Energy in Simple Harmonic Motion: Study with Video Lessons, Practice Problems & Examples

In a mass-spring system undergoing simple harmonic motion, energy oscillates between kinetic energy and elastic potential energy. At maximum amplitude, elastic potential energy is maximized while kinetic energy is zero. Conversely, at equilibrium, kinetic energy is maximized and elastic potential energy is zero. The conservation of mechanical energy can be expressed as 1/2k(a)2 = 1/2k(x)2 + 1/2mv^2. This relationship allows for the calculation of velocity at any position.

Energy in Simple Harmonic Motion

Hey, guys. So now that we've seen the forces of mass spring systems, we want to look at what the energy looks like for mass spring 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 they 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've considered, we've considered no friction. So there's no work done by non-conservative forces. And what that allows us to do is use energy conservation. The mechanical energy is conserved. And so what we're going to do is compare the energies at 2 special points of mass spring systems. The 2 special points are going to be the amplitude and the equilibrium. So we've got minus a and plus a, and then we've got the equilibrium in which x equals 0. So let's take a look at 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, 12kaa2. But the velocity at the specific point is equal to 0 at the endpoints, and so there's no kinetic energy. So that contribution actually goes away, so there's no kinetic energy. Which means that the only mechanical energy that we have is just due to the potential energy, the elastic energy at that endpoint. At the equilibrium here, this situation is reversed. Because now what happens is when x is equal to 0, the elastic energy goes away. And what does 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, well the elastic energy is equal to 0 and the kinetic energy is maximized. It's got 12mv_max2. So that means that the total mechanical energy is only made up of the kinetic energy, k zero. Well, what about at any other points? At any other points, so for instance, if I just label this point right here and I call that p. So at x equals p, the elastic energy has to do with what the deformation looks like. So how much it's 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 12kx_p2, whereas the kinetic energy is going to be 12mv_p2, and those are both not 0. 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 at 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. 12ka2, and we've got 12mv_max2, and that's equal to 12kx_p2 plus 12mv_p2. So this is the energy conservation equation for springs. It's really powerful. And what questions are going to ask you is to relate all of these energies to each other. So they'll give you one and ask you for another, and 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 kinematic quantity that we need. That's the velocity as a function of position. So to do that, I'm just going to cancel out some halves. I'm going to manipulate some numbers around, and then I get that the velocity as a function of position is equal to the square root of k over m 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 kilogram 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 v_max is, I'm just going to 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 v_max is. So v_max, if I go ahead and solve for that, I'm going to cancel out the half terms and I'm going to move the m to the other side and then take the square roots. So I'm going to get k over m, and then I've got 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 m, 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 over 5 which is k over m, multiply it by the amplitude, and what I get is a v_max of 0.98, and that's meters per second, not meter per second squared. So what does part b ask? Part b is asking us for a speed at a specific position. So for part b, we've got x equals negative 0.2 meters, and we're supposed to find the velocity. So we're going to 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? Well, we've got the square root of 30 over 5, that's k over m. And then we've got 0.4 squared minus 0.2 squared, and we've got this negative sign right here. So if you go ahead and take the square root of that, what you're going to find is that the velocity is equal to 0.85 meters per second. So that is the velocity at that specific position. So now this last part here asks us to figure out what the total mechanical energy of the system is, which means that we're going to 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, what I'm going to use is 12k30a0.42, what I get is a total mechanical energy of 2.40 joules. Right? So if I use those other two variables, so if I go down here, I've got 125v_max 0.982, so I get a total mechanical energy of 2.40 joules, which is the same exact answer. So the idea is that you can actually use any one of these to relate the energy at different positions. Alright, guys. So that's it for this one. Let's keep.

Example

Hey guys, let's check out this example here. So we've got a 0.25 kilogram mass. It's oscillating on a spring. We're told the period's 3.2 seconds. Now we're told that at a specific position, the speed is 5 meters per second. So it's a lot of numbers. I'm just gonna start writing stuff down. So I've got the mass is equal to 0.25, the period's 3.2, at x equals 0.4, I've got the velocity at that specific position is 5. And what 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 A's in them. So let's take it from the top. You only use these equations when you're told something about the force or the acceleration. We're not. So we're not gonna use these, which means we can't use their max values either. Now, in these second row equations, we can only use them if we have a time to plug in. Right? Because remember these are 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 v max or a max or anything like that, so I can't use those.

So whenever all this fails, we need to use our energy conservation. So let's take a look at which equation I'm 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'd have to figure out what k is, which I might be able to do, but I don't have the mechanical energy. So that means I can't use this first part. Right? Okay. But what about the second equation? In the second equation, I don't have what v max is. So and I if I don't have that and I don't have the mechanical energy, I can't solve it. And in this last equation over here, 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 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 as a function of x is, and I also do know what that x position is. So all I really have to do in this case is solve for omega. So let's see if I can do that all over here. So omega is equal to what? Well, I've got this big omega equation I'm gonna use over here. I've got 2 pi frequency, and then I've got 2 pi over period, and that's equal to square root of k over m. So let's take a look. I don't have anything about the frequency, the linear 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, omega is just gonna be 2 pi over the period. I have the period as 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 radian or you're gonna get 1.96 radians 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 vx is equal to 5 and the omega is 1.96. And 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 to 0.4. Okay. So, now I'm just going to go ahead and divide over the 1.96 and it'll become 2.55, and then I've got the square roots of A squared minus 0.16. And let me go ahead and write that, 0.16. Okay. So now I've got this nasty square root in here. So to get rid of it, I've just got to square both sides. So we've got 2.55 squared equals the square so the square root 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 gonna be A squared. So go ahead and plug this into your calculator and then just remember to take the square root and we'll get an amplitude that's 2.58 meters.

