okay so for this we're going to come back to this idea of restoring force which we've talked about before so we've talked about the force from a spring briefly when we started talking about elastic potential energy and we had so called Hookes law where we had that the restoring force from a spring was equal to minus K which was the spring constant times X and this was a so-called restoring force because it always points towards equilibrium so what that means is that if it's just stretch a string a little bit it wants to go back to equilibrium if you compress it it goes back to equilibrium so if we go ahead and just visualize this again to remind ourselves if we have some equilibrium position for a spring we call that x equals zero so this X is very specifically the distance from equilibrium and if you stretch the spring then there's a force here that's acting to restore it back to equilibrium and the X is the distance from equilibrium to where it's at and this negative sign denotes that the force is always opposite your displacement from equilibrium so that if you compress the spring you squish it in now the force wants to push it back to equilibrium and the X points from equilibrium to that location all right so if this is the force it's the only force acting then we can use Newton's second law and we can say that the force is equal to the spring force which is equal to the mass times the acceleration so we can write that the mass times the acceleration again if we're just talking about one dimensional motion the acceleration in X is going to be equal to this force here from the spring which is equal to negative K X so taking the first and last part of these equations we can then write that the acceleration a sub X is equal to minus K over m times X and again this X is relative to equilibrium where x equals zero and for this acceleration to now try to solve this and write out a math at fault mathematical representation of the motion we can remember that the acceleration is actually related to the position right because the acceleration is equal to DV DT so DV in the x-direction DT which then is equal to D DT of the velocity which is DX DT so that's then simply equal to d squared X DT squared so the acceleration in X is equal to d squared X DT squared the second derivative of X with a function of time so now we can rewrite and we can rewrite this as simply d squared X DT squared is equal to minus K over m times X okay so this is what's called a differential equation so you likely have not taken differential equation yet since calculus is the only calculus one is the only prerequisite for this course but we can appreciate even without having that further mathematical capabilities yet is that this is an equation where K and M are constants right and so the only real variable is X as a function of time and so you can it's not a simple algebraic you know rearrange things and solve for X but it is an equation for X where we should be able to come up if we have the mathematical capabilities for a solution of X as a function of time would solve this equation and then this is what a differential equation is where it's not an algebraic expression where you just have to rearrange things but again it has your derivatives in it and it has the second derivative of X with respect to time is equal to negative constant times X so even without knowing differential equations and having to come up with the solutions we can understand the solution to this one because if you think about types of functions where if you take a derivative and you take a derivative again you see them end up with basically the same kind of expression that you started with so you should think about cosines and sines and even e to the X for the example where if you take multiple derivatives you're not substantially changing the nature of the function so it turns out then that for this solution using just differential equations to come up with it that you can write the position as a function of time and this is again for just a mass on a spring obeying Hookes law is going to be equal to some amplitude a and then your book and I'll use that convention as well chooses to use the cosine solution you could also write it as a sine function or as an exponential eat in it X essentially function so cosine Omega T plus Phi and I'll define all of these in a second but the idea being that if you take the derivative of X as a function of time the derivative of cosine is negative sine the derivative of sine is cosine so you come back and you can imagine how this would satisfy that equation all right let's go ahead and define the expressions in here because this is the main result that we're going to use in periodic motion to describe and understand the motion so Omega here is what's called the angular frequency and Omega is actually defined to be the square root of K over m and that's specifically for a mass on a spring now the reason that this happens is because when you start trying to solve for the derivative and this is your original equation then you need to have a K over m come out from somewhere so when you take the derivative of this the first time you end up with the square root of K over m the second time you end up with another square root of K over m and that's where the K over m comes from so we can now basically say that d squared X DT squared is equal to negative times Omega squared times X and this here is a general expression for anything that obeys simple harmonic motion and so then that's basically going to be you know sinusoidal or cosine you know soil motion anything that's periodic symmetric in the way that a sine function cosine function are okay so Omega is the square root of K over m for a mass on a spring and a as I mentioned is the amplitude and so you can imagine that this then would be in order to find the amplitude sometimes you can say that this is when cosine of Omega T plus Phi is equal to 1 right so it's the maximum location that the position has away from equilibrium and Phi is what's called a phase constant and this allows you to start your clock when T equals zero at any time right a cosine function typically will have at T equals zero X is equal to a so it's at its maximum amplitude and then it comes down but maybe you started your clock differently or it's a different function and it just shifts it over it allows you basically to transform a cosine function into a sine function or other combinations there so the phase constant is going to depend on where T equals zero so specifically what is X as a function of zero another way to think about this is where did you start the clock where was it when you pressed go on your stopwatch and started measuring the motion a lot of times we'll deal with a face constant zero but it's good to know that this then allows you to have this translation between not just a cosine wave but a sine wave as well okay so if we now go ahead and just look at this so if we write again X as a function of T is equal to a times cosine of Omega T plus Phi then we can look at what this graph looks like and I'll plot this for a Phi equals zero so that I'm just going to start and draw a normal cosine wave where here is the amplitude right this would be zero and this would be negative a so top to bottom it actually goes to a but a is the displacement from equilibrium and that's as a function of time so one full cycle right it gives us the time for one full cycle so the time for one full cycle note that it's not just top to bottom but top to bottom and then back to the top is called the period and that's going to be denoted by capital T so the period of oscillation capital T is equal and related to the frequency it's actually going to be equal to 2 pi over this Omega which was our so called angular frequency so we can then say for this graph Phi equals zero so X is equal to a times cosine Omega T sorry that's a little T X evaluated at zero simply equals a and then note that X evaluated at the period T is also equal to a because it equals a time's the cosine of two PI over Omega that's evaluated