Riak pada air kolam terjadi karena perambatan getaran akibat jatuhnya batu pada air kolam tersebut.
Bab
1
Sumber: Science Encyclopedia, 2001
mendeskripsikan gejala dan ciri-ciri gelombang secara umum. Setelah mempelajari bab ini, Anda harus mampu:
menerapkan konsep dan prinsip gejala gelombang dalam menyelesaikan masalah.
Hasil yang harus Anda capai:
Gejala Gelombang
A. Pemahaman
Gelombang
B. Gelombang
Berjalan dan
Gelombang
Stasioner
C. Sifat-Sifat
Gelombang
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A. Pemahaman Gelombang
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Tes Kompetensi Awal
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Aktivitas Fisika 1.1
Jenis Gelombang Berdasarkan Arah Getarannya Tujuan Percobaan
Mengamati gelombang transversal dan gelombang longitudinal. Alat-Alat Percobaan
Sebuah slinki (pegas plastik) yang panjangnya 3 meter. Langkah-Langkah Percobaan
1. Pegang kuat-kuat salah satu ujung slinki oleh teman Anda atau diikatkan pada tiang.
2. Rentangkan sesuai panjangnya dan getarkan ujung yang satu dengan satu kali hentakan naik turun dari posisi setimbang dan kembali ke posisi setimbang (posisi saat tangan Anda diam).
3. Ulangi langkah 2, dan getarkan ujung slinki tersebut terus menerus naik turun, kemudian amati perambatan gelombang sepanjang slinki.
Tokoh
Frank Oppenheimer
(1912–1985)
Frank Oppenheimer lahir di New York pada tahun 1912. Di Exploratorium
dia mendemonstrasikan pengamatan-nya bahwa gerakan pendulum bila diproyeksikan pada sabuk kertas bergerak membentuk lintasan grafik sinusoidal.
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Gambar 1.1
Berdasarkan arah getarnya, gelombang mekanik dikelompokkan menjadi: (a) Gelombang transversal; (b) Gelombang logitudinal.
/
0 T
arah rambat
arah getar
&
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Tugas Anda 1.1
Selain dalam medium slinki, dapatkah Anda mencari medium lain untuk menggambarkan gelombang longitudinal?
Gambar 1.2 Gelombang pada tali. 4. Letakkan slinki di atas lantai licin, kemudian dengan bantuan teman Anda
pegang salah satu ujungnya.
5. Hentakkan salah satu ujung pegas dengan satu kali dorongan dan satu kali tarikan ke posisi semula. Amati rapatan dan regangan yang merambat sepan-jang slinki.
6. Ulangi langkah 5, namun dorongan dan tarikannya dilakukan secara terus-menerus. Amati perambatan dan regangan sepanjang slinki.
arah rambat
T
1. Panjang, Frekuensi, Periode, dan Cepat Rambat Gelombang
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2. Persamaan Umum Gelombang
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0
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!; @
!/?/9/<A/?/>B<1/9; Gambar 1.3
(a) Panjang gelombang pada gelombang transversal. (b) Panjang gelombang pada gelombang longitudinal.
panjang gelombang /
0 panjang gelombang rapatan renggangan puncak
lembah
Gambar 1.4 Panjang gelombang ( ) pada grafik simpangan (y) terhadap arah rambat (x).
Contoh
1.1
2 = 36 m A B'
C E
D
F' G F B
H
y
x
lembah gelombang
3. Energi Gelombang
7/D/:0/07<7<2/A3:/6;3<53A/6B70/6D/53:=;0/<5276/@7:9/< =:36 53A/?/< E/<5 27?/;0/A9/< ,<AB9 ;3<567AB<5 3<3?57 53:=;0/<5 <2/ 6/?B@ ;3<57<5/A 93;0/:7 ;/A3?7 27 "3:/@ . A3<A/<5 3<3?57 A=A/: ;/@@/E/<503?=@7:/@7>/2/>35/@<3?57A3?@30BA27AB:7@9/<@30/5/703?79BA
23<5/<
@30/5/7 9=<@A/<A/ 5/E/ 2/< /2/:/6 /;>:7AB2= *3:/<8BA<E/ >3?6/A79/< 5/;0/? 03?79BA
'+ ;3?B>/9/< @30B/6 ;=23: 53:=;0/<5 A/:7 3:=;0/<5 A/:7 A3?@30BA A3?8/27 9/?3</ /2/<E/ 53?/9/< A/<5/< +/<5/< A3?@30BA 03?53?/96/?;=<7923<5/</;>:7AB2=2/<4?39B3<@7@B2BA3?/9/< A/<5/< 7<7 ;3<E30/09/< @3A7/> 3:3;3< A/:7 ;3<5/:/;7 53?/9 6/?;=<79 /2/:/6;/@@/2/?7@35;3<9317:A/:7;3?B>/9/<?/>/A;/@@/:7<3/? A/:7 2/<";3?B>/9/< >/<8/<5 @35;3< A/:7 3<5/< 23;797/< 3<3?57 53:=;0/<5 2/>/A 27AB:7@9/< @30/5/7 03?79BA
7>3?;B9//<@B<5/7A3?2/>/A2B/0B/65/0B@E/<5A3?>7@/6@38/B61;"32B/<E/ </79AB?B<A3?0/D/53:=;0/<5/7?*/:/6@/AB5/0B@03?/2/27>B<1/953:=;0/<5 @32/<59/<5/0B@E/<5:/7<03?/2/272/@/?53:=;0/<57/<A/?/932B/5/0B@A3?@30BA A3?2/>/A@/AB0B97A!79/4?39B3<@753:=;0/<5/7?A3?@30BA/2/:/6FA3<AB9/<13>/A ?/;0/A53:=;0/<5>/2//7?
