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Current time:0:00Total duration:12:20

Video transcript

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 and so this carbon-hydrogen bond can be modeled as a spring if it's a spring that Abbes Hookes law the stretching vibration of the bond is like a spring mass oscillation and so let's look at some classical physics here some mechanics to see if we can understand this a little bit better so 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 and so if you 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 to there like that so let's say you moved the box a distance of Delta X so from here to here all right we move 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 right you're holding it because there's a there's a 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 Hookes law the force of the spring is equal to negative K X and the negative sign just means a restoring force so this negative sign here means a restoring force meaning the force the spring is back towards this 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 weak spring a decreased value for K so if there's a strong we go 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 so we displace the Box Delta X here so that's what that's we're talking about here so 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 Hookes 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 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 and 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 1 over 2 pi square root of K over m so let's think about what the frequency of oscillation is referring to so 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 all right 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 is going to keep going right because of energy and it's going to go all the way over to here and so now it's going to compress the spring and now there's a force of the spring back towards this direction like that and of course the spring is now compressed all that 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 so 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 all 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 1 over seconds and this is called 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 frequencies okay so 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 and let me write that down increase K right increase the the strength of the spring you increase the frequency what about the mass if you if you increase the mass what happens to the frequency right 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 this is what we're thinking about when we're treating 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 and 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 so let me go ahead and draw a line so we're going to pull to the right and stretch the spring out all right so we're going to pull the hydrogen here to the right so the force that we apply is in this direction and 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 and so you're going to have an increased frequency of oscillation so after you after you release this hydrogen right it's going to oscillate in a way analogous to this spring mass system here so once again if you increase the strength of the spring you increase the frequency what happens if you change the mass right so what happens if you let me use a different color here so what happens if you change the mass so instead of hydrogen if you change it carbon or oxygen or something it has more mass than hydrogen all right what would happen to the frequency of oscillation if you increase the mass you decrease the frequency of oscillation and so this is this is a pretty good model if you think about this way however we're talking about only only the hydrogen moving this time but we know that we're talking about a stretching vibration all right both of these both of these are moving so let's get some more space and let's deal with that next so we have we have the situation where we have two masses I'm just going to generalize it now so we have M 1 and then we have M 2 over here and the bond between them so M 1 and M 2 are the masses of the nuclei of the atoms that we're talking about so both masses are actually moving in in 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 so the frequency of oscillation is equal to 1 over 2 pi square root of K over m but this isn't this is assuming the only one mass is moving we have both masses moving when we're talking about a stretching vibration and so we have to do it to use something different for M we use what's called the reduced mass so let me go ahead and write it down here so the frequency is equal to 1 over 2 pi square root of K over m but we can't write M anymore because we have a slightly different situation we're going to use that symbol to represent the reduced mass so the reduced mass is equal to the m1 times m2 divided by m1 plus m2 and we're talking about the mass of the nuclei in am use worried more about units in the next video and you can use the atomic mass to get an approximate value for the mass of the nuclei so let's let's think about a carbon hydrogen bond so a carbon hydrogen bond so that would be like m1 is equal to carbon and m2 is equal to hydrogen what is the reduced mass equal to so the reduced mass is equal to well the atomic mass of carbon is 12 so B 12 times the atomic mass of hydrogen which is 1 all right divide that by 12 plus 1 so let's do that math so let's go ahead and do that really quickly so 12 times 1 is 12 divided by 12 plus 1 which is 13 and so we get 0.92 3 so we get point 9 2 3 so that's the reduced mass of our system let's uh let's do another one let's do let's do carbon carbon this time so carbon carbon so the reduced mass is equal to 12 times 12 over 12 plus 12 so what is the reduced mass equal to now so 12 12 times 12 is 144 divide that by 12 plus 12 which is 24 and so we get 6 so the reduced mass is equal to 6 here so let's think about what that does to the frequency of vibration all right so if we if we increase if we increase the reduced mass right so if we're going from a reduced mass of 0.9 to 3 to a reduced mass of 6 right we're increasing this right 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 so let's uh let's do this idea one more time except let's think about let's think about a double bond so this is 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 6 but what happens to the spring constant what happens to the force constant well for a carbon-carbon double bond right this is 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 2 K it would be twice that and so we're increasing the value of the spring constant we're increasing K because a double bond were we're assuming a dull bond is twice as strong as a single bond so what happens the 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 and so the bottom line is that the things you have to remember are stronger bonds vibrate faster right so that's what this is saying right here a stronger bond increase K you get an increased increase frequency of vibration so a stronger bond vibrates faster and what about what about a lighter atom here right so if we're thinking about a hydrogen alright so a hydrogen a lighter atom that has a smaller value for the reduced mass so that's going to that's once you vibrate faster than then a than a heavier atom so stronger bonds vibrate vibrate faster and lighter atoms vibrate faster too and this is this is what we gather from thinking about bonds and Springs these are two very important points to think about so what affects the frequency vibration would be the strength of the bond and and also and also the mass that we're thinking about here the reduced mass