1

concept

## Energy in Simple Harmonic Motion

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Hey, guys. So now that we've seen the forces of mastering systems, we wanna look with the energy looks like for mastering systems. So at any point of simple harmonic motion, you've got this mass that's moving back and forth. And so it's exchanging energy, right? The energy that's associated with its motion is the kinetic energy and the other type of energy that might have has to do with the compression and stretching of the spring. So it's got some elastic potential energy Now, in all the cases that we considered, we've considered no friction. So there's no work done by non conservative forces and without allows us to do is use energy conservation. The mechanical energy is conserved. And so what we're gonna do is compare the energies at 22 special points of mass spring systems. The two special points are gonna be the amplitude and the equilibrium, so we've got minus and plus a and then we've got to the equilibrium in which X equals zero. So let's take a look what the energy looks like at all of these points. Well, we've got elastic potential energy here when X is equal to positive or negative amplitude. So that means the elastic potential energy is maximized here, one half k square. But the velocity of the specific point is equal to zero at the end points. And so there's no kinetic energy, so that contribution actually goes away. So there's no kinetic energy, which means that the Onley mechanical energy that we have is just due to the potential energy, theological energy. At that end point at the equilibrium here the situation is reversed because now what happens is when X is equal to zero. The elastic energy goes away. What does that make? It is the kinetic energy look like? Well, at the equilibrium position. We've got this object either flying to the left or to the right, and it's got its maximum velocity here. So that means that the kinetic energy will the elastic energy is equal to zero, and the kinetic energy is maximized. It's got one half V max squared, so that means that the total mechanical energy is on Lee, made up of the kinetic energy que zero Well, what about any other points and any other points? So, for instance, find Just label this point right here and I call that p so it X equals P. The elastic energy has to do with what the deformation looks like. So how much it stretched or compressed. But it also has some non zero velocity. It's also going either to the right or to the left, with some velocity. So it's got some kinetic energy. The elastic energy is just going to be one half times k x p squared, whereas the kinetic energy is gonna be one half M v p squared. And those were both not zero. So that means that the total mechanical energy is actually the sum of all of those energies. So now, if we've compared all of these energies that these different points and we remember that the total mechanical energy is conserved, it means that these things just get exchanged. They never get lost or destroyed. And so we can actually just plug in our expressions here for our energies. One half k squared. We've got one half m v max squared. That's equal to one half k times x p squared plus one half M v p squared. So this is the energy conservation equation. First Springs. It's really powerful. And what questions they're gonna ask you is ask you to relate all of these energies to each other, so we'll give you one and ask you for another on. So you need to know how to use this equation right here. So what we can do is we can use this equation and actually solve for the last cinematic quality that we need. That's the velocity is a function of position. So to do that, I'm just gonna cancel out some halves. I'm gonna manipulate some some numbers around, and then I get that the velocity is a function of position is equal to the square root of K over em times the square root of a squared minus x p squared. So that's basically it. Let's go ahead and take a look at an example. So in here in this example, we've got a 5 kg mass. We've got the k constant and we've got the amplitude, and we're supposed to find the maximum speed. So in this first part here, if I'm asked to find what the Max is, I'm just gonna write out my conservation of energy equation. Now, this is the conservation of energy equation. So I got all these things that are equal to each other. So I'm looking for specifically V max. So I'm looking for this guy right here. So let's take inventory of all my variables, I know what the mass is. I'm given what the amplitude and the K constant is. So I can actually just take a look at these two relationships in order to figure out what the max is. So the max, if I go ahead and solve for that, I'm gonna cancel out the half terms and they're gonna move the M to the other side and then take the square roots. So I'm gonna get K over em. And then I've got, uh, the amplitude that's squared. So yeah, the amplitude. So you might recognize this equation. V. Max is equal to a times omega, and that's because these two things are the same. So when I got the square root of K over over em, you might recognize that as that omega symbol. So if I've got all I need, I've got all of these variables I need. So I just take the square root of 30/5, which is Cavor M multiplied by the amplitude and when I get is a V max of 0.98. That's meters per second meters per second per second squared. So what is part V? Ask parties asking us for a speed at a specific position. So for part B now for me. So for part B, we've got X equals negative 0.2 m and we're supposed to find the velocity. So we're gonna use our velocity as a function of position equation. So we've got V when X is equal to minus 0.2 is going to be what? What we've got the square root of 30/5. That's K over em. And then we've got 04 squared minus 0.0 point two squared, and we've got this negative sign right here. So you go ahead and take the square root of that. What you're gonna find is that the velocity is equal to 0.85 m per second. So that is the velocity as a function or at that specific position. So now this last part here asks us to figure out what the total mechanical energy of the system is. So the total mechanical energy means we're gonna use that conservation of energy equation So that conservation of energy equation is this guy right here? So now which one is the one that we can use? Well, the answer is you can actually use any of them. So notice here how I have the k. I have a I also have m and V Max, so I can use all of those. So if I use this guy right here, So if I use the first terms, but I'm gonna use is one half the cake, Constant is 30. And then I got the amplitude, which is 0.4, and I square that what I get is a total mechanical energy of 2. jewels, right? So if I use those other two variables So if I go down here, I've got one half times five and then v max was 0. squared, so I got a total mechanical energy of 2.40 jewels, which is the same exact answer. So the idea is that you can actually use any one of these to relate the energy different positions. Alright, guys. So that's it for this one. Let's keep going

