Page 42 - Rekayasa Pantai
P. 42

O  T  L  1                                       (3.14)
                            E k  =        (u 2  + w 2 ) .dzdx  .dt
                                − h  O  O  2
                                        u  =       =  agk  cosh k (h +  ) z  cos(kx −  ) t 
                                                                                  (3.15)

                               -
                                  x   w    cosh kh
                                     agk  sinh k (h +  ) z                      (3.16)
                            w  = −  =                 sin(kx −  ) t 
                                   z   w    sinh kh
                        Dari persamaan dispersi      tanh   kh =  w 2                          (3.17)
                                                         gk


                                              coshk (h +  ) z
                        Jadi             u =  aw          cos(kx − wt )               (3.18)
                                                 sinh kh

                                              sinhk (h +  ) z
                                       w =  aw            sin(kx − wt )                       (3.19)
                                                 sinhkh

                  Maka

                       1  (  aw) 2  O T L
                                                                        ).
                  E  =            cosh 2  k (h +  ) z cos 2 (kx −  wt ) sinh+  2  k( h +  z sin  2 ( kx −  wt)  dxdz.  . dt
                      k
                       2  sinh 2  kh  − hO O
                          O                    z   sinh  k 2  (h +  ) z  O
                           cosh  2 k( h +  z) dz =   +        
                          − h                 2        4 h      −   h

                                             h   sinh  . 2 .  kh
                                               =  +
                                             2        k 4


                                             h  kh sinh2  +  2 .  kh  
                                               =              
                                             2       2 kh     


                          O                  h  − kh sinh2  +  2 .  kh  
                           sinh 2  k( h + z) dz  =            
                          −h                    z    2 kh      

                                                 L T
                           cos 2 (kx − t )dx. dt =  1   ( 1+ cos  ( 2  kx − t ) dtdx.
                                                2
                                                 O O

                                                1  T                L
                                                    =   x + sin  ( 2  kx − t)  dt
                                                2  O                O



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