of three Magnetically Coupled Oscillation Circuits. 579 



By analogy from electrical practice, we can write down 

 the coefficients of coupling as : 



m 2 2 a 2 h 2 2 

 ^ — t^~^ ^- 9 between A and d, 



m 3 Wi 8 2 



A and C, 



SwiKj 2 = ??2 1 K 1 2 + m 2 a 2 + ?n 3 6 2 , 



and so on. 



To solve the equations (1), (2), and (3), assume tentatively 



# = Acos (nt + e), cj> = B cos (nt + e), and ty = Q cos (nt + e) . 



Substituting these values in (1), (2), and (3), and eliminating 

 0, (j>, and -v^, we have 



A{2 m^i 2 X n 2 + S?Ji!%} + Bm 2 ah 2 n 2 



+ Cm 3 bh B n 2 = 0, 





(7) 



Am 2 ah 2 n 2 4- B(«? 2 K 2 2 ?i 2 4 m 2 gh 2 ) = 0. 



3 



(8) 



and Am f) ^ 3 u 2 4-C(m 3 K 3 2 rA 2 4-m 3 (//i3) = 0. 



• • 



(9) 



From this we get the determinantal equation : 







(?2 2 ^??2 1 K 1 2 + <7X"'i/ii) : n 2 m 2 ali 2 : n 2 m z bh B 







n 2 m 2 ah 2 : (?i 2 ??i 2 K 2 2 4- m 2 gh^) : 





= 0. 



n 2 m z bh 3 : : (n 2 wi 3 K 3 2 4 m 



3^s) 



(10) 



On simplification this gives us a cubic in n 2 of the form 



ax?i 6 + bin 4 4- an 2 4- ^ = 0, 



having three real positive roots. 



Assuming that n b n 2 , and n 3 are the three roots of this 

 cubic equation, the most general solution would be 



6 — Ai cos (nit 4- €i) 4- A 2 cos (n 2 t + e 2 ) 4 A 3 cos (n B t 4 e 3 ) , 

 <£ = Bi cos (nit 4 €i) 4- B 2 cos (n 2 t 4- e 2 ) + B 3 cos (n B t 4- e 3 ), 

 and 



yfr = Ci cos («i^ 4- ei) + C2 cos (n 2 £ 4- e 2 ) 4 C 3 cos (n 3 t 4 e 3 ), 



where n l9 ?i 2 , and n 3 depend upon the constitution of the 

 system, and A, B, C, and e are arbitrary quantities and 

 enable us to satisfy any prescribed initial conditions. Thus 

 the three pendulums act and react upon one another, thereby 



2P2 