2

example

## Example

6m

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Hey, guys, let's check out this example here. So we've got a 0.25 kg mass. It's oscillating on a spring. We're told the periods 3.2 seconds. Now we're told that at a specific position the speed is 5 m per second. So it's a lot of numbers. I'm just gonna write starts, start writing stuff down. So I've got the mass is equal to 0.2 25 periods, 3.2 at X equals 0.4. I've got the velocity at that specific position is five. And now we're supposed to do is figure out what the amplitude of the system is. So I'm looking for capital A. So let's go to my equations. Right. But unfortunately, almost all of them have Aisin them. So let's take it from the top. So you only use these equations when you're told something about the force or the acceleration. What we're not so we're not gonna use these, which means we can't use their max is either. Now, in these second row equations, we can only use them if we have a time to plug in right, because remember these air functions of time now we don't have a time to plug in. Which means we also can't use the max values of these equations. Either we're not told anything about the max or a max or anything like that, so I can't use those. So whenever all this fails, we're going to use our energy conservation. So let's take a look at which equation and specifically gonna use. So if I'm looking for the amplitude, that's gonna be capital A and that appears in both of these equations. So in this equation, I have the mechanical energy that I'm gonna need to know. And let's take a look at all of these components here because I've got a bunch of different equations so I could figure this out. The problem is, I have to figure out what K is which I might be able to dio. But I don't have the mechanical energy, so that means I can't use this first part. Right. Okay. What about the second equation in the second equation? I don't have what v. Max is so and I don't have that. I don't have the mechanical energy. I can't solve it. And in this last equation over here, um I don't have the K, and I don't have the mechanical energy either. So let's just take a look at what I know, right? I can't use that. But I do have this velocity as a function of X. So let's take a look at my equation at the very bottom. So I've got this velocity as a function of X. So maybe I can use this one. It's also the simplest one, so let's just try it. So let me write it out. I've got that. The velocity is a function of X is equal to omega times, the square root of the amplitude minus the X at a particular position squared. So let's take a look. I've got my amplitude that I'm trying to solve for And I do know what this velocity is. A function of X is and I also do know what that exposition is. So all I really have to do in this case is solved for Omega. So let's see if I could do that all over here. So Omega is equal to what I've got this big omega equation I'm gonna use over here. I've got two pi frequency and then I've got two pi over, period, and that's equal to square root of K over em. So let's take a look. I don't have anything about the frequency, the linear, your frequency. But I do have something about time. I do have one of those variables, and I also don't You know, I don't really know anything about K, so let's just not even worry about that. If I'm trying to figure out what Omega is va Omega is just gonna be two pi over the period. I have period of 3.2 seconds. So I've got that right over here. Let's just go ahead and plug that in. If you do that, you're gonna get a radiance, or you're gonna get 1 96 radiance per second. So I'm just gonna plug it back into that formula. So now I'm just gonna start plugging numbers in because I know what all of these things are. The V X is equal to five, and the Omega is 1.96. Then I've got square root of a squared minus this particular position squared, right? That's 0.4. I'm told that X is equal 2.4. Okay, so now I'm just gonna go ahead, divide over the 1.96 it'll become 2.55. And then I've got this square roots of a squared minus 0.16 and let me go and write that 0.16. Okay, so now I've got this nasty square root in here, so get rid of it. I just got a square, both sides. So we've got 2.55 squared, equals the square. So the square, we will just go away, and I've got a squared minus 0.16. So that means that 2 55 squared right? Plus, that 0.16 is going to be a square. So I couldn't plug this into your calculator and then just remember to take the square root and we'll get an amplitude. That's 2.58 m. So that is the answer to part a. Let's take a look at part B now. So Part B is asking us to figure out what the total mechanical energy is. So let's go to our equations Now, Fortunately, this is pretty straightforward. We know we're gonna have to use this mechanical energy equation now. It's just now The question is, which part do we use? So let's take a look. I've just figured out what this amplitude is. I've just figured out what a is. So if I wanted to figure out the mechanical energy, all I have to do is figure out what the K constant is. Okay, so that might be a place to start now again, I don't know what the V Max is, so I can't use that guy. And what about the third one? Well, in the third one, let's take a look. I have what the position is. I have the mass and the velocity. But in order to solve this, I'm gonna need to figure out what K is. So if both of them I need to figure out what K is, I'm just gonna use the simplest one the one half k squared. So let's start out with that one. So, in part B, the mechanical energy I'm gonna use is one half times k a squared. I just figured out what a is. All I have to do is just go over here and figure out what K is. So how do I figure out K? Well, I've got that K. I can find out if I have the, um, the forces or the acceleration, but I don't have any of those, so I can't use that. So what I could use is I could use this omega equation, right? So I've got that Omega is equal to two pi f and two pi over tea. But I want that, but I've got that's equal to the square root of K over em. Now, in this case, I know what Omega is, and I know what em is. So I could go ahead and figure that out. So Omega I know is 1 96. We got square roots of K over 0.25. That's the mass. And then I've just gotta square both sides. So 1 96 squared is equal to, uh, k over 0.25. So that means 0.25 times 1.96 is squared is equal to K. And so, if you if you plug that in really carefully, you're gonna get 0.96 as a k constant right, Newtons per eater. So that is what K is equal to now. I could just go ahead and plug it right back into my mechanical energy formula. So I got this. Total mechanical energy of the system is one half 0.96 then the amplitude is 2.58 You gotta square that. So if you plug that in, you should get three points, 20 jewels. Alright, so that's pretty much it. So you could have also figured it out by using this equation. But it's just a little bit extra steps. So let me know if you guys have any questions, let's keep going on for now.

