MINGGUKE KE-5. Learning Outcome:

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1 1/14/1 MINGGUKE KE-5 Learning Outcome: Setelah mengikuti kuliah ini, mahasiswa diharapkan : Mampu menjelaskan konsep gaya balik Mampu menyelesaikan persamaan gerak harmonik Mampu menyelesaikan kasus harmonik teredam dan terpacu. Mampu menjelaskan dan menyelesaikan kasus resonansi. 1

2 BAB 3 mitrayana@ugm.ac.id 1/14/1 OSILATOR HARMONIK

3 3.1 Gaya Pembalik Linear. Gerak Harmonik 1/14/1 F = k ( X X ) e (Hukum Hooke) F ( x) = kx X X e x 3

4 1/14/1 m& & x + kx = x = e qt m d dt e qt + ke qt = mq + k = k q = ± i = ± m iω ω = k m 4

5 1/14/1 ( t) = C e iωt C e iω t x + + ( t) = Acosω t + B ω t x sin ( t) = R cos( ω t ϕ) x 5

6 1/14/1 6

7 1/14/1 π T = = ω π m k f 1 = = T 1 π k m 7

8 Konstanta gerak dan kondisi awal & ( t) = Acosω t + B ω t x sin ( t) = Aω sinω t + Bω ω t x cos t = x = A x & = Bω x ( t) = x cosω t + ω t x sin ω & 1/14/1 mitrayana@ugm.ac.id 8

9 Efek gaya luar yang konstan pada osilator harmonik 1/14/1 ( X X ) mg F = k + e = k ( ) X ' X mg + e e ' X = X + e e mg k x = X X ' e = X X e mg k = m& & x + kx = dan F kx 9

10 ' X e = X e + mg k x m X 1/14/1 mitrayana@ugm.ac.id 1

11 Contoh 1/14/1 1. Ketika sebuah pegas ringan menopang sebuah balok bermassa m dalam posisi vertikal, pegas diketahui terenggangkan sejauh D 1 dari posisi awalnya. Jika balok kemudian ditarik kebawah sejauh D dari posisi setimbangnya dan selanjutnya dilepaskan, katakanlah pada saat t =, carilah (a) persamaan geraknya, (b) kecepatan balok ketika ia melewati posisi setimbangnya, dan (c) percepatannya ketika balok berada di puncak gerak osilasi F = = kd + mg x 1 k mg = D 1 k ω = = m g D 1 11

12 x t = Acosω t + Bsinω t ( ) mitrayana@ugm.ac.id 1/14/1 x& = A sin t + B cos t ω ω ω ω x = D = A; x& = = Bω ; B = (a) g x ( t ) = D cos t D 1 g g x& ( t) = D sin t D 1 D 1 g g && x( t) = D cos t D 1 D 1 1

13 1/14/1 (b) & g x = D D 1 (pusat) (c) && = g x = D D1 (puncak) 13

14 3. Tinjauan Tenaga dalam Gerak Harmonik 1/14/1 F = F = kx ext x x x k W F dx kx dx x = = = 1 V ( x) = kx ext 1 1 E = mx& + kx 14

15 dx = ± E k x dt m m 1 mitrayana@ugm.ac.id 1/14/1 dx ( ) 1 t = = m m k cos 1 1 ( x R ) + C ± ( E m) ( k m) x E R = k E = mv = kr max 15

16 1/14/1 Contoh. Fungsi tenaga pendulum sederhana. θ l V = mgh h = l l cosθ V ( θ ) = mgl( 1 cosθ ) cosθ = θ 1! + 4 θ 4!... 16

17 1/14/1 ( ) 1 s = lθ V θ = mglθ ( ) 1 mg s s l V = 1 E = ms& + 1 mg l s 17

18 3.3 Gerak Harmonik Teredam mx && + cx& + kx = x kx c& x d d m e c e ke dt dt qt + qt + qt = mq + cq + k = q = ( 4 ) 1 c ± c mk m 1/14/1 mitrayana@ugm.ac.id 18

19 1/14/1 I c 4mk > (overdamping) II c 4mk = (critical damping) III c 4mk < (underdamping) 19

20 I. Kasus overdamping 1/14/1 q 1 c c k 1 = + = γ 1 m 4m m q 1 c c k = = γ m 4m m ( ) = 1t + γ x t A e γ A e 1 t

21 1/14/1 x Overdamped Critically damped O t 1

22 II. Kasus critical damping q = q = c m = γ 1 c 4m = k = γ m (&& γ & γ ) m x γ x γ x + + = d d + γ + γ x = dt dt u = γ x + dx dt 1/14/1 mitrayana@ugm.ac.id

23 1/14/1 d + γ u = dt u = Ae γ t γ t γ Ae = x + dx dt 3

24 1/14/1 dx d A = γ x e γ t + = xe dt dt ( γ t ) ( ) = ( + ) γ t = γ t + γ t x t At B e Ate Be 4

