where AX(z) is the net horizontal distance traveled by a water particle, 

 z feet below the surface, during one wave period. 



For the example problem when 



z = 



/tthV cosh(4TTd/L) C 



AX(0) = T — ; 



\ L/ sinh2(2iTd/L) 2 



(ttH)2 cosh(4iTd/L) (tt1)2( 1 .899) 



= 0.347 m (1.14 ft) 



2L sinh2(2ird/L) 2(60)(0. 6705)2 

 (d) The first-order approximation for pressure under a wave is 



P = 



pgH cosh[2iT(z + d)/L] 



cos 6 - pgz 



when 



and when 



Therefore, 



cosh(2iTd/L) 

 9=0 (i.e., the wave crest), cos 9=1 

 2tt(z + d) 



z = -d, cosh 



= cosh(O) = 1.0 



P = 



(1025)(9.8)(1) 1 



- (1025) 9.8(-6) 



2 1.204 



= 4171 + 60,270 = 64,441 N/m^ (1,345 lbs/ft^) 



at a depth of 6 meters (20 feet) below the SWL. The second-order terms 

 according to equation (2-56) are 



3 ttH^ tanh(2TTd/L) (cosh[4Tr(z + d)/L] l) 



- Pg — ; — < ) cos 28 



8 L sinh2(2TTd/L) [ sinh2(2TTd/L) 3 J 



1 irH^ tanh(2Tid/L) ( 4n(z + d) | 



pg . — (cosh 1} 



8 L sinh2(2Trd/L) [ L J 



Substituting in the equation: 



3 Tr(l)2 (0.5569) 



- (1025)(9.8) — 



8 60 (0.6705)2 



(0.6705)2 3 



(1) 



1 Tr(l)2 (0.5569) 



-- (1025)(9.8) (0) = 462 N/m2 (10 lbs/ft2) 



8 60 (0.6705)2 



2-43 



