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Persamaan Gelombang Berjalan- Pada sebuah tali yang panjang diregangkan di dalam arah x di mana sebuah gelombang transversal sedang berjalan. Pada saat t = 0, bentuk tali dinyatakan: y = f (x) ……………………………………………………… (1.11) dengan y adalah pergeseran transversal tali pada kedudukan x. Bentuk gelombang tali yang mungkin pada t = 0 ditunjukkan pada Gambar. Pada waktu t gelombang tersebut berjalan sejauh vt ke kanan, dengan v menunjukkan besarnya kecepatan gelombang, yang dianggap konstan. Maka persamaan kurva pada waktu t adalah: y = f (x – vt) ……………………………………………. (1.12)
Persamaan (1.12) adalah persamaan umum yang menyatakan sebuah gelombang yang berjalan ke kanan, di mana x akan semakin besar dengan bertambahnya waktu, dan secara grafis ditunjukkan pada Gambar 1.11(b). Apabila kita ingin menyatakan sebuah gelombang yang berjalan ke kiri, maka: y = f (x + vt) ……………………………………………….. (1.13)
Untuk sebuah fase khas dari sebuah gelombang yang berjalan ke kanan berlaku: x – vt = konstan
Maka dari diferensiasi terhadap waktu akan diperoleh: dx/dt = v
Dengan v adalah kecepatan fase gelombang. Untuk gelombang yang berjalan ke kiri kita memperoleh kecepatan fase gelombang adalah -v. Persamaan gelombang tali pada waktu t = 0 dinyatakan: y = A sin (2π/λ)x ……………………………………………… (1.15) Gambar 1.6 Gelombang yang merambat pada seutas tali beruva kurva sinus
Bentuk gelombang tersebut adalah sebuah kurva sinus, ditunjukkan pada Gambar 1.12. Pergeseran maksimum, A, adalah amplitudo kurva sinus tersebut. Nilai pergeseran transversal y adalah sama di x seperti di x + λ , x + 2 λ , dan sebagainya. Panjang gelombang λ menyatakan jarak di antara dua titik yang berdekatan di dalam gelombang tersebut yang berfase sama. Jika gelombang tersebut bergerak ke kanan dengan kecepatan fase v, maka persamaan gelombang tersebut pada waktu t adalah: y = Asin(2π/λ)(x – vt)
Waktu yag diperlukan gelombang untuk menempuh satu panjang gelombang ( λ ) disebut periode (T ), sehingga: λ = v .T ……………………………………………………… (1.17)
Dengan mensubstitusikan persamaan (1.17) ke persamaan (1.15), maka akan diperoleh: Pada konsep gelombang berlaku suatu bilangan gelombang (wave number), k dan frekuensi sudut ( ω ), yang dinyatakan: k= 2π/λ; dan ω=2π/T
Sehingga, dari persamaan (1.18) akan diperoleh: y = A sin (kx –ωt) ………………………………………. (1.20)
Persamaan tersebut berlaku untuk gelombang sinus yang berjalan ke kanan (arah x positif). Sementara itu, untuk arah x negatif berlaku: y = A sin (kx +ωt) ………………………………………. (1.21)
Dari persamaan (1.17) dan persamaan (1.19), akan diperoleh nilai kecepatan fase (v) dari gelombang adalah: v=λ/T = ω/k
Persamaan (1.20) dan (1.21) menunjukkan pergeseran y adalah nol pada kedudukan x = 0 dan t = 0. Pernyataan umum sebuah deret gelombang sinusoida yang berjalan ke kanan adalah: y = Asin(kx −ωt −φ) ……………………………………. (1.23)
Dengan φ adalah konstanta fase. Jika φ = -90o, maka pergeseran y di x = 0 dan t = 0 adalah ym, yang dinyatakan: y = Acos( kx −ωt ) ………………………………………. (1.24)
Hal ini disebabkan fungsi cosinus digeser 90o dari fungsi sinus. Jika sebuah titik pada tali berlaku x = π k , maka pergeseran di titik tersebut adalah: y = A sin(ωt + φ) ………………………………………… (1.25)
Persamaan tersebut menunjukkan bahwa setiap elemen khas dari tali tersebut mengalami gerak harmonis sederhana di sekitar kedudukan kesetimbangannya pada waktu gelombang berjalan sepanjang tali tersebut.
Contoh soal Gelombang Berjalan
1. Persamaan gelombang berjalan pada seutas tali dinyatakan dengan y = 0,02 sin (20 π t – 0,2 π x). Jika x dan y dalam cm dan t dalam sekon, tentukan:
a. amplitudo, d. bilangan gelombang, dan
b. panjang gelombang, e. frekuensi gelombang!
c. kelajuan perambatan,
Penyelesaian:
Persamaan umum gelombang y, seperti yang diperlihatkan pada persamaan (1.20) adalah: y = ym sin(kx −ωt ) y = -ym sin(ωt − kx ) diberikan: y = 0,02sin(20π t – 0,2πx)
Jadi,
a. Amplitudo, A = 0,02 cm
b. Panjang gelombang ( λ ), λ=2π/k = 2π/0,2π=10 cm
c. Kelajuan perambatan (v) v=ω/k = 20π/0,2π= 100 cm/s
d. Bilangan gelombang (k), k=2π/λ=2π/10=0,2π e. Frekuensi ( f ), ω = 2πf
20 π = 2 π f f= 10 Hz
2. Fungsi gelombang pada suatu medium dinyatakan sebagai: y = 0,1 sin (5t – 2x), dengan x dan y dalam meter dan t dalam sekon. Tentukanlah frekuensi dan panjang gelombang tersebut!
Penyelesaian:
Diketahui: gelombang berjalan, y = 0,1 sin (5t – 2x)
Ditanyakan: f = . . .? λ = . . .?
Jawab:
Dengan menggunakan persamaan 1.5 dapat kita ketahui bahwa: A = 0,1 m dan ω = 2 π f = 5, sehingga: f=5/(2π) Hz
Dengan persamaan gelombang berjalan kita ketahui bahwa k = 2, sehingga: k = 2π/λ λ = 2π/k λ = π m
3. Sebuah gelombang merambat pada tali yang memenuhi persamaan : Y = 0,4 sin 2π (60 t – 0,4 x) di mana Y dan x dalam meter dan t dalam sekon, tentukanlah:
a. amplitudo gelombang,
b. frekuensi gelombang,
c. panjang gelombang,
d. cepat rambat gelombang, dan
e. beda fase antara titik A dan B pada tali itu yang terpisah sejauh 1 m.
Penyelesaian :
Untuk menyelesaikan persoalan gelombang berjalan yang diketahui persamaan gelombangnya, kita mengubah bentuk persamaan gelombang tersebut ke dalam bentuk persamaan gelombang umum.
Diketahui : Y = 0,4 sin 2π (60 t – 0,4 x)
Ditanyakan : a. A = … ?
b. f = … ?
c. λ = … ?
d. v = … ?
e. Δφ = … ?
Jawab :
Y = 0,4 sin 2π (60 t – 0,4x) diubah menjadi bentuk
Y = 0,4 sin (120π t – 0,8πx)
YP = A sin (ωt – kx)
a. A = 0,4 m
b. ωt = 120 πt ω = 2πf → 2πf = 120 π → f= 60 Hz
c. k = 0,8π → 2π/λ = 0,8π → λ = 2,5m
d. v = f x λ = 60 x 2,5 = 150 m/s
e. Δφ = Δx/λ = 1/2,5 = 2/5