25 III. Kasus underdamping 1/14/1 q c = + i k c m m 4m 1 1 q c = i k c m m 4 m 1 ω d = k m c 4m 1 = ( ω γ ) 1 dengan γ = dan ω = ( ) 1 c m k m 5

26 1/14/1 q = γ + iω 1 d q = γ iω d ( ) ( γ + ω ) ( γ ω ) i t i d t + x t = C e d + C e ( iω ) dt iωd t + γ t = e C e + C e 6

27 1/14/1 ( ) = γ t ( cosω + sinω ) x t e A t B t ( ) = γ t Rcos ( ω ϕ ) x t e t d d ( ) 1 T = π ω = π ω γ d d d 7

28 R t x = Re γt t x = Re γt 1/14/1 mitrayana@ugm.ac.id 8

29 Mechanical Suspensions 1/14/1 ( ) c 4km = 4k m m crit Tinjauan Tenaga 1 1 E = mx& + kx de mxx kxx ( mx kx ) x dt = &&& + & = && + & 9

30 1/14/1 m && x + cx& + kx = de dt = cx& 3

31 3.4 Gerak Harmonik Terpacu. Resonansi 1/14/1 mx && + cx& + kx = F ext Fext = F e i ω t mx && + cx& + kx = F e i ω t x t = Ae i( ω t ϕ ) ( ) 31

32 1/14/1 d ( ) d ( ) ( ) m Ae c Ae kae F e dt dt i ωt ϕ + i ωt ϕ + i ωt ϕ = iωt ( ) ω ω ϕ ϕ m A + i ca + ka = F e = F cos + isin ( ) cos A k m ω F ϕ iϕ = (real) cω A = F sinϕ (imajiner) 3

33 cω tanϕ = k mω mitrayana@ugm.ac.id 1/14/1 ( ) A k mω + c ω A = F A ω ( ) F = 1 ( ω ) k m + c ω γ = c m ω = k m dan 33

34 1/14/1 tan γω ϕ = ω ω A ω ( ) F = 1 ω ω + 4 γ ω ( ) m A ω ω ω ω γ ω ( ) F m = ; D( ) = ( ) + 4 D ω ( ) 1 34

35 A( ω) 1/14/1 ω r ω ω 35

36 1/14/1 r ( ) 1 = ω ω γ r ( ) 1 d ω = ω γ Jika γ = ω A ( ω) = [( ) ] 1 ( 4 4 ) 1 ω ω + ω ω ω + ω F m = F m 36

37 Amplitudo osilasi pada puncak resonansi 1/14/1 A max = γ F ω m γ = c F ω d F A = max F γmω cω (untuk kasus damping yang lemah) 37

38 Ketajaman Resonansi. Faktor Kualitas 1/14/1 A ( ω) A max γ ( ) ω ω + γ Ketika ω = ω ± = γ A = 1 A max γ = ukuran lebar kurva resonansi 38

39 1/14/1 Q = ω d γ (Faktor Kualitas) ω ω = f 1 Q f 39

40 1/14/1 Kriteria Assessment: Kognitif dan skill Metode Assessment: PR Bobot Nilai: 1,5 % 4

41 PR 1/14/1 Soal di Buku Fowles&Cassiday fifth editions No. 3.1 No.3. No. 3.4 No. 3.7 No. 3.1 PR dikumpul di loker Dr. Mitrayana di Jurusan Fisika FMIPA UGM (MIPA Utara) Paling lambat Jumat at tanggal Maret 1 pukul

42 1/14/ mm mm 7 1 mm 5 mm 4

43 1/14/1 Komputer Sumber arus laser Cermin datar Lock-In amplifier Laser diode Chopper Sel FA Cermin datar Pengontrol aliran Cermin datar 3 43

44 1/14/1 Sumber arus Pemayar Modulator Pengontrol suhu qcl L1 BS QCL CP1 Pengkombinasi CP CP3 CP5 L Sel multipass Laser HeNe L CP4 Pengontrol aliran gas Pengontrol tekanan gas Detektor Inframerah Komputer Lock-In Amplifier 44

45 .8.3 Sinyal fotoakustik (mv/w) Sinyal fotoakustik (mv/w) Frekuensi chopper (Hz) Frekuensi chopper (Hz) 8 Sinyal fotoakustik (mv/w) /14/1 Frekuensi Chopper (Hz) mitrayana@ugm.ac.id 45

46 1/14/1 46

47 1/14/

48 1/14/ A B 15 48

49 Resonansi kecepatan 1/14/1 x x& ( t) = A( ω) cos( ωt ϕ) ( t) = ωa( ω) sin( ωt ϕ) v ( ω) = [( ) ] 1 ω ω + 4γ ω ω F m 49

50 1/14/1 Kriteria Assessment: Kognitif dan skill Metode Assessment: PR Bobot Nilai: 1,5 % 5

51 PR 1/14/1 Soal di Buku Fowles&Cassiday fifth editions No. 3.1 No. 3.3 No. 3.5 No. 3.7 No. 3.9 PR dikumpul di loker Dr. Mitrayana di Jurusan Fisika FMIPA UGM (MIPA Utara) 51