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 )¿