0
793A/6B7F
3<5/<;3;3?6/A79/<5/;0/?27>3?=:36
1;
1;
! 1;IF 1; @
; @
!/2713>/A?/;0/A53:=;0/<5>/2//7?/2/:/6; @
Contoh
1.2
m x
Gambar 1.5
Gerakan tangan dirambatkan oleh tali dalam bentuk gelombang.
1,5
Pembahasan Soal
Gelombang air laut menyebabkan permukaan air naik turun dengan periode 2 s. Jika jarak antara dua puncak gelombang 5 m, gelombang akan mencapai jarak 10 m dalam waktu ....
a. 1 s b. 2 s c. 3 s d. 4 s e. 5 s
UMPTN 2001 Pembahasan:
T = 2 s
jarak antara dua puncak = satu gelombang
v = 5 m 2 s
T = 2,5 m/s
s = vt
10 m = (2,5 m/s)t t = 10 m
2,5 m/s = 4s
Jawaban: d
5 m
/
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0 !
!
F
!/274?39B3<@753:=;0/<5A/:7/2/:/6
F
Tugas Anda 1.2
Embusan angin kencang membuat
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0 3?/>/9/62/E/E/<527?/;0/A9/<A/:7A3?@30BA
0
793A/6B7 1; ;
1; 5?/;
! ; @ /
P1 P1
t1
t1 + t
vt
Gambar 1.6 (a) Pada waktu t1, gelombang tali mencapai titik P1 dan memiliki energi karena gerak harmonik elemen-elemennya. (b) Pada saat waktu bertambah sebesar t, gelombang tali menjalar sejauh vt.
"
"
H
*3:/<8BA<E/ >3?6/A79/< '+ 03?79BA
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"
!
H
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!
H
/?7 +,'( 1 <2/ 2/>/A ;3:76/A 0/6D/ 2/E/ E/<5 27A?/<@ ;7@79/< 03?0/<27<5 :B?B@ 23<5/< 9B/2?/A /;>:7AB2= 9B/2?/A 4?39B3<@7 2/< :/8B 53:=;0/<5
Contoh
1.3
Tantangan
untuk Anda
Grafik simpangan terhadap waktu sebuah gelombang yang memiliki kecepatan 0,05 m/s seperti pada gambar berikut.
Berdasarkan gambar tersebut, tentukan:
a. amplitudo (A); b. periode (T);
c. frekuensi gelombang (f).
t(s)
y(cm) 1 0
150 cm
0
Kata Kunci
• gelombang mekanik • gelombang elektromagnetik • gelombang transversal • gelombang longitudinal • panjang gelombang • periode gelombang • frekuensi gelombang • simpul
• laju energi gelombang • daya gelombang • cepat rambat gelombang • rapat massa linear • energi total gelombang )/>/A;/@@/:7<3/?A/:7/2/:/6
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@-B. Gelombang Berjalan dan Gelombang
Stasioner
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Tes Kompetensi
Subbab
A
1. Gelombang Berjalan
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Gelombang merambat ke arah sumbu-x positif.
O P
y
v
x
y = Asintkx
+A = awal getaran gelombang ke atas
–A = awal getaran gelombang ke bawah
+k = gelombang merambat ke kiri
–k = gelombang merambat ke kanan
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a. Sudut Fase, Fase, dan Beda Fase
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Tugas Anda 1.3
Gelombang air laut merupakan gelombang berjalan. Energi gelombang air laut ini dapat digunakan untuk menggerakkan generator listrik. Bukan hal yang mustahil teknologi ini suatu saat diterapkan di Indonesia. Menurut Anda, kira-kira di daerah laut manakah teknologi ini dapat diterapkan? Coba Anda diskusikan alasannya bersama teman-teman.
