assumption, the horizontal length 2A of the path moved by the water particle 

 as a vave passes in shallow water is approximately 



and the maximum horizontal water velocity is 



u = ^ (4-17) 



max zd 



The term under the radical is the vvave speed in shallow water. 

 *************** EXAMPLE PROBLEM 1*************** 



GIVEN : A wave 0.3 meters (1 foot) high with a period of 5 seconds is 

 progressing shoreward in a depth of 0.6 meter (2 feet). 



FIND: 



(a) Calculate the maximum horizontal distance 2A the vater particle 

 moves during the passing of a wave. 



(b) Determine the maximum horizontal velocity ^yf,^^ of ^ water particle. 



(c) Compare the maximum horizontal distance 2A with the vavelength in 

 the 0.6-meter depth. 



(d) Compare the maximum horizontal velocity u with the wave speed C . 



'^ ■' max '^ 



SOLUTION: 



(a) Using equation (4-16), the maximum horizontal distance is 



2A = iiI-^ 



2ird 



OA 0.3 (5) V9.8 (0.6) „ „, ^ a M f .^ 



2A = „ ,- . , ^ = 0.96 meter (3.17 feet) 



2iT (0.6) 



(b) Using equation (4-17) the maximum horizontal velocity is 



HT Vjd 

 max -id 



0.3 V9.8 (0.6) n c^ . A r-^ n f .\ 



u = „ -^ - > = 0.61 meter per second (2.0 feet) 



max 2 (.0.6) 



(c) Using the relation L = T Vgd to determine the shallow-water 

 wavelength. 



L = 5\/9.8 (0.6) = 12.12 meters (39.78 feet) 



4-4 7 



