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# Bonds as springs

## Video transcript

- [Voiceover] In the last video, we said that certain frequencies of IR radiation can cause the bond to stretch. And we can think about the bond as a spring. So this carbon-hydrogen bond can be modeled as a spring. If it's a spring that obeys Hooke's law, the stretching vibration of the bond is like a spring-mass oscillation. Let's look at some classical physics here, some mechanics, to see if we can understand this a little bit better. Here I have a box. So here we have a box of mass m, so this box is mass m here. And there's no friction between the box and the ground here. But the box is attached to a wall right here by a spring. So, if you grab this box and you pull this box to the right, so you're going to apply a force to the right, and so you're going to stretch the spring. So you move the box over to here, and the spring would obviously stretch out tho there, like that. Let's say you moved the box a distance of delta x. From here to here, we moved the box a distance delta x. The box is going to feel a force pulling it to the left. So if you're just holding it there, you're holding it because there's this force of the spring that wants to pull the box back to the left. So this is the force of the spring. And according to Hooke's law, the force of the spring is equal to negative kx. And the negative sign just means a restoring force. So this negative sign here means a restoring force, meaning, the force of the spring is back towards this original position right here. K is called the spring constant. So this is the spring constant right here, and it depends on how strong or weak the spring is. So if you have a very stiff spring, or a strong spring, you have an increased value for k. If there's a loose spring, or a weak spring, a decreased value for k. Let me go ahead and write that down. Strong spring or stiff spring increased value for k. If you have a weak or loose spring, a decreased value for k. X refers to the displacement from the original position here. We displaced the box delta x here, so that's what we're talking about here. Let's think about what these terms mean for the force of the spring. If you have a strong spring, the box is going to experience a stronger force, right, according to Hooke's law. If you increase k, you increase the force of the spring. And also, the more you stretch it ... All right, so if you increase the value for delta x, also, the stronger the force of the spring would be. And so, if you get into the physics of ... We're not going to get too far into it, but you could set this equal to mass times acceleration, because negative kx, F is equal to ma, that's equal to ma, and the acceleration is the second derivative of the position, so you could write the acceleration as being the second derivative of the position. I won't get into the physics of it, but eventually, you can solve for the frequency of oscillation for this spring-mass system. And when you do that math ... Let me go ahead and write down what you would get ... You're going to get the frequency of oscillation is equal to one over two pi square root of k over m. Let's think about what the frequency of oscillation is referring to. If you, once again, you pull the box to the right, so we're starting with our box at this position right here, and then, if you release it, think about what happens. What's the motion of the box? Well, the force of the spring is going to cause the box to go towards this direction, right? It's going to keep going, right, because of energy, and it's going to go all the way over to here. So now it's going to compress the spring. And now there's a force of the spring back towards this direction, like that. Of course the spring is now compressed, all the energy is stored in the compression of the spring, it pushes on the box and the box, therefore, goes back towards the equilibrium position, the center. But it's going to keep on going until it reaches this position where we started, so the original position. That's like one oscillation. And the time it takes for one oscillation is called the period. So that would be the period. The period is measured in seconds. So the period is measured in seconds, the time it takes for one oscillation. Right, so one over the period ... One over the period is equal to the frequency. So you could write frequency like that, or you could write frequency like this. And so the frequency would be equal ... The units would be one over seconds. And this is talking about the number of oscillations per second. So frequency is number of oscillations, number of oscillations per second. So what affects the frequency? Well, once again, the spring constant affects the frequency, so k. If you increase the value for k, you're going to increase the frequency. If you have a strong spring, it's going to cause that mass to oscillate faster. Let me write that down. Increase k, all right, increase the strength of the spring, you increase the frequency. What about the mass? If you increase the mass, what happens to the frequency? If we increase this number, just think about it mathematically, right? That would decrease this number. So an increase in the mass, right, would decrease the frequency. You wouldn't get as many oscillations per second, you get a slower oscillation. And so this is what we're thinking about when we're treating this bond as a spring. So let's go back up here to this diagram, right, where we have the bond as a spring. So let's think about keeping the carbon stationary for now. So we keep the carbon stationary, and we're going to pull, we're going to pull on the hydrogen here. So we're going to pull to the right. Just let me go ahead and draw a line. So