by means of a Double Pendulum. 569 



Then, if T represents the KE.of the whole system, 

 2T = m 1 K 1 2 0' 2 + m 2 IL 2 2 cf> 2 + (m 2 h 2 2 <i> 2 + m 2 a 2 6 2 



+ 2m 2 ali 2 0<p cos (</> — 0)) ; 

 and if W = work done on the system, 



W = — m^hi cos 6 — m 2 g (a cos + li 2 cos (/>) + C. 



Since there are no extraneous forces acting on the system, 

 equations of motion of the system are given by : 



d 



dt 



d0~ 





aw 



B0 



d 

 dt 



BT_ 

 'df 



?3T 

 "B0 



aw 



a*' 



Thus for small oscillations we have : 



(miKj 2 + m 2 a 2 ) + 0%/^ + m 2 a)gd + m 2 ak 2 (f) = 0, 

 ■a/i s 6f+^A 2 ^ + (K 8 2 + /i 2 s )</i = 0. 



(1) 

 (2) 



By analogy with the electrical problem, we can write for 

 the coefficient of coupling between the two circuits, 



.2 



,a 2 7v 



To solve these equations, assume : 



= A 1 cos (nt + a) and <£ = B x cos (ft£ + a), 

 .*. = — n 2 Ai cos (nt + a) 

 and (£ = — n 2 B 1 cos (nt + a). 