Gambar 1.8
Titik A yang berjarak x1 dan titik
B yang berjarak x2 dari O. Kedua titik memiliki beda fase .
y
x
B
x1 x2
A O
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;3:B:B6:/<A/99/<0B;7136+B5/@<2/1/?7:/67<4=?;/@7;3<53</753:=;0/<5 A@B</;7A3?@30BA@3>3?A70/5/7;/</A3?8/27<E//>/>3<E30/0>3<E30/0<E/9/>/< 2/<27;/</>3?</6A3?8/27+@B</;72/<7<4=?;/@7:/7<<E/A3<A/<5A@B</;75/? :3076;3</?79A/;0/69/<5/;0/?5/;0/?E/<5?3:3C/<932/:/;AB:7@/<<2/
Tantangan
untuk Anda
Sebuah gelombang merambat dari sumber S ke kanan dengan laju 8 m/s, frekuensi 16 Hz, dan amplitudo 4 cm. Gelombang tersebut melalui titik P yang berjarak 91
2 m dari S. Jika S telah
bergetar 11
4 S dan arah gerak
pertamanya ke atas, hitunglah simpangan titik P pada saat itu.
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0
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Contoh
1.5
Satuan frekuensi adalah hertz (Hz), diambil dari nama Heinrich Hertz, orang yang pertama kali mendemonstrasikan gelombang radio pada 1886. Satu getaran per sekon disebut 1 hertz; dua getaran per sekon dinamakan 2 hertz, dan selanjutnya.
The unit of frequency is called the
hertz (Hz), after Heinrich Hertz, who demonstrated radio in 1886. One vibration per second is 1 hertz; two vibrations per second is 2 hertz, and so on.
Informasi
untuk Anda
Information for You
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2. Gelombang Stasioner
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Gambar 1.10
Gelombang pantul dari seutas tali yang terikat.
gelombang datang Gambar 1.9
Gelombang pantul dari seutas tali yang bergerak bebas (tidak ada perubahan fase).
gelombang
pantul perut
simpul
gelombang pantul
gelombang datang gelombang
pantul
O
P A
x
Gambar 1.11
Simpangan gelombang datang
y1 dan simpangan gelombang pantul y2 pada ujung bebas tidak mengalami beda fase.
y1 y2
x 23<5/< B8B<5 A72/9 A3?79/A '+ ;/B>B< 23<5/< B8B<5 A3?79/A
'+
a. Gelombang Stasioner pada Tali dengan Ujung Bebas
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b. Gelombang Stasioner pada Tali dengan Ujung Terikat
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Gambar 1.12
Simpangan gelombang datang
y1 dan simpangan gelombang pantul y2 pada ujung terikat memiliki beda fase 180°.
A
gelombang pantul
gelombang datang
x
P y1
Tantangan
untuk Anda
Gelombang stasioner dapat terjadi karena superposisi gelombang datang dan gelombang pantul oleh ujung bebas. Titik simpul yang kesepuluh berjarak 1,52 m dari ujung bebasnya. Jika frekuensi gelombang itu 50 Hz, berapakah cepat rambat gelombangnya?
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Tugas Anda 1.4
Rumus syarat terbentuknya simpul dan perut dapat juga ditulis seperti berikut.
x = (2n – 1)41 untuk simpul dan
x = n 21 untuk perut.
Coba Anda diskusikan dengan teman, bagaimanakah penurunannya?
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Jawab: 793A/6B7
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Contoh
1.6
Kata Kunci
• bilangan gelombang • amplitudo
• frekuensi
• kecepatan sudut getar • sudut fase
• fase • beda fase
• cepat rambat gelombang • simpul gelombang • perut gelombang
Contoh
1.7
2,4 m P
O Q
4 m
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3. Kecepatan Gelombang Stasioner
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Aktivitas Fisika 1.2
Percobaan Melde Tujuan Percobaan
Mengamati gelombang stasioner. Alat-Alat Percobaan
1. Kawat tipis (2 buah) yang berbeda massanya 2. Beban
3. Pembangkit getaran (vibrator) 4. Katrol
Langkah-Langkah Percobaan
1. Susunlah alat-alat seperti pada gambar berikut. vibrator
gelombang
beban P S P S
2. Hidupkan pembangkit getaran dengan menghubungkannya ke sumber tegangan sehingga pada tali terbentuk gelombang seperti pada gambar. 3. Jika belum terbentuk gelombang stasioner, ubahlah berat beban yang
tergantung pada ujung katrol sehingga pada suatu saat terbentuk gelom-bang stasioner.
