 Now let's have a look at this problem. I'm going to do it on paper because if I use the board We'll just not be enough space. It's going to take us a couple of pages So what we're discussing here is coupled oscillators and we're going to use the Laplace transform To solve this problem. So what we have here is a mass Hanging from a spring There's the spring and it has a spring constant k1 There's a mass for the mass m1 But that hangs from the bottom of that another spring with another mass m2 and this spring has its constant so if this hangs at equilibrium and We displace mass 1 from equilibrium. We displace mass 2 from equilibrium and we can give it an initial Velocity in any direction Now we just have to set up our equations of motion here. We have f equals ma So there's my mass 1 and a sub 1 goodness the second derivative of position. So let's put this as the Let's put this as the positive x direction Good. So what are we going to have here if we just have f equals ma So there's m sub 1 and a sub 1 there Well, first of all, there's going to be a force two forces on this mass 1 the spring pulling in that direction and the spring pulling in This direction So the first of all it's going to be negative k sub 1 x sub 1 Because we're taking down as positive up as negative. So that's going to be restoring force But there's also going to be a force in this direction In the opposite direction uses a positive direction and by how much well, we just have to look at it It's not We have to look at the net displacement if we have a look at the net displacement for this Pull down. It's the difference between this x 2 and x 1 It's gonna be x 2 minus x 1 the same is going to go for The same is going to go for for a mess number 2 it's going to be a negative case of 2 and again the difference between the difference between those remember This is now pulled down This x 2 direction so from equilibrium down. So that's x 2 But not all of that is just due to x 2 some of that Pull down the difference from there to there is due to x 1. So you've got to subtract that So that's going to leave me here with a if you can see there negative k 1 k sub 1 x sub 1 That's k sub 2 x sub 2 minus k sub 2 x sub 1 And this is going to be in k sub 2 x sub 2 plus k sub 2 x sub 1 So let's put some values in here. We have m sub 1 and m sub 2 both being 1 for this example problem and Let's have k sub 1 equals 6 and k sub 2 equals 4 We also have this initial conditions x sub 1 at time 0 Let's have that at 0 x sub 2 also at 0 so they actually both at the Resposition, but we give it initial velocity x sub 1 we're going to give an initial velocity of 1 Downward and except to we're going to give an initial velocity at 1 upward So we're going to throw these two together and now these two oscillators are going to be coupled So if we just if we just put these values in let's see what we are What we are left with so m sub 1 is just 1 so we have x sub 1 double prime equals negative k the negative k So it's going to be negative 6 x sub 1 and we're going to have a positive We said that's 4 x sub 2 minus a 4 x sub 1 And if I put those together, it's going to be a negative 10 times x sub 1 plus 4 times x sub 2 And if we look at the second one, that's going to be x sub 2 double prime equals negative 4 x sub 2 plus 4 x sub 1 Okay, so let's put down our two equations. Let's just get Everything on the left hand side. So we have two homogeneous differential equations. Let's do that So where are we going to end up with? Let's have a look we're going to have We're going to have an x sub 1 double prime Plus 10 times x sub 1 and we're going to have minus 4 times x sub 2 equals 0 And we're going to have an x sub 2 double prime plus 4 times x sub 2 minus 4 times x sub 1 equals 0 So we have this linear system of differential equations and they are They are coupled in as much as that x1 definitely is that x sub 1 and all these are functions of time So one way to solve this would just be for us to take the in the Laplace transform of both sides What would be the Laplace transform of this? Remember the Laplace transform of this left hand side would be equal to the Laplace transform of each of these So what is the Laplace transform of this a second a second derivative? So that's going to be an s squared x sub 1 of s minus s times the x sub 1 of 0 minus x sub 1 prime of 0 We're going to be have plus 10 times. What is the Laplace transform of of x? Well, that's just x sub 1 of s Minus 4 times x sub 2 of s in the Laplace transform of a constant is that constant over s a 0 over s Which is this 0 I Think we can take out a common factor here that would be x sub 1 of s So we'll be left with s squared And we're going to be left with that there we go a positive 10 We're going to have x sub x sub 1 of 0 0 so that falls away This was 1 remember so that's negative 1 and then minus 4 times x sub 2 of s equals 0 And if we just clean this up a bit that's going to be x sub 1 of s with an s squared plus 10 and The minus 4 times x sub 2 of s and that's going to equal 1. Let's call this our equation number 1 This is clean up with a second one or let's take the Laplace transform here So we're going to be level with x sub 2 well s squared x sub 2 of s Minus s times x sub 2 of 0 minus x sub 2 prime of 0 Plus we have 4 times x sub 2 of s minus 4 times x sub 1 of s That is going to equal a 0. Let's clean this up. We have an x up to s Here is a common factor. So we have s squared there and we have a plus 4 there Again, this is 0. This is minus This is negative 1 negative 1 negative positive 1 taking to the other side will be negative 1 And we are left with minus 4 times x sub 1 of s So look at this. We have two equations and two unknowns Let's call this Equation number 2. So here we have two equations and two unknowns x sub 1 capitals x sub 1 and x sub 2 Two equations to unknowns. We just have to solve this Let's take let's take equation number 1 here And just solve for x sub 2 Let's take equation 1. We're going to solve for x sub 2. So 4 times x sub 2 of s It's going to equal x sub 1 of s s squared plus 10 Minus 1 in other words x sub 2 of s is going to equal a quarter x sub 1 of s is S squared plus 10 minus a quarter. We'll call that equation number 3 Now let's take equation number 2 and we put equation number 3 into equation number 2 So instead of x sub 2 s we're going to have all of this So we're gonna have a quarter times x sub 1 of s s squared plus 10 minus a quarter and Now this s squared plus 4 minus 4 times x sub 1 of s and that's going to equal negative 1 Okay, so the whole reason behind I hope you couldn't see there The whole reason behind doing the Laplace transform is I'm now stuck with normal algebra now These are long algebraic problems, and it's easy to make a little mistake some way. So let's just go at it Let's distribute this s squared plus 4 in there. So I'm going to have a Quarter x sub 1 of s s squared plus 10 and then s squared plus 4 Minus a quarter s squared plus 4 minus 4 times x sub 1 of s Equals negative 1 so I'm just trying to solve for x sub 1. I'm going to multiply throughout by 4 So I'm left with x sub 1 of s And an s might as well multiply these out So it's going to be s to the power 4 plus 14 is squared plus 40. It seems Minus a quarter. I can take over to the other side already minus I'm just multiplying 3 up by 4 so that is going to be 16 x sub 1 of s and that's going to equal a negative 4 and I'm taking that over to the other side, which is now just an s Square that's the negative. I'm taking to the other side So it's an s squared plus 4 on that side easy to make a mistake s Just taking this out as a common factor. So I'm left here with s to the power 4 I'm just going to have to change the battery pack after this 14 s squared and 40 so that's going to leave me with negative 24 and that equals just s squared Let me stop here and I'll change the battery pack