436 Prof. A. W. Porter and Mr. H, E. Gibbs on 

 We are only concerned with the case of two roots complex, 

 say f 2 = A-MB, 



6=A-tB." 



Hence f v +2A=-<7 (i.) 



A 2 + B 2 + 2Af 1 = 7 tj (ii.) 



?i(A 2 + B 2 )=-; (iii.) 



From (iii.) f i is of tlie opposite sign to ^ and is therefore 

 essentially negative. fi=— Pi, 



2A is positive when P : >^. 



From (i.) A pusses throngh zero when f x = —5/, and at the 

 same time B 2 = /i from (ii.), thus h is then positive. 



Further, f x B 2 is then equal to —j from (iii.) and therefore 



B 2 =^=/i. 

 9 

 If A>0, %i is more negative than — g. In terms of a and 



b, for A to be positive 



> t-» 7 r R 



_f 1 = P 1 >a+/,>-+ . 



It follows that* if A=0, f x = — #, and j—gh, so that A is 

 positive. 



Hence the solution for the current is, 



0=g+Fe-^ + Gcos (B*-^) 



where B= v 7 /* = Vi/<7 • 



Also 



y = - |g + Fir-* + G' cos (Bf - V) • 



„ n . /K/L . K/m 



The value of V" is / =p- 



V £ + L 



Hence a disturbance such as in the above critical case 

 if once excited will be maintained constant. It will be noticed 



that a/ is the frequency-constant of the free undamped 