4. Amatilah dengan saksama gelombang yang terjadi dan tulislah kesimpulan Anda dari kegiatan tersebut.
5. Lakukan langkah 1 sampai 4 untuk kawat kedua yang berbeda massanya.
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Pembahasan Soal
Tali yang panjangnya 5 m ber-tegangan 2 N digetarkan sehingga terbentuk gelombang stasioner. Jika massa tali 625 × 10–3 kg, maka cepat rambat gelombang tali adalah ....
a. 2 m/s b. 5 m/s c. 6 m/s d. 10 m/s e. 40 m/s
UMPTN 1996 Pembahasan:
Diketahui:
= 5 m; F = 2 N; m = 625 × 10–3 kg. Menurut Melde kecepatan rambat gelombang tali adalah
v = F = Fm
= -3
2 N 5 m 6,25 10 kg = 40 m/s
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Contoh
1.9
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Contoh
1.10
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Tantangan
untuk Anda
Kawat untuk saluran transmisi listrik yang massanya 40 kg diikat antara dua menara tegangan tinggi yang jaraknya 200 m. Salah satu ujung kawat dipukul oleh teknisi yang berada di salah satu menara, sehingga timbul gelombang yang merambat ke menara yang lain. Jika gelombang pantul terdeteksi setelah 10 s, berapa tegangan kawat?
Tugas Anda 1.5
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Tes Kompetensi
Subbab
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C. Sifat-Sifat Gelombang
1. Pemantulan
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a. Pemantulan Gelombang Lingkaran oleh Bidang Datar
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/ 0
Gambar 1.13
Pemantulan gelombang pada tali: (a) ujung terikat; dan (b) ujung bebas.
Gambar 1.14 Hukum pemantulan
Gambar 1.15 Pemantulan gelombang lingkaran oleh bidang datar.
Sumber: Physics for ‘O’ Level, 1999
Aktivitas Fisika 1.3
Pemantulan Gelombang Tujuan Percobaan
Mengamati pemantulan gelombang pada tangki riak. Alat-Alat Percobaan
1. Wadah bak air 4. Lampu atau OHP
2. Papan 5. Air
3. Kaca
Langkah-Langkah Percobaan
1. Buatlah semacam bak air (tangki riak) dengan kaca sebagai alasnya. 2. Letakkan bak tersebut di bawah sinar lampu OHP, kemudian letakkan sebuah
kertas putih di bawah bak.
3. Buatlah riak kecil dengan cara mencelupkan salah satu jari Anda di tengah-tengah bak. Perhatikan yang terjadi.
4. Masukkan papan tersebut di ujung bak, lalu gerakkan secara perlahan dalam arah horizontal sehingga terbentuk riak.
5. Jelaskan hasil yang diperoleh dan sebutkan perbedaan kedua riak tersebut. Kemudian, jelaskan pula bagaimana dengan pemantulan gelombang pada dinding.
i
r normal
gelombang datang
sinar gelombang
datang sinar gelombang pantul
muka gelombang pantul
1
2
Gambar 1.16
Tangki riak yang ditempatkan di atas OHP.
Sumber: Physics for ‘O’ Level, 1999
Gambar 1.17 Pembiasan pada tangki riak yang kedalamannya berbeda.
Gambar 1.18 (a) Pembiasan terjadi pada bolpoin yang dicelupkan ke dalam air; (b) Skema perjalanan sinar pada proses pembiasan. Sumber: Dokumentasi Penerbit
2. Pembiasan Gelombang
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gelombang datang batas
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dibiaskan
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sinar bias
n2 > n1 n1
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;/0;+5<3=3.+57/8A/,+,5+8:/7,/695+8<38+;(38+;.+=+81+5+8.3=/;><
5+8=+8:+:/7,/6955+8+;+2</:/;=3=/;632+=:+.+%)
35+6/,32,/<+;.+;3:+.++5+86/,325/-36.+;3:+.+%+.+ 5/+.++8 383 <38+; .3,3+<5+8 7/8./5+=3 1+;3< 89;7+6 .+8 7/84+>2 1+;3< :/;:+84+81+8 <38+; .+=+81 </:/;=3 :+.+%) 35+6/,325/-36.+;3:+.++5+86/,32,/<+;.+;3:+.+%+.+
5/+.++8383<38+;.3,3+<5+87/84+>231+;3<89;7+6.+81+;3<:/;:+8
4+81+8 <38+; .+=+81 </:/;=3 :+.+%)
Gambar 1.19
Pembiasan sinar bergantung pada indeks bias medium.
1
sinar datang
2
garis normal
n1
n2 sinar
bias
n2 = n1
sinar datang
n1
n2
sinar bias
n2 > n1
sinar datang
n1
n2
sinar bias
n2 < n1
+ ,
-garis normal garis normal
1
2
1
2
Contoh
1.8
(/,>+21/697,+81.+=+81:+.+,3.+81,+=+<+8=+;+.>+7/.3>7./81+8<>.>=.+=+81 C 35+:/;,+8.381+8+8=+;+38./5<,3+<7/.3>7:/;=+7+.+87/.3>75/.>++.+6+2
Tugas Anda 1.6
Menurut Anda, mengapa gelombang air laut yang begitu besar ketika sampai di pantai menjadi kecil? Apakah terjadi hal yang sama untuk laut yang berbatasan dengan daratan yang bertebing karang curam?