here for T is equal to two PI over Omega times Omega and so then that's just the cosine of two pi and that's where the a comes from okay so getting comfortable with this cosine function mathematically and how it you know translates into this motion is you know kind of why I drew this out so that you can picture what hopefully you're you know encountered your cosine functions before in your math courses and just having a refresher about some of their characteristics now back to this idea of the period is equal to two PI over Omega T again is equal to two PI over Omega and the period is equal to the inverse of the frequency not the angular frequency but just the normal frequency so frequency F is equal to cycles per second which is equal to a unit of a Hertz and it turns out that the period is equal to 1 over the frequency so I'll box some of the main equations that you'd be then provided with on your equation sheet so the period is equal to 1 divided by the frequency and the period is also equal to 2 PI over Omega and we know that Omega equals this word of K over m so we have a lot of different combinations that we can write down I'll write down some of the most popular ones so we have that here we have that T is equal to 2 PI over Omega which is equal to 2 pi over and then Omega is equal to the square root of K over m so that's one way which again is equal to 1 over F so you have various combinations here you can also relate the frequency and Omega 2 PI over Omega is equal here to the period which equals 1 of F and so you can write more naturally that the frequency F is equal to the angular frequency divided by 2 pi so that's also a common one okay so we can now basically move in between anything any expression of the angular frequency and the regular frequency and the period and then use that as part of our expression for the motion that's given here so if we now go ahead and go back to our general expression so if I say in general we can write again X as a function of T is equal to a times the cosine Omega T plus Phi what we know is that the maximum location is the amplitude and we should be very comfortable with this but we also know that the velocity is simply DX DT so what that allows us to do then is write an expression for the velocity as a function of time which is just simply going to be the derivative of this so that's going to be equal to when Omega gets pulled out here cosine negative sign so negative a omega times the sine of Omega T plus Phi and that's just the derivative of X as a function of time and so what we can see from this now is that the maximum velocity V Max is equal to negative a oh may go so again no you take your expression for X as a function of time you take the derivative of that with respect to time and then that amplitude gives you your maximum velocity so that happens to be equal to negative a oh may go and if we then to write the speed the maximum speed would just be the actual absolute value of this and we can do a similar thing for the acceleration because we know the acceleration is equal to DV DT so the acceleration then as a function of time sine is cosine so that would be equal to negative a times another Omega so Omega squared here times cosine Omega T plus Phi and we can again write the maximum acceleration equaling to a Omega squared so now we have an expression for the maximum acceleration the maximum speed and the maximum position from equilibrium just all from this one base equation again by taking the derivative and finding that value there so what we can also do is note that we can go back to our original expression for a differential equation which was that the acceleration as a function of time was equal to negative Omega squared times X as a function of time which remember worked out because we had that d squared X DT squared equals negative K over m which is omega squared times X so that then also shows that this expression satisfies this differential equation ok for the last part of this video let's just look at a graph that's shown and it has numbers provided for here on the x axis that says 2 meters and negative 2 meters 1.5 seconds 3 seconds and 4.5 seconds for different points and one point P marked on this graph as a function of X of T and then there's questions what is the sign of the velocity at Point P what is the sign of the acceleration at Point P describe the motion at Point P and then write the general equation X as a function of T for this graph so if we go ahead and we think about the first question what is the sign of the velocity at Point P well this then brings us all the way back to the beginning of the course where we remember that the velocity right is equal to DX DT so it's basically equal to the the derivative at that point so we can see very closely very clearly that the velocity is negative so the velocity at Point P is negative and that's going to be from the slope right and that tells us remember a negative velocity means it's headed when we get to this step about describing the motion it's headed in the negative x direction so it's headed in the negative x direction and that also makes sense just from looking at this graph okay for the second part here the acceleration is equal to DV DT so it's equal to the rate change of the velocity so that's one way to picture this is how is the slope changing and then the next one is that we also have that the acceleration at this point we can relate it to the position right so there's multiple ways to kind of think about this if we do it mathematically because that tends to work best for most students if we fall back on the idea that the acceleration is equal to negative Omega squared X so we want to recall that the acceleration then from this equals negative Omega squared X is always going to be opposite X and that makes sense right because again when you have this basic picture of you know the equilibrium position and you stretch your spring the force is always going to be pointed opposite your displacement oh sorry that went off the screen let me show that again the force is always going to be opposite your displacement so that means the acceleration is always opposite X right this is going to be the idea of a restoring force so that makes this very easy right this is the mathematical representation of that because if Point P is here where X is positive then that tells us that the acceleration has to be negative so X is positive so the acceleration is negative so what that tells us then is that if it's headed in the negative direction and the acceleration is negative then at Point P it must be speeding up because it tells us that the acceleration and the velocity are in the same direction the acceleration and the velocity are in the same direction so again going back to the very beginning of this course we know that that means that the object is speeding up and it's headed in the direction of the velocity so in this case we would describe the motion as headed in the negative x-direction and speeding up so it's headed back towards equilibrium in the negative x direction and it's speeding up now for Part C we can then write out the equation X as a function of time for the graph and for this we note that Phi is equal to zero and the reason I know that Phi equals zero is because I'm starting at my positive displacement right here so a Phi equals zero then we can say that X as a function of T is equal to a times the cosine of Omega T and we know what a is because a is going to be equal to the amplitude given in the problem which was just 2 meters and then we see that the period of oscillation is 3 seconds so we can write this as a times the cosine of Omega which is 2 pi over the period times T so that's going to be equal to 2 meters times the cosine of 2 pi over 3 that was 3 seconds times T and that gives us then our expression of X as a function of time for that particular graph just plugging in the values from the graph we now have for any given time what is the position ok that concludes this discussion of periodic motion will continue this in class any questions let me know