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 going to 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 1 half k A squared. So let's start out with that one. So in part b, the mechanical energy I'm gonna use is 1 half times kA 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, 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 2 pi f and 2 pi over t, but I don't want that. But I've got that that's equal to the square root of k over m. Now in this case, I know what omega is and I know what m is, so I can 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 got to square both sides, so 1.96 squared is equal to k over 0.25. So that means 0.25 times 1.96 squared is equal to k. And so if you plug that in really carefully, you're gonna get, 0.96 as a k constant. Right? Newtons per meter. So that is what k is equal to. Now I can just go ahead and plug it right back into my mechanical energy formula. So I've got that this total mechanical energy of the system is 1 half, 0.96, and then the amplitude's 2.58, and you got to square that. So if you plug that in, you should get 3.20 joules.

Alright. So that's pretty much it. You could have also figured it out by using this equation, but it's just a little bit of extra steps. So let me know if you guys have any questions. Let's keep going on for now.

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/s² 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

Example

Hey, guys. Let's take a look at this. We're not given any numbers in this problem. It's all going to be conceptual, right? So we're saying that we're increasing the amplitude of an oscillation, and we're asked which of these statements are correct. So let's just take a look at 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 giant 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 k is just a property and then mass is just mass. So that means the period of oscillation does not increase. So that's wrong. 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 2 of them. This one, a max, is when kma. So what happens is that in this equation for a max, if a increases and k and m are just properties of the 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. So I'll write 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 this v max is equal to aω. 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 v 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, d, which I guess we'll call the maximum kinetic energy? So k 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 k max is when 12vmax2. So we just said that v max squared increases if you increase the amplitude, so that means that k max 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 going to look at that same equation. We said that the maximum potential energy, the elastic potential, was 12ka2. So for E, we've got that U max is equal to 12ka2. 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.

And now for this last one, the maximum total energy. So maximum total energy is just going to be this whole entire mechanical energy formula. So what happens is, again, if you increase the maximum amplitude, right, if a goes up, then the whole entire mechanical energy goes up. So if mechanical energy is 12ka2 and a goes up, that means mechanical energy also goes up. So that means that that is also a true statement.

Here’s what students ask on this topic:

What is the relationship between kinetic energy and potential energy in simple harmonic motion?

In simple harmonic motion, the total mechanical energy is conserved and oscillates between kinetic energy (KE) and elastic potential energy (PE). At maximum amplitude, the potential energy is maximized, and kinetic energy is zero. Conversely, at the equilibrium position, kinetic energy is maximized, and potential energy is zero. The conservation of mechanical energy can be expressed as:

$\frac{1}{2}k{a}^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$

This equation shows that the sum of kinetic and potential energy remains constant throughout the motion.

How do you calculate the maximum speed in a mass-spring system undergoing simple harmonic motion?

To calculate the maximum speed (v_max) in a mass-spring system undergoing simple harmonic motion, you can use the conservation of energy principle. The maximum speed occurs at the equilibrium position where all the energy is kinetic. The equation is:

$v$_max = a√km

Here, a is the amplitude, k is the spring constant, and m is the mass. This equation shows that the maximum speed is directly proportional to the amplitude and the square root of the spring constant divided by the mass.

What is the formula for the velocity of a mass at a specific position in simple harmonic motion?

The velocity of a mass at a specific position x in simple harmonic motion can be calculated using the following formula derived from the conservation of energy:

$v=\surd \frac{k}{m}\left(a^2-x^2\right)$

Here, k is the spring constant, m is the mass, a is the amplitude, and x is the position. This equation shows that the velocity depends on the position within the oscillation and the system's parameters.

How is the total mechanical energy of a mass-spring system calculated?

The total mechanical energy (E) of a mass-spring system undergoing simple harmonic motion is the sum of its kinetic energy (KE) and elastic potential energy (PE). It can be calculated using the following formula:

$E=\frac{1}{2}k{a}^2$

Here, k is the spring constant, and a is the amplitude. This formula shows that the total mechanical energy is constant and depends only on the spring constant and the amplitude of the motion.

What happens to the energy in a mass-spring system when there is no friction?

In a mass-spring system with no friction, the total mechanical energy is conserved. This means that the energy oscillates between kinetic energy (KE) and elastic potential energy (PE) without any loss. At maximum amplitude, the energy is entirely potential, while at the equilibrium position, it is entirely kinetic. The conservation of energy can be expressed as:

$\frac{1}{2}k{a}^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$

This equation shows that the total mechanical energy remains constant, and energy is continuously exchanged between kinetic and potential forms.