3

Problem

A block of mass 0.300 kg is attached to a spring. At x = 0.240 m, its acceleration is a_{x}= -12.0 m/s^{2} and its velocity is v_{x}=4.00 m/s. What are the system’s (a) force constant k and (b) amplitude of motion?

A

9 N/m; .35 m

B

15 N/m; .61 m

C

15 N/m; 1.22 m

D

9 N/m; .11 m

4

example

## Example

3m

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Hey, guys, let's take a look at this. We're not giving any numbers, and this problem is all gonna be conceptual, right? So we're saying that we're increasing the amplitude of oscillation and were asked which of these statements are correct? So let's just take a look at the first one. The first one, the period of oscillation increases. Okay, so let me just go into my equations and figure out What do I have for equations of period? Well, that's going to be the big Omega equation right here. So in this equation for Omega and T and all that stuff, is there anything that involves amplitude? No, there's not right. It's just the frequency. The period and then K doesn't change with a and the cases of property. And then masses just mass. So that means the period of oscillation does not increase. So that's wrong. So let's take a look at Yeah, so that's wrong. Um, so let's take a look at the second one. The maximum acceleration increases. Let's look at our formulas for a max. We've got two of them. This one, a max is when, uh, K over M times A. So what happens is that in this equation for a Max? If k over M times A. If this a increases and K and M are just properties of spring and the mass of the object, then that means that a max also has to increase. So that means that this is actually a correct statement. I think I've got a repeat between C and D, but whatever. Okay, so that is actually true s all right that there Part C is asking for the maximum speed. So what happens to the maximum speed if we increase the amplitude? So just like we did over here, the V max is what they're looking for. So with this V, Max is equal to eight times omega. So we've got that V. Max is a Omega. Now let's see what happens if I increase this amplitude. Does anything happen to omega? Well, Omega is equal to just you know, all of this stuff over here, and we said that that doesn't change with amplitude. So if a goes up, what happens is the max goes up. Sometimes you have to check if one of these variables will, like decrease if you increase the amplitude because sometimes there might be that kind of relationship. So that's just like an extra question we have to ask. Okay, so that means that that is actually true, right? So the V Max does actually increase, So that is good. So what about this? This d which I guess we'll call the maximum kinetic energy. So Okay, Max. So let's look at our energy conservation equation. Well, so energy conservation is like the maximum elastic potential energy, whereas this is the maximum kinetic energy. So KMAX is when one half v max squared. So we just said that V max squared increases if you increase the amplitude. So that means that KMAX also has to increase. So that means Yes, it does. So sometimes you also might have those, like, indirect relationships as well, so that means that does increase. So what about the max potential energy? Okay, what does that mean? So again, we're gonna look at that same equation. We said that the maximum potential energy theological potential was one half k k squared. So for E, we've got that you Max is equal to once. We've got one half k a squared, so it's pretty obvious that if this a just goes up, that means the maximum potential energy also has to increase. So that means that that is true. Okay. And now for this last one, the maximum total energy. So maximum total energy is just gonna be this whole entire, um, mechanical energy formula. So what happens is again if you increase the maximum amplitude, right, If it goes up, then the whole entire mechanical energy goes up. So if mechanical energy is one half k a squared and a goes up, that means mechanical energy also goes up. So that means that that is also a true statement. Boom books. All right, so let me know if you guys have any questions about this and let's keep going.

Additional resources for Energy in Simple Harmonic Motion

PRACTICE PROBLEMS AND ACTIVITIES (8)

- A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 cm and a frequency of 0.850 Hz. Find (a) the ...
- A thrill-seeking cat with mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and osci...
- A mass is oscillating with amplitude A at the end of a spring. How far (in terms of A) is this mass from the e...
- For the oscillating object in Fig. E14.4 , what is (b) its maximum acceleration?
- For the oscillating object in Fig. E14.4 , what is (a) its maximum speed?
- A 0.500-kg glider, attached to the end of an ideal spring with force constant k = 450 N/m, undergoes SHM with ...
- A 0.500-kg glider, attached to the end of an ideal spring with force constant k = 450 N/m, undergoes SHM with ...
- A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The am...