to add that to my sketch by calling this my X axis. Here’s my zero, my X equilibrium position. This snapshot I’ve shown would be the object at the beginning here, which is my X position coordinate of +0.1 meters.
I pull it to the spot and then I let it go. It now is going to oscillate back and forth between plus 0.1 meters and minus 0.1 meters. If this is where I pull it to and let it go, this is the amplitude.
That’s the maximum and then the minimum X coordinates it is going to oscillate back and forth between. If I were to draw my little cycle that I draw here, this object’s going to oscillate back and forth.
It asked me then about the speed of this block when it’s 0.05 meters to the left of equilibrium. That would be here, at this coordinate of minus 0.05 meters. I want to know what the speed is at that point.
Let me emphasize, this is the basic question I’m being asked to answer. It’s a comparison of speed and position. Remember I suggested whenever you see that comparison of speed and position to try and apply energy conservation. That’s exactly the right tool to apply.
Our basic energy conservation equation for mass on a spring says that our mechanical energy for our object always has to be a combination of 1/2 MV squared, plus our potential. One-half KX squared.
How big is this mechanical energy, the total amount? Well, it would be the maximum potential energy, or I’ll get the same number if I knew information about the maximum kinetic energy, 1/2 MV squared.
Which of these would make the most sense to apply to this problem?
Well, what I want is V, the speed, for X equals minus 0.05 meters. To look at my basic equation then, this is what I want. I’m highlighting that in yellow. I have the mass. That’s the number that I have. It’s 0.2 kilograms.
I have the value of X of interest. I need to think more about what I would want to plug into the left side of this equation. I also need to think more about K. What do I know about that?
Well, if you read this last section here, it says that the same block hanging from the same spring would stretch the spring by 0.2 meters. That gives me information about K. Here’s what I mean. Let’s draw that other scenario here off to the left.
This time I’m hanging, from this spring, this same mass. From equilibrium, I know that it is stretched because of the spring force countering the force of gravity. It’s stretched by 0.2 meters. To emphasize with forces, I have the force of the spring and the force of gravity balancing.
My force analysis here of this scenario tells me that the force of the spring balances the force of gravity. Or the force of the spring is equal to MG. How big is the force of the spring? Well, the magnitude is just K times X.
Do a little algebra rearranging here then to get K, and I see that K then has to be MG divided by X, or K will have the value of 0.2 kilograms times 9.8 meters per second squared, divided by the 0.2 meters.
I’m not worried about sign much in this problem. Sign would just capture direction. I’m already taking the direction information into account by this visual mathematical equation that captures that the size of the spring force just has to equal the size of MG. I’m just putting in sizes of these quantities or magnitudes.
This of course calculates to a pretty simple number. Those two values of 0.2 cancel each other out. I end up with a K that happens to have the numeric value of 9.8, with units of newtons per meter.
I have now my value of K. I’m going to check that off. Now the only thing between me and solving for what I need is thinking about the left side. What is our total mechanical energy?
Well, as I say, I can think of that, or I can calculate that either based on knowing what the maximum potential is or kinetic. I was given in the problem this information about the amplitude. I was told that I stretch it to this point when I let it go. That has to be then my amplitude. I’m going to oscillate back and forth between plus that position and minus that position.
I do have K and A. This is a good choice to use for the left side of this equation. Let’s use the expression for all of the potential, with all of the mechanicals in the form of potential energy for the left side.
Then my basic equation becomes 1/2 KA squared is equal to 1/2 MV squared, plus 1/2 KX squared. I’m solving for V. Let me multiply through by two on both sides. That will get rid of a factor of a half.
When you do some algebra and rearrangement, you’re going to end up with V squared is equal to KA squared minus KX squared, divided by M. I then need to take the square root of both sides.
I end up with a final expression that the speed that I want is equal to the square root of KA squared minus KX squared, divided by M. The numbers I plug in for that are 9.8 newtons per meter, times my amplitude of 0.1 meters squared, minus 9.8 newtons per meter. Now it’s times 0.05 meters squared. Then I have to divide all that by my mass, which was 0.2 kilograms.
When you multiply that out, divide, subtract, square root, etc., you end up with 0.61 meters per second. That’s my speed at that location.
Last example. We have a block sliding along a frictionless horizontal surface with velocity V, and it collides with and compresses a spring. The maximum compression is 1.4 centimeters. If the block then collides with the spring while having a velocity 2V, how will the spring’s maximum compression compare? What will it be?
Let me start you out in solving this problem. We really want to compare two scenarios. In the top scenario on my frictionless surface, my block is sliding along with this velocity V. It runs into a spring that I’m going to attach to the table here rigidly. When it runs into this spring and of course compresses the spring, I’m told that it compresses it by an amount 1.4 centimeters.
That’s meters. This spot would be my equilibrium. 0.014 meters is my X coordinate when this spring is at its maximum compression.
In terms of thinking of this motion when it hits the spring, I want you to appreciate that when that block hits the spring, at that snapshot, it has kinetic energy, 1/2 MV squared, and no potential.
My maximum energy then has to be calculated as being 1/2 M times V squared. That’s my total mechanical energy. As the spring starts to compress, this kinetic decreases and I start converting that into potential energy. It all gets converted into potential at the instant that this block reaches its maximum compression, or it comes to this brief stop here at the turning point.
Then 1/2 MV squared, my incoming speed, which is going to be my max right at the beginning, all is converted into potential. One-half K, my amplitude squared.
All I want to know then is if I double V, how does my maximum compression change? How does the variable A change? That’s what A would be, the spring’s maximum compression.
Let me rearrange this equation then to solve for what we want. I’m going to get rid of the two factors of two. They cancel. I’m going to rewrite it as A squared then equals M over K times Vmax squared.
Do my next algebra step here. I’m going to take the square root of both sides. I see that A depends on the square root of M over K, times the square root of V squared. That’s just then Vmax.
Now I have the clear mathematical connection between my maximum compression and how it depends on my maximum speed. If I double my incoming speed and I leave this the same, I’m going to double my compression. If I double my incoming velocity, I’m going to double my compression. Instead of 1.4, the spring compresses to 2.8.
That then brings us to the end of Lecture 24.