=/8=>5+8<>.>=,3+<8A+.+86>53<5+8:/7,3+<+81/697,+81=/;</,>=
-35/=+2>3 <>.>=.+=+81C
38./5<,3+<
3. Polarisasi
%96+;3<+<3 7/;>:+5+8 :/;3<=3@+ A+81 7/7:/;632+=5+8 -3;3 52+< 1/697,+81 =;+8<?/;<+6 %/;3<=3@+ :96+;3<+<3 .+:+= =/;4+.3 5+;/8+ :/;3<=3@+ :/7+8=>6+8 :/7,3+<+8 ,3+< 5/7,+; </6/5=30 .+8 :/;3<=3@+ ,3.+81 1/=+; %/;3<=3@+ :96+;3<+<3 .+:+= .3?3<>+63<+<35+8 ./81+8 7/7,+A+815+8 1/697,+81 =;+8<?/;<+6 :+.+ </>=+< =+63 %/;2+=35+8%)
gelombang terpolarisasi celah
Gambar 1.20 Gelombang tali yang terpolarisasi.
+ (>.>=,3+<
<38
<38
<38
<38
<38
(/>=+<=+63.31/=+;5+8./81+87/6/@+=3</,>+2-/6+2</7:3=?/;=35+6 )+63 =/;632+= 7/8A37:+81 </:/;=3 <:3;+6 (/=/6+2 1/697,+81 =+63 7/6/@+=3 -/6+2 2+8A+ +;+2 1/=+; ?/;=35+6 <+4+ A+81 7+<32 =/;<3<+ </.+815+8 +;+2 1/=+; 29;3B98=+6 .3;/.+7 +=+> .3</;+: 96/2 -/6+2 </7:3= =/;</,>= /697,+81 A+81 5/6>+; .+;3 =+63 .3</,>= 1/697,+81 =/;:96+;3<+<3 638/+; *8=>5 6/,32 7/7+2+73 :96+;3<+<3 +7+=36+2 %)
%) 7/7:/;632+=5+8 1/697,+81 =;+8<?/;<+6 A+81
.36/@+=5+8:+.+.>+-/6+2A+81:9<3<38A+<+7+.+81/697,+81=+63=/;8A+=+
6969< :+.+ 5/.>+ -/6+2 =/;</,>=%) 7/7:/;632+=5+8
1/697,+81 =;+8<?/;<+6 A+81 .36/@+=5+8 :+.+ .>+ -/6+2 A+81 <+6381 7/8A36+81 =/1+5 6>;>< )/;8A+=+ 1/697,+81 6969< :+.+ -/6+2 A+81 :/;=+7+ A+3=> -/6+2 A+81 </4+4+; ./81+8 +;+2 1/=+; =/=+:3 =3.+5 6969< :+.+ -/6+2 A+81 6+38 A+3=> -/6+2 A+81 =/1+5 6>;>< ./81+8 +;+2 1/=+;
/;.+<+;5+8%) .+8%) .+:+=
.3<37:>65+8,+2@+2+8A+1/697,+81=;+8<?/;<+6A+81.3:/81+;>23-/6+2 /697,+81 A+81 .+:+= 6969< .+;3 -/6+2 2+8A+ 7/736353 <+=> +;+2 1/=+; /697,+81 A+81 ./7353+8 .3</,>= 1/697,+81 :96+;3<+<3 (38+; +6+73 </:/;=3 <38+; 7+=+2+;3 ,>5+8 =/;7+<>5 <38+; =/;:96+;3<+<3 (37,96 >8=>5 <38+; A+81 =3.+5 =/;:96+;3<+<3 +.+6+2 </.+815+8 <37,96 >8=>5
<38+; A+81 =/;:96+;3<+<3 +.+6+2
Gambar 1.21 Polarisasi gelombang tali.
+
gelombang terpolarisasi
,
gelombang tidak terpolarisasi
37°
53° medium 1
medium 2
Gambar 1.22 Cahaya tak terpolarisasi dilewatkan pada sebuah kristal.