we're going to pull to the right and stretch the spring out, right? So we're going to pull the hydrogen here to the right. So the force that we apply is in this direction. So the hydrogen feels a restoring force in this direction. That's the force of the spring. And so, once again, if you have a really strong bond, right, so a really strong bond, that means you have an increased value for k, right? So a stronger bond means a stronger spring constant, if you will. So you're going to have an increased frequency of oscillation. So after you release this hydrogen, it's going to oscillate in a way analogous to this spring-mass system here. Once again, if you increase the strength of the spring, you increase the frequency. What happens if you change the mass, right? What happens if you ... Let me use a different color here. What happens if you change the mass? Instead of hydrogen, if you change it to carbon or oxygen, or something maybe has more mass than hydrogen? Right, what would happen to the frequency of oscillation? If you increase the mass, you decrease the frequency of oscillation. This is a pretty good model, if you think about it this way. However, we're talking about only the hydrogen moving this time. But we know that, when we're talking about a stretching vibration, both of these, both of these are moving. So let's get some more space and let's deal with that next. We have this situation where we have two masses. I'm just going to generalize it now. So we have m one, and then we have m two over here and the bond between them. So m one and m two are the masses of the nuclei of the atoms that we're talking about. So both masses are actually moving in our situation, so we have to amend the equation for frequency slightly. So let's go back up here and let's look at this equation again. The frequency of oscillation is equal to one over two pi square root of k over m. But this is assuming only one mass is moving. We have both masses moving when we're talking about a stretching vibration. We have to use something different for m. We use what's called the reduced mass. Let me go ahead and write it down here. The frequency is equal to one over two pi square root of k over m, but we can't write m any more because of the slightly different situation. We're going to use that symbol to represent the reduced mass. The reduced mass is equal to the m one times m two divided by m one plus m two. And we're talking about the mass of the nuclei in AMUs, we'll worry about units in the next video. And you can use the atomic mass to get an approximate value for the mass of the nuclei. Let's think about a carbon-hydrogen bond. So a carbon-hydrogen bond, that would be like m one is equal to carbon, and m two is equal to hydrogen. What is the reduced mass equal to? The reduced mass is equal to ... Well, the atomic mass of carbon is 12, so it'd be 12 times the atomic mass of hydrogen, which is one, all right? Divide that by 12 plus one, so let's do that math. Let's go ahead and do that really quickly. 12 times one is 12, divided by 12 plus one, which is 13, and so we get .923. We get .923, so that's the reduced mass of our system. Let's do another one. Let's do carbon-carbon this time. So carbon-carbon, the reduced mass is equal to 12 times 12 over 12 plus 12. So what is the reduced mass equal to now? 12 times 12 is 144, divide that by 12 plus 12, which is 24, so we get six. So the reduced mass is equal to six here. Let's think about what that does to the frequency of vibration, all right? So if we increase, if we increase the reduced mass, so if we're going from a reduced mass of .923 to a reduced mass of six, we're increasing this. We're increasing, we're increasing the reduced mass. What happens to the frequency of vibration? Well, if we increase this value, of course we're going to decrease the frequency of vibration. So we would expect a carbon-carbon single bond to have a lower frequency of vibration than a carbon-hydrogen single bond. Let's do this idea one more time, except, let's think about ... Let's think about a double bond. This was a carbon-carbon single bond, right? What about a carbon-carbon double bond? Well, the reduced mass would be the same, right? It would still be six. But what happens to the spring constant? What happens to the force constant? Well, for a carbon-carbon double bond, we can pretend like this is twice as strong as a single bond. So if the value for the single bond ... If the spring constant was k for the single bond, for the double bond, it would be two k. It would be twice that. So we're increasing the value of the spring constant. We're increasing k because a double bond, we're assuming a double bond is twice as strong as a single bond. So what happens to that frequency of vibration if you increase k? If you increase this value, you're going to increase this value. So if you increase k, you're going to increase the frequency of vibration because you have a stronger bond. So the bottom line is, the things you have to remember, are, stronger bonds vibrate faster. All right, so that's what this is saying right here. A stronger bond, increased k, you get an increased, increased frequency of vibration. So a stronger bond vibrates faster. And what about a lighter atom here? So if we're thinking about a hydrogen, all right ... So a hydrogen, a lighter atom, that has a smaller value for the reduced mass. So that's going to vibrate faster than a heavier atom. So it's, stronger bonds vibrate faster, and lighter atoms vibrate faster, too. And this is what we gather from thinking about bonds and springs. These are two very important points to think about. What affects the frequency of vibration would be the strength of the bond and also the mass that we're thinking about here, the reduced mass.