soal ttg gelombang berjalan 1. sebuah gelombang merambat dari sumber S ke kanan dengan laju 8 m/s, frekuensi 16H, amplitudo 4 cm. gelombang itu melalui titik P yang berjarak 9,5 meter dari S. jika S telah bergetar 1,5 detik, dan arah gerak pertamanya ke atas, maka berapakah simpangan titik P pasa saat itu..? jawab : v = 8 m/s f = 16 Hz
A = 4 cm x = 9,5 m t = 1,5 s

ω = 2πf = 2π(16) = 32π rad/s λ = v/f = 8/16 = 1/2 m k = 2π/λ = 2π/(1/2) = 4π m^-1

y = A sin (ωt - kx)
..= 4 sin (32πt - 4πx)

untuk t = 1,5 = 3/2 s dan x = 9,5 = 19/2 m
y.= 4 sin ( 32π(3/2) - 4π(19/2) )
y.= 4 sin ( 48π - 38π )
y.= 4 sin ( 10π )
y.= 4 sin ( 5×2π )
y.= 4 sin ( 5×360º ) y = 4 sin 0º
...= 4 (0)
...= 0 cm

Kalau kita lihat gelombang berdasarkan berubah atau tidaknya simpangan/amplitudo gelombang, maka gelombang yang merambat dengan ampitudo tetap kita sebut Gelombang berjalan, sedangkan gelombang yang merambat dengan amplitudo berubah kita sebut Gelombang stationer.

Kalau kita perhatikan videonya. ketikan gelombang laut telah merambat dalam waktu t tertentu, maka kita dapat menentukan simpangan terntentu dari si peselancar (di gambar di wakili dengan suatu titik)
Jika gelombang merambat dengan kecepatan v maka untuk mencapai titik P sepanjang x dibutuhkan waktu . Jika dari titik 0 gelombang telah berjalan t detik maka waktu di titik P adalah
Dengan konsep persamaan getaran harmonis y = A sin wtp kita peroleh persamaan umum gelombang berjalan yaitu:

Ket: A = amplitudo gelombang (m); l = v /.T = panjang gelombang (m); v = cepat rambat gelombang (m/s);
= bilangan gelombang (m’); x = jarak suatu titik terhadap titik asal (m)
Ketentuan tanda:
a. tanda ± di depan amlplitudo positif (+) => arah getar pertama kali ke atas negatif ( – ) => arah getar pertama kali ke bawah
b. tanda ± di depan bilangan gelombang positif (+) => arah rambat gelombang ke kiri negatif ( – ) => arah rambat gelombang ke kanan
Rahasia!!
rumus gelombang berjalan diatas, hanya salah satu bentuk rumus, kita bisa memvariasikan menjadi bentuk rumus yang lain. Saran saya pahami rumus yang satu di atas, dan wajib hapal rumus-rumus sponsor/ pendukung berikut: 1. v = λ f dan f = 1/T 2. dan 3. 4. Sudut fase : besar sudut dalam fungsi sinus (dinyatakan dalam radian)

