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



EXAMPLE PROBLEM 4************** 



GIVEN : An incident wave with a height of 1 meter (3.28 feet) and a period 

 of 30 minutes approaches a coastline through water 2,500 meters (8,200 

 feet) deep, and passes onto a shelf where the water depth is 100 meters, 

 at an angle of incidence 9j = 30° 



FIND : 



(a) The angle at which the transmitted wave propagates onto the shelf, 



(b) the height of the reflected wave, and 



(c) the height of the transmitted wave. 

 SOLUTION : 



(a) From equation (164) 



( d A l/2 



5111 9 2 = \d^) Sin 9 1 



2 = sin -1 



7 100 V /2 

 I2.500 sin 30 



■- " 



6 = 5.74° 





•]■ 



sin -1 (0.1) 



(b) From equation (163) 



J&7 cos 9 - v r dl cos 9„ 



H = — : - (H.) 



vd cos 9 + /d cos 8 



„ /2,500 cos 30° - /iW cos 5.74° ,.. _ ,., \_ ,- n _ . t . 



H = —^^2 ; CI) = 0.626 meter (2.05 feet) 



r /2.500 cos 30° + t/lOO cos 5.74° 



(c) From equation (158) 



H x = H. + H 

 t % r 



H = 1 + 0.626 = 1.626 meters (5.33 feet) 

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



When the initial angle of incidence 9^ > 0, the distance between 

 adjacent wave rays is different for the incident and transmitted waves. 

 As the energy equations are written for a unit length of wave crest, 

 from conservation of energy, 



71 