4. Interferensi Gelombang
(/6+38 :/7+8=>6+8 .+8 :/7,3+<+8 <30+= 6+38 .+;3 1/697,+81 +.+6+2 38=/;0/;/8<3 (30+= 38=/;0/;/8<3 :+.+ 1/697,+81 +.+6+2 :96+ :/81>+=+8 38=/;0/;/8<3 598<=;>5=30 .+8 :96+ :/81236+81+8 38=/;0/;/8<3 ./<=;>5=30 7>5+1/697,+81.+;3.>+1/697,+81A+81<+6381,/;=/7>*8=>57/632+= :96+ 383 6/,32 4/6+< :/;2+=35+8%) ,/;35>=
Gambar 1.23
Interferensi gelombang air interferensi konstruktif
Sumber: www.physics.umd.edu
interferensi destruktif
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
5. Difraksi Gelombang
30;+5<3.+6+7,+2+<+</2+;32+;3.+:+=.3+;=35+8</,+1+3:/6/8=>;+8 %/;3<=3@+ .30;+5<3 +=+> :/6/8=>;+8 1/697,+81 .+:+= .3+7+=3 ./81+8 7/811>8+5+8 =+8153 ;3+5 .+8 :+:+8 ,/;-/6+2 $+7>8 1/697,+81 A+81 =37,>6.3+3;=3.+5.32+<365+896/21/=+;+8=+81+87/6+385+8.+;31/;+5+8 7+4> 7>8.>; </,>+2 :+:+8 </23811+ 7>5+ 1/697,+81 ,>5+8 ,/;,/8=>5 63815+;+8 35+ 8.+ 7/6+5>5+8 2+6 =/;</,>= 8.+ +5+8 7/8.+:+=5+8
Gambar 1.24
Interferensi dua gelombang (a) interferensi konstruktif (menguatkan) dan (b) interferensi destruktif (menghilangkan).
+ ,
+ +
= =
getaran horizontal diserap sempurna oleh polaroid
cahaya alami yang datang
getaran vertikal diserap sebagian
6. Efek Doppler
35+ 8.+ 7/7,>+= 1/=+;+8 :+.+ :/;7>5++8 +3; 7/811>8+5+8 </,>+2 +6+= :/811/=+; :+.+ :/;7>5++8 +3; =/;</,>= +5+8 =/;,/8=>5 1/697,+811/697,+81 A+81 7/;+7,+= 5/ </1+6+ +;+2 </:/;=3
.3=>84>55+8%) :+ A+81 =/;4+.3 435+ ?3,;+=9; .3=+;35 5/
5+8+8 35+?3,;+=9;.3=+;355/5+8+8+5+8=/;,/8=>563815+;+81/697,+81
+3; </:/;=3 :+.+%)
+;+5 1/697,+81 .3 </,/6+2 5+8+8 6/,32 ;+:+= .3,+8.3815+8 ./81+8 4+;+5 1/697,+81 .3 </,/6+2 53;3 )/;632+= ,+2@+ :><+= .+;3 </=3+: 7>5+ 1/697,+81 A+81 .32+<365+8 ,/;1/;+5 5/ 5+8+8 </23811+ 4+;+5 +8=+;+ :>8-+5 5/ :>8-+5 1/697,+81 A+81 ,/;./5+=+8 .3 </,/6+2 5+8+8 6/,32 :/8./5 .+;3:+.+ :+84+81 1/697,+81 .3 </,/6+2 53;3 ;=38A+ <>7,/; 1/697,+81 A+81 ,/;1/;+5 7/8A/,+,5+8 :/;>,+2+8 :+84+81 1/697,+81 %+84+811/697,+81A+81,/;>,+2+5+87/8A/,+,5+8:/;>,+2+80;/5>/8<3 %/;3<=3@+ :/;>,+2+8 0;/5>/8<3 1/697,+81 +53,+= :/81+;>2 1/;+5 ;/6+=30 +8=+;+ <>7,/; 1/697,+81 .+8 :/81+7+= .38+7+5+8 /0/5 9::6/;
8.+ .+:+= 7/81+7+=3 /0/5 9::6/; 5/=35+ ,/;.3;3 .3 :38113; 4+6+8 .+8 +.+ 79,36 7/8>4> +=+> 7/8./5+=3 8.+ ,>8A3 79,36 =/;</,>= =/; ./81+; </7+538 =38113 /7353+8 :>6+ 435+ 79,36 =/;</,>= 7/84+>23 8.+,>8A379,36=/;</,>==/;./81+;</7+538;/8.+2/897/8+=/;</,>= 5+63 :/;=+7+ .3+7+=3 96/2 </9;+81 03<35+@+8 ><=;3+ )!*+!& '(($)
D </23811+ .38+7+5+8 /0/5 9::6/;
