Triple Pendulums ivith Mutual Interaction, 



29 



Then, substituting (31) in (30) and its differentiation with 

 respect to t, we have 



A 2 sin « 2 + ^3 sin ot 3 = x , 



A 



Bi sin ^ ^- 2 sin a 2 -\- A z sin a 3 = 0, 



A 



— B 1 sin aj_ s^sin a 2 + A z sin a 3 = 0. 



m 2 A 2 cos a 2 + m 3 JL 3 cos a 3 = 0, 

 •mucosal ^ J. 2 cosa 2 + w 3^-3 cos «;!~0? 



—m 1 B 1 cos «! ^ ^1 2 C0S a 2 + ^3^ 3 c° s a 3 = 0. 



The six equations of (32) are satisfied by 



. (32) 



-£i = 0, ^2=3^0, A= j, | 



*i.= 0, 



7T 



2 "J 



k . . (33) 



That is to say, for these initial conditions only two of the 

 three possible vibration components occur, the quickest 

 (involving m l5 whose square is g/c) being absent. 



Thus, putting (33) in (30), we may write as the solution 

 for the present initial state, 



3<r = 2%q cos m 2 t + x cos m s t, 

 3?/ = — x cos m 2 t + x cos m^t 

 ?>z =- — x Q cos m 2 t -f x cos m%t. 



(34) 



<7as£ 77. — Consider now the initial state in which the 

 bob P is pulled aside and held displaced a distance x while 

 the bobs Q and R are hanging freely at rest in equilibrium 

 displaced, say y and z respectively. Then it is seen at once 

 that y Q = z = iv (where w is the displacement of the points 

 C , where the suspensions of Q and R produced vertically 

 upwards meet the lines C'E', C"E"). 



Comparing the present initial state with the general case 

 treated in equations (3)-(12), Ave see that the equations 

 which held for the points B, B', C, C" still hold, for. owing 

 to the negligible masses of the bridges, the former dynamic 



