Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. The motion of a mass attached to a spring serves as a relatively simple example of the vibrations that occur in more complex mechanical systems. We consider a body of mass M attached to one end of an ordinary spring that resists compression as well as stretching. The other end of the spring is attached to a fixed wall such as in this figure. We have a restorative force of F of South equaling negative KX, where K is a positive and X is a direction force opposite to the instantaneous direction of the mash. And of the mass is going to be F of R, which is equaling negative CV, which equals negative C, where V is velocity or DXDT, which is actually negative CX prime. And then we're going to have some external force Fe. So our F, our total force is going to be a combination of these forces FS plus Fr plus Fe. Remember that F equals MA, where a is acceleration. So that's M equaling the second derivative of X in terms of T or MX double prime. So the total force is going to be MX double prime plus CX prime plus KX. And this is a second order linear differential equation that governs the motion of the mass. If we have dampening occur, we get MX double prime plus CX prime plus KX equaling zero and then we can solve that without dampening. That just means that our C is going to equal 0, so MX double prime plus KX equaling 0. Recall that frequency is written as Omega, some not equaling sqrt K / m. The period is just 2π divided by Omega and given the equations Y equal a sine, the quantity BX plus C + d or Y equal a cosine, the quantity BX plus C + d The absolute value of a is amplitude 2π / b is a different way of saying the period negative C / b is a phase shift and D is a vertical translation that's in relationship to a graph of sine or cosine. So undamped motion MX double prime plus KX equaling 0. Recall that Omega is just sqrt K / m, so the Omega squared is K / m. If we took the original MX double prime plus KX equaling zero and divided by M, we'd get X double prime plus K over MX equaling 0. But K / m is just Omega squared, so we get X double prime plus W not squared X or Omega squared X equaling 0. So r ^2 plus Omega squared equals 0. Or R is positive negative Omega times I. So our U of T equation would be a cosine Omega t + b sine Omega T. Now if we thought about using a triangle, we could have the A being on the adjacent side, the B being on the opposite, and the C being the hypotenuse. So now we have a relationship C would equal sqrt a ^2 + b ^2. Cosine of alpha is a / C and sine alpha is b / C So we could rewrite U of T equaling C * A / C, cosine Omega t + b / C sine Omega T Now recall the cosine of alpha minus beta is just cosine alpha, cosine beta plus sine alpha sine beta. But this a over CA moment ago based off that triangle we said was just cosine A. And this B over CA moment ago we said was sine alpha. So we get U of T equaling C cosine alpha cosine Omega T plus sine alpha sine Omega T, or U of T equaling C cosine alpha minus Omega T Now cosine's an even function, so we can actually factor out a -1. And the cosine of a negative angle is the same thing as the cosine of the positive. So we can get U of T equaling C cosine Omega T minus alpha. And then our amplitude would be this value right here in front of the cosine, the absolute value of it. And we could find the period 2π divided by the Omega. And the phase shift, if we wanted, would be alpha divided by Omega. So let's look at an example. If we're given that M equal, 3C equal thirty, K = 63, X sub not equal 2, and V sub not equal 2 with dampening we'd get an equation that says three r ^2 + 30, R Plus 63 = 0. If we divide everything through by three and factor, we can see R equal -3 and R equal -7. So our X of T equation is going to be C1 E to the negative 3T plus C2 E to the -7 T We take our first derivative. We get -3 C 1E to the -3 T -7 C 2E to the -7 T Putting in our initial values of X not equaling 2 and V not equaling 2, we can see 2 = C one plus C22 equal -3 C 1 - 7 C 2. Solving we get C1 equal 4 and C2 equal -2. So X of T equal 4 E to the -3 T -2 E in the -7 T That's with dampening. Now if we look at without dampening, we get an equation that says three r ^2 + 63 = 0. So r ^2 + 21 = 0 R equal positive negative sqrt 21 I. So undamped U of t = C one cosine sqrt 21 T plus C2 sine sqrt 21 T or U prime T equals negative sqrt 21 C one sine sqrt 21 T plus sqrt 21 C two cosine sqrt 21 T. Putting in the initial conditions, we call that cosine of 0 is 1 and sine of 0 is 0. So 2 = C one and two also equals sqrt 21 C 2. So our U of T equation is 2 cosine sqrt 21 T plus 2 / sqrt 21 / 21 sine sqrt 21 T. If we thought about making our triangle to find our alpha and our hypotenuse length, we would get 2 is the opposite side 2 sqrt 21 as the adjacent side. So we get U of T equaling approximately 2 sqrt 22 / 21. For a hypotenuse length cosine sqrt 21 T minus, our alpha is going to be 2 / 2 sqrt 21 or approximately .2148. So our amplitude here would be this 2 square roots of 22 / 21 and our period would be 2π / sqrt 21, our phase shift .2148 / sqrt 21. Thank you and have a wonderful day.