depths is a linear slope S = 0.1 and the wave is at a zero angle of 

 incidence with the slope transition, i.e., 6-j = 0. 



FIND : 



(a) The height of the reflected wave, 



(b) the height of the transmitted wave, and 



(c) show that energy is conserved, i.e., that the total energy in 

 the reflected and transmitted waves equals the incident wave energy. 



SOLUTION : 



(a) L x = Cj T = >/gd^ T 



L'j = /9.8 x 100 (40 x 60) = 75,100 meters 

 From equation (172) , 

 4iTd, 



LjS 



477 X 10 ° = 0.167 



1 75,100 x 0.1 

 From Figure 20, where 



d l 100 



— = — = 0.033 and Z, = 0.167 



d 2 3,025 1 



it is found that 



K - -0.62 (the negative sign indicates that the reflected wave 

 is it radians out of phase with incident wave) 



H^ = 0.62 W i = 0.62(0.5) = 0.31 meter (1.02 feet) 



(b) From Figure 20, 

 K « 0.32 



H^ = 0.32 H i = 0.32(0.5) = 0.16 meter (0.52 foot) 



( c ) Y = Pg = (1,026 kilograms per cubic meter) (9.8 meters per 

 second squared) 



Y = 10.055 kilograms per square meter - second squared 



75 



