﻿138 Mr. C. Rodgers on the Vibration 



Then u =« — &©, 



and the accelerations are 



along A : u — vo) = a — 2/;<y — aw 2 , 



along OB : v + uco = b + 2aco — bay 2 . 



If the force required to produce unit deflexion in the shaft 

 is <r + e along OA and a — e along OB, and we resolve along 

 OA. and OB, the equations are : 



M(« — 2bco — aco 2 ) 4- (<r + e)a = — M<? sin g>£, 



M("i + 2aco — bco 2 ) + (o— e)6= — 1% cos cot ; 

 that is, 



{M(D 2 -ft) 2 ) + (o- + e)}a-2Mr«)D/>=-M^sinft)^ 



{M(B 2 - co 2 ) + (a -6)}b + 2McoI)a=-Mc/ cos cot, 

 giving 



[{M(D 2 -a) 2 ) + cr} 2 -6 2 + 4MVD 2 ] a 



= — M.g(a — e— 4&> 2 ) sin o>£, 

 [{M(D 2 -o) 2 ) + cr} 2 -€ 2 + 4M 2 a) 2 D 2 ] 6 



== — M#(y + e— 4« 2 ) cos cot, 

 The solution is (neglecting e 2 in comparison with cr) 



This gives the position with respect to the rotating axes ; 

 the position with respect to the fixed axes is 



that is 



x = a cos cot — b sin cot, 

 y = b cos cot -{-a sin cot, 



This result is the same for y as obtained in (34) by the 



