28 Prof. Barton and Miss Browning on 



x = A 1 sin [m^t + ot 1 )-\-A 2 sin (m 2 t -f « 2 ) 



-M 3 sin (??? 3 £ + a 3 ), 



( y = ^) 1 (J_ 1 ) sin (»? 1 f + a 1 )+ </> 2 (-4 2 ) sin (??? 2 £ + * 2 ) 



+ (/) 3 (^ 3 ) sin (m 3 £ + a 3 ), 



^ = ^(^4!) sin (m^ + ai) + ^ 2 (J. 2 ) sin (m 2 t + a 2 ) 



+ ^(A 3 ) sin (m 3 t + u 3 ). J 



,(28) 



Substituting the values of the m's from (22) in (25) and 

 (26), we have the following values of the functions </> and i|r, 



Hence, by (29) and (28) we may write the general solu- 

 tion in the final form, 



x — A 2 sin (m 2 t -f c/. 2 ) + ^L 3 sin (m 3 t + a 3 ) 



?/ = jB 1 sin (??i^ + «i) — o- 2 sin (m 2 £ + a 2 ) + JL 3 sin (m 3 £ + a 3 ) 



-4 

 z——B x sin (jji^ + «j) ~ sin (??i 2 £ + « 2 ) 



-VA Z sin (w 3 ^ + a 3)- 



(30) 



It is seen that this set of equations involves only six 

 arbitrary constants, viz. : — A 2 , A 3 , B x , «i, a 2 , and « 3 . The 

 values of these six constants may be obtained for any given 

 initial state, since that involves the specification of the 

 three displacements of, y, and z, and the three corresponding 

 velocities. 



(d) Initial Conditions. 



Case /.—Take first (for the sake of its theoretical 

 simplicity) the initial state in which two pendulum bobs, 

 say Q and R, are held still in their equilibrium positions 

 while the other bob P is held displaced, and then all three 

 are let go. This may be represented by the following- 

 equations 



x = x , ?/ = 0, 2=0,1 



f °"- %-o, *=», £',o. ■ • ' (m 



dt dt ' dt > 



