Lecture 4 Normal Modes

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1 Lecture 4 Noral Modes Coupled driven oscillators Double pendulu

2 The daped driven pendulu = g/l +k y+fcost y = y gy/l k y d dt + d dt + g + k l k k d dt + d dt + g + k l y = F 0 Re eit

3 y =Re X Y eit CF d dt + d dt + g l + k k k d dt + d dt + g l + k y = F 0 Re eit Substitute cople eigenvalues in to obtain eigenvectors +i + g l + k k k +i + g l + k X Y = 0 0 +i + g l + k k k +i + g l + k =0 Eigenvalue eq.

4 Siple case = = l =l =l i + g l + k k k i + g l + k =0 i + g l =0 or i + g l +k =0, =i ±, Eigenvalues = g l or = g l + k = 0 eigenvalues c.f. previous result

5 Eigenvectors i + g l + k k X k i + g l + k Y = 0 0 i + g l =0 or i + g l +k =0, =i ±, Eigenvalues = = k k k k k X Y =0 k k k X Y =0 X Y = A ei X Y = A ei Eigenvectors

6 i + g l + k k k i + g l + k X Y = 0 0 i + g l =0 or i + g l +k =0, =i ±, Eigenvalues y =Re A ei e t e it * +Re A ei e t e it * =e t A cos t * + +A cos t * + Eigenvectors

7 Decoupling ethod = g l +ky y=g y l kyy +y= g l +y+y y= g l kyy q + q + q =0 q + q + q =0 Two nd order differential equations - techniques discussed last ter give q = A cos t + e t q = A cos t + e t = = =

8 *+,,-.-/03-.4,05. = g/l +k y+fcost y = y gy/l k y d dt + d dt + g + k l k k d dt + d dt + g + k l y = F 0 Re eit

9 d dt + d dt + g l + k k k d dt + d dt + g l + k y = F 0 Re eit y =Re P Q eit +i + g l + k k k +i + g l + k P Q *MP= F 0 P P Q =M F 0 M= +i + g l + k k k +i + g l + k y =Re M F 0 eit

10 y =Re F M 0 eit +i + g + k l k M= k +i + g + k l e.g. = = l =l =l =0 = g l, = g l + k M= + g + k l k k + g + k l M = DetM + g + k l k k + g + k l DetM = + g l + k k = y = F cost + g l k + k = F cost = F cost / /

11 **+,-./0*-34 = = l =l =l ++ g l y+y+ g l +k y=fcost y+k y=0 +y + +y + g l y + y + g l +y= F cost y+ k y = F cost q q g l q = F cost q + q + g l +k q = F cost q = +y q = y 5+*+6-789:

12 q + q +g l q = F cost =Re F ep it, q =+y q =Re A ep it +i +g l A =F/ A = g F/ep i l + * = F/ep i / + + * / where tan = / and = g l the undaped noral ode frequency q = F/ + * * cos t++ /

13 q + q + g l +k q = F cost q =y q =Re A ep it F/ep i A = g l +k + * = F/ep i / + + * /, tan = / q = F/ + * * cos t++ / q = F/ + * * cos t++ / Finally = q +q, y= q q c.f. y = F cost +, * =0

14 d dt + d dt + g + k l k k d dt + d dt + g + k y = F 0 l Re eit The case = =, l l, no daping, no driving force. *+,-* + g + k l k k + g + k l =0 A=g/l +k/= +k/ B=k/ C =g/l +k/= +k/ y =Re 0 * y 0 eit +, -, = + +k/± + k/ 0 y 0 = k ± + k/ y 0 0 = / y 0 0 r

15 y 0 0 = / y 0 0 r 0 = De i, 0 = De i t = y = r Dcos t+ + r Gcos t+ Initial conditions : 0 = y = a 0, t=a cos t+r cos t / +r y t=ar cos tcos t / +r =0 t=acos tcos t/ a r +r y t=arsin tsint// +r = +, = sin tsint/ r +r a a r 0y +r a

16 Diagraatic Representation of Noral Modes v, = 0 i+y 0 j/ 0 +y 0 0 y 0, = k ± + * + k/,. - a For k/ 0 0 /y 0. or 0 b For k/ 0 /y 0. or c Interediate k/ y 0 0 = tan = k/ ± + k/ y 0 y 0 y 0 v v v v 0 v v 0

17 Double pendulu sall angle approiation θ, φ << l φ l θ l θ.. = g θ + g φ θ l θ.. + φ.. = g φ d dt θ φ = g l + / / + / + / θ φ

18 Noral odes: = θ θ0 = e iωt φ φ 0 ω +g/l + / g/l / g/l + / ω +g/l + / θ0 φ 0 = 0 det ω +g/l + / g/l / g/l + / ω +g/l + / = 0 ω g/l + / = ± g/l + / i.e., ω ± = g l / + noral frequencies

19 Eaple [CP4 June 006]. Two identical asses = = are connected by a assless spring with spring constant k. Mass is attached to a support by another assless spring with spring constant k. The asses and springs lie along the horizontal -ais on a sooth surface. The asses and the support are allowed to ove along the -ais only. The displaceent of the support in the -direction at tie t is given by ft and is eternally controlled. Write down a syste of differential equations describing the evolution of the displaceents and of the asses fro their equilibriu positions. [5] Deterine the frequencies of the noral odes and their aplitude ratios. [8] The displaceent of the support is given by ft = Asinωt with ω = k/ and constant aplitude A. Find epressions for t and t assuing that any transients have been daped out by a sall, otherwise negligible, daping ter. [7] k k ft

20 Coupled oscillators with a driving ter: k k ft ẍ = k ẍ = k ft k Hoogeneous case f = 0: noral frequencies ω = k +, ω = k aplitude ratios : X /X NM =, X /X NM = +

21 Driving ter ft = Asinωt, ω = k/: t = C I e iωt, t = C I e iωt = ω C C 3 = ω C C +ω A 0 Thus : C = 3C +C +A C = C C = C = 0,C = A t = 0, t = Asinωt

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