45, 46] 



STRING OF BEADS. 



165 



with the segment of the string, which is the difference of two con- 

 secutive displacements divided by a. Accordingly the force on the 

 r ih bead is 



from which we obtain the potential energy, 



83) W = {y-^ -\- (?/ 2 2/ A ) 2 -1- (?/ 3 ?/ 2 ) 2 H h y n 2 }- 



It will be seen that the ?/'s are not normal coordinates since product 

 terms appear in W. 



Forming the differential equations of motion we obtain 



dt* 



84) 



m 



dt* a 



-( 

 (y 



ft W ^ 



d*y 8 



m ~dt^ + (y n - y n - 1 + y n - 0) = o. 



Now putting y r = A r e* t and collecting according to the J.'s we 

 obtain 



/ , 2 . 2S\ A S A 



(ml? -\ ) A* - A = 0, 



\ a ] l a 2 



86) s , / . 2/s\ , 5 



A. 4- m A 2 H Mo A* = 0, 



a x \ a / ^ a 



; 



o 



or, dividing through by J - and putting 



l2 r 

 =, 



- =0, 



0- -- = 0, 



^ + + ...==0, 



The determinantal equation for A is 



C,-l, 0, 0, 0,... 

 -1, G , -1, 0, 0,... 



88) J> (1 = o , - 1, C , - 1, 0. , . . . = 0; rows. 



, , - 1, C ,-!,... 



