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# Harmonic motion part 2 (calculus)

## Video transcript

so where I left off in the last video I just rewritten the spring equation and I just wrote you know force is mass times acceleration and I was in the process of just saying well if if X is a function of T what's acceleration well velocity is the derivative of X with respect to time right your change in position over change of time and acceleration is is the derivative velocity or the second derivative of position so you take the derivative twice of X of T right so let's rewrite this equation in those terms let me erase all of this I actually want to keep all of this just so we remember what we're talking about this whole time let me see if I can erase it too cleanly that's pretty good let me erase all of this all of this I'll even erase this I think that's pretty good all right now back to work so we know we know that or hopefully we know that acceleration is the second derivative of X as a function of T so we can rewrite this as mass times the second derivative of X so I'll write that is well I think the easiest notation would just be X prime prime that's just the second derivative of X as a function of T I'll write I'll write the function notation just so you remember that this is a function of time is equal to minus K times X of T and what you see here what I've just written this is actually a differential equation a differential equation and so what what is a differential equation well it's an equation where in one expression in one equation on both sides of this you not only have a function but you have derivatives of that function and the solution to a differential equation isn't isn't just a number right a solution to the equations that we've done in the past our numbers essentially or a set of numbers or maybe a line but the solution to a differential equations is actually going to be a function or a class of functions or a set of functions so that's a it'll take a little time to get your head around it but this is as good an example as ever to be exposed to it and we're not going to solve this differential equation analytically we're going to use our intuition behind what we did earlier in the previous video we're going to use that to to guess at what a solution to this differential equation is and then if it works out then we'll have a little bit more intuition and then we'll actually know what the position is at any given time of this mass attached to the spring so this is exciting this is a differential equation we when we drew the position our intuition for the position over time our intuition tells us that it's a cosine function with amplitude a so we said it's a cosine Omega T you know where this is the angular angular velocity of well I don't to go into that just yet what we'll get a little bit more intuition a second and now now let what we can do is let's test this expression this function to see if it satisfies this equation right so what is if we say that X of T X of T is equal to a cosine of WT what is the derivative of this X prime of T and you could review the dirt the derivative videos to remember this well it's the derivative of the inside so it'll be that Omega times the outside scalar a Omega and then the derivative I'm just doing the chain rule the derivative cosine of T is minus sine of whatever is in the inside so it's put the up of the minus outside so it's minus sine of WT and then if we want the second derivative so that's X prime prime of T X prime prime of T let me do this in a different color just so it doesn't get monotonous that's derivative of this right so what's the derivative of we take the derivative these are just scalar values right these are just constants so the derivative of the inside is an Omega I multiply the Omega times the scalar constant I get I get minus a Omega squared and then derivative of sine is just cosine but the minus is still there cuz I have the minus to begin with minus cosine cosine of Omega T now let's see if if this is true so if if this is true I should say I should be able to say that M M times the second derivative of X of T which is in this case is this x minus a w squared cosine of WT that should be equal to that should be equal to minus K times my original function times X of T and X of T is a cosine WT so a cosine I'm running out of space WT hopefully you understand what I'm saying I just substituted X prime prime the second derivative into this and I just substituted X of T which I guessed as that in here and now I got this and let me see if I can rewrite maybe I can get rid of the spring up here I'm trying to look for space I don't want to get rid of this because I think this gives us some intuition of what we're doing one of those days that I wish had a larger blackboard let me erase the spring hopefully you can remember that image in your mind and I check and erase that I can erase that erase all of this just so I have some space without getting rid of that nice curve I took the time to draw in the last video almost there ok back to work to work make sure my pen tools right ok so what all I did neat is as I took we set the by the spring constant if you rewrite force as mass times acceleration you get this which is essentially a differential equation I just rewrote acceleration as the second derivative then I took a guess that this is X of T just based on our intuition of the drawing I took a guess and then I took the second derivative of it right this is the first derivative this is the second derivative and then I substituted the second derivative here and I substituted the function here and this is what I got and so let me see if I can if I can simplify that a little bit so if I rewrite there I get minus M a w squared cosine of WT is equal to minus K a cosine of WT well it looks good so far let's see we can get rid of the minus signs on both sides get rid of the minus signs on both sides get rid of the A's on both side right we can derive both divide both sides by a we just in black just so it really erases it so if we get rid of a on both sides we're left with that and then so let's see we have MW squared cosine of Omega T is equal to K cosine of Omega T so this equation holds true if what is true this equation holds true if if M W squared or Omega squared I think that's Omega I always forget my is equal to K or another way of saying it if Omega squared is equal to K over m or Omega is equal to the square root of K over m so there we have it we have figured out what X of T has to be we said that this differential equation is true if this is X of T and Omega is equal to this so now we've we figured out that the the actual function that describes the position of that spring as a function of time X of T X of T is going to be equal to we write about D a and that's just intuition right because the amplitude of this cosine function is a a cosine instead of writing W because we didn't you know we can write now write the square root of K over m the square root of K over m T that to me is amazing we have now using not too sophisticated the calculus solve the differential equation and now can if you tell me at five point eight seconds whereas X I can tell you and I just realized that I am now running out of time so I will see you in the next video