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Normal Mode Coordinates / Initial Value Problem Overview And Lecture 4 Phys 3750 Normal Mode Coordinates / Initial Value Problem Overview and Motivation: We continue to look at the coupled-oscillator problem. We extend our analysis of this problem by introducing functions known as normal- mode coordinates. These coordinates make the coupled-oscillator problem simple because they transform the coupled equations of motion into two uncoupled equations of motion. Using the normal modes, we then solve the general initial-value problem for this system. Key Mathematics: We gain experience with linear transformations and initial value problems. I. Normal Mode Solutions A. Summary from Last Lecture The problem that we studied last time is shown in the following diagram. There are two objects, each with mass m and three springs. The springs on the ends have spring constant ks and the one in the center has spring constant ks′ . ks m m ks ks′ q1 q2 q1 = 0 q2 = 0 Last time we found the two normal-mode solutions, which can be written as q ()t 1 1 iΩ1t −iΩ1t = ()A1e + B1e (1) q2 ()t 1 1 and D M Riffe -1- 2/1/2013 Lecture 4 Phys 3750 q1 ()t 1 iΩ2t −iΩ2t = ()A2e + B2e , (2) q2 ()t 2 −1 ~ ~ 2 ~ 2 where Ω1 = ω and Ω 2 = ω + 2ω′ are the normal-mode frequencies of oscillation. As we will show below, any solution to this problem can be written as a linear combination of these two normal modes. Thus, we can write the most general solution to this problem as q1 ()t 1 1 iΩ1t −iΩ1t iΩ2t −iΩ2t = ()()A1e + B1e + A2e + B2e (3) q2 ()t 1 −1 B. Normal Mode Coordinates 1 Let's now consider the following linear transformation of the displacements q1 (t) and q2 ()t , Q ()t 1 1 q ()t 1 = 2 2 1 1 1 . (4) Q2 ()t 2 − 2 q2 ()t Calculating the rhs of Eq. (4) produces2 Q1 ()t 1 q1 ()t + q2 ()t = . (5) Q2 ()t 2 q1 ()t − q2 ()t Written more pedantically, we have q ()t + q ()t Q ()t = 1 2 (6a) 1 2 and q ()t − q ()t Q ()t = 1 2 . (6b) 2 2 For reasons that will soon become apparent, the functions Q1 (t) and Q2 ()t are known 1 Any linear transformation of an N -vector can be represented as lhs multiplication of that vector by an N × N matrix. 2 Note that the normal-mode coordinate Q2 as defined here is the negative of Q2 as defined in Dr. Torre’s text FWP. We define it here with this change so as to be more consistent with the later treatment of N coupled oscillators. To be honest, it also makes some of the equations look prettier! D M Riffe -2- 2/1/2013 Lecture 4 Phys 3750 as normal-mode coordinates. (Why aren't they called normal-mode functions? Don't ask me!) Let's now apply the linear transformation in Eq. (4) to the rhs of Eq. (3) and see what it tells us. Applying the transformation and equating the result to the lhs of Eq. (4) yields Q1 ()t 1 0 iΩ1t −iΩ1t iΩ2t −iΩ2t = ()()A1e + B1e + A2e + B2e . (7) Q2 ()t 0 1 This equation may look complicated, but, in fact, it is very simple. It says that Q1 (t) harmonically oscillates at the first normal-mode frequency Ω1 and that Q2 (t) harmonically oscillates at the second normal-mode frequency Ω2 . Pretty cool! In fact, if Eq. (3) is the general solution to this problem (more on this below), Eq. (7) say that no matter what the motion, the sum q1 (t)+ q2 (t) always oscillates at Ω1 , and the difference q1()t − q2 ()t always oscillates at Ω2 . C. Equations of Motion Let's now go back to the coupled equations of motion, ~2 ~ 2 q&&1 + ω q1 + ω′ ()q1 − q2 = 0 (8a) and ~2 ~ 2 q&&2 + ω q2 + ω′ ()q2 − q1 = 0 , (8b) and see what happens if we write them in terms of the normal-mode coordinates Q1 ()t and Q2 ()t . To do that we need the inverse of the transformation in Eq. (4). Using −1 1 1 1 1 2 2 = 1 1 (9) 2 − 2 1 −1 (make sure that you understand this!) we apply this inverse transformation to Eq. (4), which gives us (after switching the rhs and lhs) q1 ()t Q1 ()t + Q2 ()t = . (10) q2 ()t Q1 ()t − Q2 ()t D M Riffe -3- 2/1/2013 Lecture 4 Phys 3750 That is, q1 ()t = Q1 ()t + Q2 ()t and q2 (t) = Q1 (t)− Q2 (t). Substituting