solution: 



(a) Using Equation 4-10, the maximum horizontal distance is 



HT Vgd^ 



2A = 



2ffd 



1 (5) ^32.2 (2)^ 



2A = = 3.2 feet . 



27r(2) 



(b) Using Equation 4-11, the maximum horizontal velocity is 



2d 

 1^32.2(2)^ 

 2(2) 



u = TTT; = 2.0 feet per second 



max T '^^^ ^ 



(c) Using the relation L = Ti/gd to determine the shallow-water 

 wavelength. 



L = 5v/32.2(2)' = 40.1 feet . 



From (a) above, the maximiam horizontal distance 2A is 3.2 feet 

 therefore the ratio 2A/L is 



2A 3.2 



— = = 0.08 . 



L 40.1 



(d) Using the relation C = v'^ (Equation 2-9) to determine the 

 shallow-water wave speed 



C = y^32.2 (2)" =8.0 feet per second . 



From (b) above the maximum horizontal velocity ^ax> ^^ ^'^ 

 feet per second. Therefore the ratio ^ax/^ i^ 



^max 2.0 



-^^^ = = 0.25. 



C 8.0 



************************************* 



Although small -amplitude theory gives a fair understanding of many 

 wave-related phenomena, there are important phenomena that it does not 

 predict. Observation and a more complete analysis of wave motion show 

 that particle orbits are not closed. Instead, the water particles 

 advance a little in the direction of the wave motion each time the wave 



4-41 