Fase gelombang

Beda fase antara titik B dan A
_______

Penyelesaian Soal Persamaan gelombang berjalan
06-08-2011 21:05:48, pada Soal
Sebuah gelombang berjalan mempunyi persamaan simpangan sebagai berikut :y = 0,04 sin π ( 50t + x ) m. Dari persamaan gelombang tersebut.Tentukan
a. arah rambatan gelombang
b. frekuensi gelombang
c. Panjang gelombang
d. cepat rambat gelombang
e. beda fase antara dua titik yang berjarak 25 m dan 50 m.
Penyelesaian
Diketahui : y = 0,04 sin π ( 50t + x ) m ; Persamaan simpangan gelombang berjalan : y = A sin ( ωt ± kx ) y = 0,04 sin (50πt + πx) y = simpangan, A = Amplitudo, ω = 2πf = kecepatan sudut , k = 2π/λ k = bilangan gelombang, A = 0,04 m, tanda ± → tanda +, bila arah rambat ke kiri, tanda -, bila arah rambat ke kanan
Ditanyakan : a. arah rambat b. f = ? c. λ = ? d. v = ? e. Δφ = ?
Jawaban : a. kekiri b. f = ω/2π = 50π/2π = 25 Hz c. λ = 2π/k = 2π/π = 2 m d. v = λ f = 2. 25 = 50 m/s e. Δφ = Δ x/λ = (50-25)/2 = 12,5

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...TH E ECONOMI C WEEKL Y September 23, 1961 Inflatio n No t a Monetar y Phenomeno n ? Promod e K Mukherje e The cause o f inflation, according t o Keith B Griffi n (Th e Economi c Weekly . Special Number 1961 ) lies i n the structura l imbalances i n a n economy . Whil e granting this, i t i s contended here that i t does not follo w that monetary policy has n o role t o play in combating inflation. I t i s not enough t o achieve a n overall balance between output and expenditure. Wha t i s importan t is to secure an increase in the output of basic consumption and investment goods. A well-conceived monetary policy is of crucial importance in discouraging outlay and consequent income creation in wrong directions. Further, an increasing proportion of the income generated in the process of development will go to augment bank deposits, thereby increasing the capacity o f the bank s t o create credit, In the absence of a restrictive monetary policy, a serious credit inflation may consequently ensue and distort the ideal relationship envisaged by Griffin between output and expenditure. Thus, a phenomenon not primarily monetary may ultimately become largely monetary in character. ACCORDIN G t o M r Griffin , infla - tio n result s "i f eithe r desire d con - sumptio ...

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...Requested by: Jianan Sun -------------------------------------------Undergraduate Academic Record SCHOOLS ATTENDED Secondary Schools: Xi'an Gaoxin Tangnan High School Higher Education Institutions: Xi'an University of Finance & Econom 08/2009 - 05/2010 Issued to: HOURS 3.0 3.0 3.0 3.0 3.0 QPTS 45.00 84.00 QPTS 3.00 9.00 12.00 9.00 12.00 GPA 3.000 2.800 U N O F F I C I A L Test Scores: -------------------------------------------------------Transfer Credit Applied to 2011 Fall Semester Xi'an University of Finance & Econom 08/2009 - 05/2010 XICN001 FUNDAMENTALS OF COLLEGE C 3.00 NEEDEVAL XICN002 ADVANCED MATHS 1 (ECONOMI 3.00 MA 1--XICN003 INTRODUCTION TO LAW OF EC 3.00 ECO 4--XICN004 MORAL CULTIVATION AND BAS 3.00 GEED2--XICN005 PE (1) 1.00 KHP 1--XICN006 MICRO-ECONOMICS 3.00 ECO 201 XICN007 COLLEGE CHINESE 2.00 GEED1--XICN008 ADVANCED MATHS 2 (ECONOMI 3.00 MA 123 XICN009 MANAGEMENT 3.00 MGT 301 XICN010 MACRO-ECONOMICS 2.00 ECO 202 XICN011 PRINCIPLE OF MARXISM 3.00 GEED2--XICN012 DATABASE APPLICATION 2.00 NEEDEVAL XICN013 PE (2) 1.00 KHP 1--XICN014 OUTLINE OF CHINESE MODERN 2.00 GEED2--Total 34.00 2011 Fall Semester Program: College of Business and Economics Bachelor of Science in Accounting Major: Pre-Accounting CRS NUM COURSE TITLE GRADE STA 210 INTRO TO STATISTICAL A REASONING CIS 110 COMP AND COM I B GLY 110 ENDANGERED PLANET INTR TO C ENVRNMNTL GEOL BIO 102 HUMAN ECOLOGY B ACC 201 FINANCIAL...

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