these results into Eq. (8) produces ~2 ~ 2 Q&&1 + Q&&2 + ω ()Q1 + Q2 + 2ω′ Q2 = 0 (11a) and ~2 ~ 2 Q&&1 − Q&&2 + ω ()Q1 − Q2 − 2ω′ Q2 = 0 . (11b) Looks pretty ugly, eh? Well , it is about to get much simpler. If we take the sum and difference of Eqs. (11a) and (11b) we get the following two equations, ~2 Q&&1 + ω Q1 = 0 (12a) and ~2 ~ 2 Q&&2 + ()ω + 2ω′ Q2 = 0. (12b) First, notice that these two equations are uncoupled: the equation of motion for Q1 doesn't depend upon Q2 and vice-versa. Furthermore, you should now able to recognize each of these equations as the equation of motion for a single harmonic ~ oscillator! Thus, Eq. (12a) tells us that Q1 harmonically oscillates at ω ( = Ω1 ), and Eq. ~2 ~ 2 (12b) tells us that Q2 harmonically oscillates at ω + 2ω′ ( = Ω2 ). Of course, this is exactly what was expressed earlier by Eq. (7). We can also infer something very important from this transformation. Because the normal coordinates are governed by Eq. (12), which is simply an harmonic oscillator equation for each coordinate, we know that the general solution for Q1()t and Q2 (t) is given by Eq. (7). Thus, the general solution for q1(t)and q2 (t) is given by the inverse transformation of Eq. (7), which is simply Eq. (3). This proves that the general solution for q1()t and q2 ()t is, indeed, a linear combination of the normal mode coordinates Q1()t and Q2 ()t . This is a general result that we will use throughout the course. II. Initial Value Problem Let's now solve the initial-value problem for the coupled-oscillator system. That is, we want to write Eq. (3), the general solution to the coupled oscillator problem, in terms of the initial conditions q1 (0) , q&1(0) , q2 (0) , and q&2 (0) (which are all real quantities). D M Riffe -4- 2/1/2013 Lecture 4 Phys 3750 Now Eq. (3), q1 ()t 1 1 iΩ1t −iΩ1t iΩ2t −iΩ2t = ()()A1e + B1e + A2e + B2e , (3) q2 ()t 1 −1 uses the complex form of the harmonic oscillator solution. Looking back on p. 8 of the Lecture 2 notes, we see that we have three choices about how to deal with making the IVP solution real. Let's use the first approach and make Eq. (3) manifestly real by * * setting B1 = A1 and B2 = A2 . Then we have q1()t 1 * 1 * iΩ1t −iΩ1t iΩ 2t −iΩ 2t = ()()A1e + A1 e + A2e + A2 e . (13) q2 ()t 1 −1 This expression is real because each term in parenthesis is the sum of a complex number and its complex conjugate. To rewrite Eq. (13) in a form that is explicitly real we use the relationship Aeix + A*e−ix = 2Re(Aeix )= 2[]Re(A)cos(x)− Im(A)sin(x) , (14) so that Eq. (13) becomes q ()t 1 1 1 = 2 []Re()A1 cos (Ω1t )− Im ()A1 sin (Ω1t )+ 2 []Re()()A2 cos Ω2t − Im ()()A2 sin Ω2t . (15) q2 ()t 1 −1 We now apply the initial conditions to Eq. (15), which gives us q ()0 1 1 1 = 2 Re()A1 + 2 Re ()A2 , (16) q2 ()0 1 −1 and q ()0 1 1 &1 = −2Ω1 Im()A1 − 2Ω 2 Im ()A2 . (17) q&2 ()0 1 −1 The easiest way to solve for the four unknowns [ Re(A) , Im(A), Re()C , and Im(C)] is 1 1 2 2 to apply the normal-mode transformation 1 1 to each side of these two 2 − 2 equations, which produces D M Riffe -5- 2/1/2013 Lecture 4 Phys 3750 1 q ()0 + q ()0 1 0 1 2 = 2 Re()A1 + 2 Re ()A2 , (18) 2 q1 ()0 − q2 ()0 0 1 and 1 q ()0 + q ()0 1 0 &1 &2 = −2Ω1 Im()A1 − 2Ω 2 Im ()A2 . (19) 2 q&1 ()0 − q&2 ()0 0 1 From these last two equations we immediately see that q ()0 + q ()0 Re()A = 1 2 , (20a) 1 4 q&1()0 + q&2 ()0 Im()A1 = − , (20b) 4Ω1 q ()0 − q ()0 Re()A = 1 2 , and (20c) 2 4 q&2 ()0 − q&1 ()0 Im()A2 = . (20d) 4Ω2 If we now substitute Eq. (20) into Eq. (15) we finally obtain the solution to the IVP, q ()t 1 1 q (0)+ q (0) 1 &1 &2 = []q1 ()0 + q2 ()0 cos()Ω1t + sin()Ω1t q2 ()t 2 1 Ω1 . (21) 1 1 q&1 ()0 − q&2 ()0 + []q1 ()0 − q2 ()0 cos()Ω 2t + sin()Ω 2t 2 −1 Ω 2 The graphs on the following page illustrate the motion that results for two sets of initial conditions. For both graphs m = 1 , ks = 1 , and ks′ = 1 (so that Ω1 = 1 and Ω2 = 3 ). In the top graph q1 ()0 = 1, q2 (0) = −1, q&1(0) = 0 , and q&2 (0) = 0 .
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