﻿

1 H 



9—f—* C0S 



1 7?T 



742 Prof. H. Lamb and Miss L. Swain on 



At the middle point of the canal we have 



/ m sin 2a\ nft * 



-. cos 2n£ 1 ; — s — . . [lb) 



1 \ sm 2ma/ 



As in the equilibrium theory, the range is very small if a 

 be small, but there is not a true node. 

 At the ends <9 = ±« we find 



1 H f/m sin 4a \ , -v 



2 m~ — 1 L^ sin4???a / 



, ?n(cos4???a — COS 4a) . n/ , . J /in\ 



H : : ; - Sill 2(>tf + *) }■ . . (ly) 



~~ sm 4???a J 



= 2 HR 1 cos2(wi±anF^i), ( 20 ) 



if r?isin 4a — sin 4wa "^ 



lii cos i<bi— —■ — « — ^r— . — . , | 



1 V1 (m 2 — 1) sm4ma ' 



!> . . (21) 

 _. . _ . mfcos 4ma- cos 4a) i 



K x sm 2(p! = ., o — ^v— : — . x~. 



1 ^ ; (wr — l)sin4wza) J 



The significance of the quantities R a , (b 1 is the same as in 

 the equilibrium formula (13) *. When a is small we have 



B 1 =2«, £!=-].„■+?«, .... (22) 



approximately, as before. 



The values of R x become infinite for sin 4ma = 0. This 

 determines the critical lengths of the canal for which there 

 is a free period equal to 7r/n, or half a lunar day. The 

 limiting value of cj> l in such a case is given by 



OJ m(cos imx — cos 4a) ja _ ,„„x 



tan 2d>! = v . , ^—. = — cot 2a, or tan 2a. . (23) 



r m sin 4a— sm4ma • ' 



according as 4wa is an odd or even multiple of it. 



For purposes of numerical illustration we have taken 

 m = 2'5. If 7r/ft=12 lunar hours, this implies a depth of 

 10820 feet, which is of the same order of magnitude as the 



mean depth of the ocean. The corresponding wave-velocity 

 c is 360 sea miles per lunar hour. The first critical length 



is 2160 miles ( «= t^tt), and the second is 4320 miles. 

 * See the footnote on p. 741. 