%/;>,+2+8 ,>8A3 79,36 7/84+.3 =38113 +=+> ;/8.+2 =/;</,>= .3</,+,5+8 96/2 :/;>,+2+8 0;/5>/8<3 !/=35+ 79,36 =/;</,>= 7/8./5+=3 8.+0;/5>/8<38A+</7+538,/<+;(/,+6358A+5/=35+79,36=/;</,>=,/; 1/;+5 7/84+>23 8.+ 0;/5>/8<38A+ </7+538 5/-36 %/;>,+2+8 0;/5>/8<3 =/;</,>= .+:+= .34/6+<5+8 </,+1+3 ,/;35>=
!/=35+ <>7,/; ,>8A3 .+8 :/8./81+; .3+7 1/697,+81 ,>8A3 A+81 .3:+8-+;5+8 96/2 <>7,/; ,>8A3 7/736353 :+84+81 1/697,+81 A+81 <+7+ ./81+8 1/697,+81 ,>8A3 A+81 .3=/;37+ :/8./81+; ,/;6+5> :/;<+7++8
$+7>8:+.+<++=<>7,/;,>8A3,/;1/;+5.+8:/8./81+;.3+77+5+ :+84+81 1/697,+81 A+81 .3=/;37+ :/8./81+; +.+6+2
Gambar 1.26 Sumber gelombang yang bergerak pada permukaan air menyebabkan panjang gelombang lebih pendek sesuai arah gerak. 1 2 3 4 5
s
P
vp vs
+
s
5 4 3 2 1
,
Gambar 1.25 Difraksi pada gelombang air
gelombang air setelah melewati celah gelombang air
celah
Sumber: Physics for Scientist & Engineers with Modern Physics, 2000
,+2@+ </=/6+2 1/697,+81 5/6>+; .+;3 :+:+8 ,/;-/6+2 ,/8=>5 7>5+ 1/697,+81=3.+5</:/;=3,/8=>51/697,+81A+81:/;=+7+%96+A+81+5+8
</23811+ 0;/5>/8<3 A+81 .3=/;37+ :/8./81+; +.+6+2 </,+1+3 ,/;35>=
D+
*8=>55+<><</,+6358A+:+.+<++=<>7,/;,>8A3.3+7.+8:/8./81+; ,/;1/;+5 :+84+81 1/697,+81 A+81 .3=/;37+ :/8./81+; +.+6+2
</23811+ 0;/5>/8<3 A+81 .3=/;37+ :/8./81+; +.+6+2 </,+1+3 ,/;35>=
D,
(/-+;+ >7>7 :/;<+7++8 /0/5 9::6/; >8=>5 <>7,/; ,>8A3 .+8 :/8./81+; ,/;1/;+5 ,/;6+5> :/;<+7++8
D
!/=/;+81+8
0;/5>/8<3 A+81 .3=/;37+ :/8./81+; B
0;/5>/8<3 .+;3 <>.>= ,>8A3 B
5/-/:+=+8 1/697,+81 ,>8A3 7<
5/-/:+=+8 1/;+5 :/8./81+; 7<
5/-/:+=+8 1/;+5 <>7,/; ,>8A3 7<
%/;4+843+8 =+8.+ A+81 .31>8+5+8 .+6+7 :/;<+7++8 +.+6+2 </,+1+3 ,/;35>=
35+ :/8./81+; 7/8./5+=3 <>7,/;:/8./81+; :9<3=30 35+ :/8./81+; 7/84+>23 <>7,/;:/8./81+; 8/1+=30 35+ <>7,/; 7/8./5+=3 :/8./81+;<>7,/; 8/1+=30 35+ <>7,/; 7/84+>23 :/8./81+;<>7,/; :9<3=30
Contoh
1.11
(/,>+279,36,/;1/;+5./81+85/-/:+=+8574+7<+7,367/7,>8A35+856+5<98 ./81+80;/5>/8<3 B+;3+;+2,/;6+@+8+8</9;+81:/81/8.+;+79=9;,/;1/;+5 ./81+85/-/:+=+8574+7)/8=>5+80;/5>/8<3,>8A356+5<98A+81.3./81+;96/2 :/81/8.+;+</:/.+79=9;=/;</,>=435+-/:+=;+7,+=,>8A3.3>.+;++.+6+2 7<
-35/=+2>3
79,36 574+7 7<
:/8./1+; 574+77<
>.+;+ 7<
56+5<98 79,36 B
:/8./81+; >.+;+ :/8./81+;
56+5<9879,36 >.+;+ 79,36
7< 7<7< 7< B
B
+.3:/81/8.+;+</:/.+79=9;7/8./81+;,>8A356+5<98./81+80;/5>/8<3B
Tokoh
Christian Johann Doppler
(1803–1853)
Christian Johann Doppler adalah fisikawan yang lahir di Salzburg, Austria. Dia belajar di Vienna dan menjadi seorang profesor Fisika (1851). Dia terkenal dengan pengamatannya tentang variasi frekuensi gelombang suara dan gelombang cahaya karena pengaruh kecepatan gerak relatif antara sumber dan pengamat.
