94 Mr. P. F. Ward on the Transverse 



From (12) 



u 



=A0 1 O) + B^ 1 (» (28) 



or 



A B 



u= — I 2 (2 s/x) + — J 2 (2 \/x). 



From (23) the equation to determine x 1 and therefore X is 



= 



a?' 



+ 



3C q 



.. ., . . (29) 



2! 3! 1! 3! 5! ' 2! 4! 7! 

 the lower roots of which are found to be 



8-7192, 21-146, 38-453. 

 The ratio of the constants A and B is from (18) found to be 



— A/^r (*i) = B/0o(O = (say), 

 whence 



u = C{0o(#i)'<h(#) — To(#i)0i (*)}• 

 In the first mode #! = 8*7192, 



m«C{ -0996438 ^(aj) +19-0308 ^i(#)}. 

 In the second mode # 1 = 21*146, 



m = C{270-325^i(*)--0473844^ 1 (*)}. 

 We then find as corresponding values: 



2 V*. 



x. 



0i(#). 



*i(*)- 



u/C. 



T5-12 

 15-14 



6-5536 

 6-6049 







+ •076685 

 -•205068 



Hence the node is at the point 



#=6-56756, or */Z = -310582. 

 Again we find : 



2V^. 



x. 





dx ' 



— [C. 



dx\ 



re-40 



[6- 12 



10-24 

 10-3041 



1-414192 



+•000358052 



-•0175107 

 +•0297799 



The point of maximum excursion is then given by 

 # = 10-26373 or ^ = -485373. 



