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Lecture

# L16_normalmodes.pdf

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School
University of Toronto St. George
Department
Physics
Course
PHY354H1
Professor
Erich Poppitz
Semester
Winter

Description
Lecture 16: Coupled Oscillators & Normal Modes ▯ relevant reading: Morin 4.5 ▯ Reminder: November break next Monday & Tuesday, so no tutorial next week. But tutorial questions still available. TA’s will have o▯ce hours during their Wednesday tutorial times. Finding the Natural Frequencies or "Normal Modes" ▯ Last time we derived the following equation for the coupled pendulums’ motions: d xA g k 2 + x A (xA▯ x B = 0 dt l m d x g k B + x B (xA▯ x B = 0 dt2 l m ▯ By setting ! 2 = g=l, ! 2 = k=m and trying solutions of the form: i!t 0 c i!t x A = Ae and x B = Be (notice we gave them the same !), we found equations for the natural frequencies. (Note: You could also ▯t try any of the other forms for sinusoids we’ve seen, e.g A = Ae or xA = Acos(▯t + ▯) or x A = Acos(▯t) + B sin(▯t):::). In fact, I recommend trying the derivation to the equations below using these other forms yourself. ▯ These natural frequencies are called \NORMAL MODES" of the sys- tem. The equations we found were: 2 2 2 ▯! A + ! 0A + ! cA ▯ B) = 0 (1) 2 2 2 ▯! B + ! 0B ▯ ! c(A ▯ B) = 0 (2) If we can ▯nd an ! that satis▯es these equations, then we have found states of the system where all the components oscillate with the same frequency. We said there were 2 ways to solve these equations. Last time we did the ‘substitution’ method to ▯nd two natural (or normal mode) frequencies: 1 ! = ! ▯ ! 2 0 1 and ! = ! +02! ▯ c 2 NOTE: Technically there are 4 frequencies here, the ▯!’s for each case. When I say there are only 2, I’m talking about those with distinct magnitudes. But remember when we put together our solution later we will need to consider both the ▯!’s for each case. ▯ Now let’s go back to equation (1) or (2) and plug in each of these solutions for ! (both equations will give you the same results so you only need to plug into 1 of them): (1) For !1we have ▯! A + ! A + ! (A ▯ B) = 0 ▯ 0 ▯ 0 c which is only true if A = B. (2) For !2, we have 2 2 2▯ 2 ▯(!▯0 2! )c + ! ▯ 0 ! (Ac▯ B) = 0 which is only true if A = ▯B. ▯ Now A and B are the amplitudes of pendulums, so these equations determine their relative positions: again, ! = ! = g=l goes with A = B;symmetric mode 1 0 ! = ! + 2! = g=l + 2k=m goes with A = ▯B; antisymmetric mode 2 0 c ▯ This means we can write the solution for our normal modes as: p 1. Symmetric normal mode: frequencies: ! = ▯ g=l. Notice that 1 we get 2 solutions here (the positive and negative square roots just like we did when we were considering single oscillators). We also know that the amplitudes of the two pendulum are the same: 2 p A = B. However, the the component with frequency + g=l can generally have a di▯erent amplitude than the component with fre- p quency ▯ g=l so we will give these amplitudes di▯erent variable names. For +! we’ll use \C" and for ▯! we’ll use D. The 1 1 important thing is that the amplitude in the x A solution is the same as the amplitude in the xBsolution. I can write the general solution for the motion as: x = Ce i!1t+ De ▯i1 t A i!1t ▯i!1t xB= Ce + De This is the same result we got from our intuition before (i.e. that the pendulums move exactly the same way). We still haven’t found C and D which is GOOD because we need some free param- eters to use for the initial conditions. C and D will be determined by the initial conditions. p 2. Antisymmetric normal mode: frequencies: ! = ▯ g=l + 2k=m. 2 Amplitude: A = ▯B. So I can write the general solutions for the motions as: xA= Ce i!2t+ De ▯i2 t x = ▯Ce i!2▯ De ▯i2 t B 3. This is the same result we got from our intuition before (i.e. that the pendulums move in exactly the opposite way. Again C and D are found from the initial conditions. ▯ The 2 normal mode frequencies are known as the eigenvalues of the problem. ▯ The pairs of coe▯cients (A;B) = C(1;1) and (A;B) = C(1;▯1) are known as the eigenvectors of the problem. This nomenclature will make more sense when we use our second method to solve equations (1) and (2) above. 3 Second: You can use matrix algebra to solve the problem ▯ To ▯nd the solutions, we can write the pair of equations (1,2) above in matrix form: ▯ 2 2 2 ▯▯ ▯ ▯ ▯ !0+ ! c ▯! c A 2 A ▯! 2 ! + ! 2 B = ! B c 0 c ▯ This system of equations is a classical eigenvalue problem of the form G~x = ▯x This picks out special sets of solutions in which the operator G acting on the vector x has the e▯ect of multiplyingx by a constant ▯. ▯ Any set of coupled linear di▯erential equations can be expressed in this way, and so the analysis is often similar. ▯ The matrix equation can be rewritten in the form (G▯▯I)~ x = 0 where I is the identity matrix: ▯ ▯▯ ▯ !2 + !2 ▯ !2 ▯! 2 A 0 c2 2 2c 2 = 0: ▯! c ! 0 ! ▯c! B The only way this pair of equations can be true, for A and B non-zero, is if the determinant of th
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