Sound between Parallel Walls. 309 



whence 



, -998 + ix 6-695 1-A , 8lOQ1u .. 810 „ 1M 



tan,g= <Q ,^.7^9 " =1 (cos 81° 31 + esin81° 31'), 



so that 



2-1 2 . tan z= 1-5 (cos 6° 31'+ 1 sin 6° 31'). 



The course of the calculation makes it clear that as z in- 

 creases, tan z approaches the limit i, so that ultimately = ^tt,. 

 or the angle reduces from 67^° to 45°. Hence 



z 2 =iy' 2 n*yi 2 /a*, (47) 



and 



m s =h 2 /a 2 -in 3 y' 2 /a\ (48) 



independent of y v 



In order to obtain u as a function of y, we have now to 

 interpret (42) for the case where y x is great. It may be 

 written 



-7^. m 



where z f given by (47), is 



-%w ^ 



By (30), since rj is large, we have approximately 



cos z=\e? (cos f +i sin?), .... (51} 

 where 



*-*-$& ^ 



Accordingly, cos z is large. If, as near the middle of the 

 layer of gas, y be not large, cos (2y/y 1 ) = l, and 



u=— l/cos£, (53) 



a small quantity. When y is so large that 2y/y x is large, a& 

 well as z, we may write 



u=— &'-* .<&?-&, (54) 



where 



*-v-S3 • (55 > 



As the walls are approached u rapidly increases, and at last 

 e*'" 71 becomes nearly equal to unity. We must bear in mind, 

 however, that (42) and therefore (54) must not be applied 

 within the frictional layer lying quite close to the walls, so 

 that we are not at liberty to suppose y actually equal to y Y . 

 Under normal conditions the thickness of the frictional 



