All-101 5^ 



The perturbation contribution vanishes at that instant since 

 its coefficient is cos %. Thus, no matter what the value of 6 

 (essentially, the frequency), as long as it lies within the 

 limits necessary for the validity of the above method of solu- 

 tion, the maximum value of the transport will occur at t = m'ii/2, 

 m = 1, 5} 9 .♦. J and its value is given by 1 + a times the 

 steady transport value. 



The interval of 9 days between the time at which the 

 wind reaches its mean amplitude and the time at which the trans- 

 port reaches its mean amplitude is also independent of the fre- 

 quency. To show this let Vq = (1 + a sin t;)Q and V-j^ = aL cos t. 

 Then V = (1 + a sin -r )Q + 6a L cos t. Since the mean value of 

 the transport is V := Q, we can find the time at v;hich this occurs 



by setting 



(1 + a sin t)Q + 6aL cos -c = Q 



0^ , L6 



tan t: = - __ . 



Since t Is small, we can write tarn; ~ i and therefor© 



T « - 



L6 



Substituting t = wt and 6 = w/Ps, we have finally 



^ Q ps 

 which is independent of frequency and a , 



It is apparent from Fig. 2 that the out-of-phase effect 

 is of relatively greatest importance in the counter-current 

 rather than in the main stream. The graph show?? the various 