(/,>+21/697,+81.+=+81.+;3>.+;+./81+8<>.>= .+=+81C7+<>55/.+6+7+3;.+8.3,3+<5+8./81+8 <>.>=,3+<C 35+:+84+811/697,+81=/;</,>=.3>.+;+ -7=/8=>5+8:+84+811/697,+81.3.+6+7+3; (/,>+21/697,+81.+=+81:+.+,3.+81,+=+<+8=+;+
.>+7/.3>7./81+8<>.>=.+=+81 C 35+38./5<,3+<
;/6+=307/.3>7.>+=/;2+.+:7/.3>7<+=>+.+6+2
=/8=>5+8 <>.>= ,3+<8A+ .+8 6>53<5+8 :/7,3+<+8 1/697,+818A+
/697,+81.+=+81.+;3:/;7>5++8+3;A+81.+6+75/ :/;7>5++8+3;A+81.+815+6./81+8<>.>=.+=+81 C 35+-/:+=;+7,+=1/697,+81:+.+:/;7>5++8+3;A+81 .+6+7.+8:/;7>5++8+3;A+81.+815+67<.+87< =/8=>5+8<>.>=,3+<8A+
/6+<5+8+:+A+81.37+5<>.:96+;3<+<3
(/,>+2+7,>6+8<,/;1/;+5./81+85/-/:+=+87< <+7,367/7,>8A35+8<3;38/8A+:+.+0;/5>/8<3 B (/9;+81 :/81/7>.3 =;>5 ,/;1/;+5 ,/;6+@+8+8 +;+2 ./81+85/-/:+=+87<7/8./81+;,>8A3<3;38/
+7,>6+8< )/8=>5+8 0;/5>/8<3 A+81 .3./81+; :/81/7>.3<++=
+ 5/.>+79,36<+63817/8./5+=3 , 5/.>+79,36<+63817/84+>23
!/=35+,/;.3;3.3=;9=9+;</9;+81+8+57/8./81+;,>8A3 ,/;0;/5>/8<3 B .+;3 <3;38/ 79,36 :963<3 A+81 7/8./5+=38A+ (/=/6+2 79,36 :963<3 7/6/@+=38A+ 0;/5>/8<3,>8A3A+81=/;./81+;7/84+.3 B/;+:+ 5/6+4>+879,36:963<3=/;</,>=
!/-/:+=+8,>8A3.3>.+;+ 7<
(/,>+2 <>7,/; ,>8A3 7/736353 0;/5>/8<3 B ,/;1/;+5./81+85/-/:+=+8 7<7/8./5+=3</9;+81 :/81+7+=A+81.3+7)/8=>5+8,/<+;0;/5>/8<3A+81 .3./81+;96/2:/81+7+=435+
+ =3.+5+.++8138
, =/;.+:+=+8138A+81,/;1/;+5./81+85/-/:+=+8 7<</+;+2./81+8+;+2<>7,/;,>8A3.+8 - =/;.+:+=+8138A+81,/;1/;+5./81+85/-/:+=+8
7<,/;6+@+8+8+;+2./81+8+;+21/;+5<>7,/; ,>8A3/:+=;+7,+=,>8A3.3>.+;+ 7<
Tes Kompetensi
Subbab
C
)"#&$ $%,#,$+! &
Rangkuman
/;.+<+;5+8<>7,/;8A+1/697,+81.35/697:955+8 7/84+.3.>+7+-+7A+3=>1/697,+817/5+835.+8 1/697,+81/6/5=;97+18/=35
/;.+<+;5+8 +;+2 1/=+;8A+ 1/697,+81 .3,/.+5+8 7/84+.3 1/697,+81 =;+8<?/;<+6 .+8 1/697,+81 69813=>.38+6
/<+; -/:+= ;+7,+= 1/697,+81 ,/;,+8.381 6>;>< ./81+8:+84+811/697,+81.+8,/;,+8.381=/;,+635 ./81+8@+5=>>8=>57/8/7:>2<+=>1/697,+81
/<+;8A+ .+A+ A+81 .3=;+8<73<35+8 7/6+6>3 <>+=> 1/697,+81 ,/;,+8.381 6>;>< ./81+8 5>+.;+= +7:63=>.95>+.;+=0;/5>/8<3.+86+4>1/697,+81
(37:+81+8 1/697,+81 ,/;4+6+8 .3 <>+=> =3=35 .38A+=+5+8./81+8:/;<+7++8
<38 <38
%/;<+7++8 5/-/:+=+8 </,>+2 =3=35 .+6+7 <>+=> 1/697,+81=/;2+.+:@+5=>+.+6+2=>;>8+8:/;=+7+ %A+81.38A+=+5+8./81+8:/;<+7++8
<38
-9<
Contoh
1.12
!/;/=++:3,/;1/;+57/83811+65+8<=+<3>8./81+85/-/:+=+8 574+7(/9;+81 7+<383<7/7,>8A35+8:/6>3=8A+./81+80;/5>/8<3 B 35+-/:+=;+7,+=,>8A3 .3>.+;+ 7<,/;+:+0;/5>/8<3A+81.3./81+;96/2:/=>1+<.3<=+<3>8=/;</,>=
-35/=+2>35/;/=++:3 574+77<>.+;+ 7<:/8./81+; 7<
:/8./81+;
>.+;+ :/8./81+;
:/6>3= >.+;+ 5/;/=++:3
7< 7< B
7<7<
B
+
