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Diketahui :
∅=Ae ik (x1+r x3−ct )
ψ=Be ik (x1+s x3−ct )
σ=0,25 ; s=0,39i
r=0,848 i ; B=−1,47 i A
Ditanya : Tuliskan Persamaan untuk mencari U 1 dan U 3 ! Gambaran Plotingnya.
Jawaban :
Mencari U 1
U 1=∂∅∂ x1
− ∂ψ∂ x3
U 1=∂∂ x1
(Ae ik (x1+r x3−ct ))− ∂∂ x3
(Be ik (x1+s x3−ct ))
U 1=Aik eik (x1+ r x3−ct )−Biks eik (x1+s x3−ct )
U 1=Aik eik (x1+ r x3−ct )− (−1,47 i A ) (ik )(0,39 i)eik (x1+s x3−ct )
U 1=Aik eik (x1+ r x3−ct )−(0,5733)Aik eik (x1+s x3−ct )
U 1=Aik eik (x1+ r x3−c t )−0,5733 Aik e
ik (x1+s x3−ct )
U 1=Aik eik (x1−ct )(eikr x3−0,5733e iks x3)
Mencari U 3
U 3=∂∅∂ x3
+ ∂ψ∂ x1
U 3=∂∂ x3
(Ae ik ( x1+r x3−ct ) )+ ∂∂ x1
(Be ik (x1+ s x3−ct ))
U 3=Aikr eik (x1+r x3−ct )+Bik e ik (x1+ s x3−ct )
U 3=Aik (0,848 i ) eik (x1+r x3−ct )+ (−1,47 i A )ik eik (x1+s x3−ct )
U 3=−0,848 Ak eik (x1+r x3−ct )+1,47 Ak e
ik (x1+ s x3−ct )
U 3=Ak eik (x1−ct )(−0,848e ikr x3+1,47e ik s x3)
U 3=−Ak e ik (x1−ct )(0,848e ikr x3−1,47e ik s x3)
Nama : Endah Juniarti
NIM : 146090300011003
MK : Seismologi
Karena gerakan gelombang Rayleigh harus merupakan bilangan real, maka kita gunakan :
e iω( px1−t)=cos [ω (p x1−t ) ]+i sin [¿ω ( p x1−t )]¿
Maka persamaan U 1 dan U 3 menjadi :
U 1=Aik eik (x1−ct )(eikr x3−0,5733e iks x3)
U 1=Aik ¿
U 1=¿
U 1=¿
U 1=−Ak sin [¿ω ( p x1−t )](eikr x3−0,5733eiks x3)¿
U 1=−Ak sin [¿ω ( p x1−t )]¿¿
U 1=−Ak sin (k x1−ωt )¿
dan
U 3=−Ak e ik (x1−ct )(0,848e ikr x3−1,47e ik s x3)
U 3=−Ak ¿
U 3=(−Ak cos [ω ( p x1−t ) ]−Akisin [¿ω (p x1−t )]¿(0,848eik r x3−1,47eik s x3))
U 3=−Ak cos [ω (p x1−t ) ](0,848eik r x3−1,47eik s x3)¿
U 3=−Ak cos [ω (p x1−t ) ](0,848ik e(−0,848 i) x3−1,47 eik (−0,39 i) x3)¿
U 3=−Ak cos (k x1−ωt )(0,848e(−0,848)k x3−1,47e(−0,39) k x3)¿
Note : kita ambil bahwa
Ak=konstan=a
x1=0
x3=0
Sehingga ;
U 1=−Ak s∈(k x1−ωt )¿
U 1=−a sin (−ωt )(e0−0,5733 e0)
U 1=−0,563a sin (−ωt )
dan
U 3=−Ak cos (k x1−ωt )(0,848e(−0,848)k x3−1,47e(−0,39) k x3)¿
U 3=−acos (−ωt )(0,848e0−1,47e0)¿
U 3=−0,194acos (−ωt )
Ploting :
kct sin kct cos kct U1 U3
0 0 1 0 -0,194
45 -0,85090,52532
20,47905
9-
0,10191
90 -0,894-
0,44807 0,503320,08692
6
135-
0,08837-
0,996090,04975
20,19324
1
1800,80115
3-
0,59846-
0,451050,11610
1
2250,93009
50,36731
9-
0,52364-
0,07126
2700,17604
60,98438
2-
0,09911-
0,19097
315-
0,745130,66691
6 0,41951-
0,12938
360-
0,95892-
0,28369 0,539870,05503
6
0 50 100 150 200 250 300 350 400
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
kct
U1, U
3
-0.6 -0.4 -0.2 0 0.2 0.4 0.6
-0.25-0.2
-0.15-0.1
-0.050
0.050.1
0.150.2
0.25
U3
U1
Agar tidak Retrograde harus berada pada kedalaman tertentu, maka U 1=0
U 1=−Ak sin (k x1−ωt )¿
0=−Ak sin (k x1−ωt )¿
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