PROCEEDINGS OP SECTION A. 



31 



Substituting from these equations for Az, B2, P2, and Qz in the 

 expression for C9, we find that 



C^ = -r^ < {s-b) {Ai cos ojit + Bi sin u)it)-{s-a) (Pi cos 



w^t + Qi sm W2O I" 



Now as Ci = C2 = when t = 0,it follows that 

 .4i = and Pj = 



hence Ci = B^ sin wif + Q^ sin wj. 



C2 =-rTf\ {^ -^) Bi sin uj^t - {s - a) Qi sin uy^ji \ 



and as K^V^ = D'^^, K^V^ = D-'^C^, 



KJ\ 



cos lO^t - -^^ COS W2^ 



^2 ^^2 = ~ rnV) (s -&) — COS wi^ -(s -a) — cos W2^!- 

 But when t = 0, }\ = ^, and V^ = 0, • ,'<f' \ C A. ?V 



therefore ^^ + 9l ^ - K^E 





Pi 



(?1_ 



which give iis 

 B 



H ^ _ 



s - a 



K^E, 9l 



C w.> 



where a, b, c, and s refer to the triangle. 



Hence for the initial conditions Fi = P, V^ = 0, Ci = C2 — 

 when t — 0, the complete solution is given by— 



Fj = - ] (* ~ ^) cos Wx< + (s-b) cos W2^ ( 



P ( ) 



F- = p2^ — ] COS Wit - cos oiot > 



Ci = - £^1 — < oji {s -a) sin toit + wg (s -6) sin w^t > 



P f -I 



(/2 = P2<^ -^2 — "! - wi sin wi^ + W2 sin o/of ?■ 



where p2 = —^ = b K^M =t.^, (See Eq. (Ill) § 3.) 



(I) 



K2L2 



a Li 



