iii the Use of Bi '.filar Suspensions. 437 



the " boundary conditions.-" This involves a knowledge of 

 all the subsequent periods of reversal. In the case of our 

 problem the solution is simple. If the force causing the 

 deflexion is kept the same in magnitude, no matter which 

 way it acts, and if the reversals are made every T minutes 

 for n times, we get, from equation (2), that 



y = T) + ae- kt } . (5) 



where a is the constant of integration, and solving equation 

 (4) we get 



y=be-**, (6) 



where b is the constant of integration. Hence 



x=D-{-ae~ M + be~ M 



=0D + <*-**, 



whore c = a + b. 



Let Xi be the distance from the origin at the time T of the 

 first reversal; let x 2 be the distance from the alternative 

 origin at the time of the second reversal, and so on up to x n , 

 which is the distance from the origin at the time of the nth 

 reversal. 



If the deflexion has been reversed n — 1 times, we have 



(7) 



and ^ = 0(1 -«-**), 



wherefore, by repeated transformations, we get 



^_ 1 = D(l-<r* T ) [l- e - kT -e- 2kT - + (-1)" «-<»-*>"]. 



Since k and T are both positive the series is convergent, 

 and we have 



l_(_£-AT)n-l 





x— —x n -i, when 



* = 0, 



therefore 

 Hence 



x=D(l-e- kt )- 



tV n -l 



e~ k \ 



where x n -i has still to be determined. 

 Now from equation (7) we see that 







# w =D(l-*-* T ) 



*Vn- 



i e~ kT 



Xn _ l = J)(l-e- k ^ 



1 + e 



■AT 



and since we are considering the case nT large compared with 

 T, we may take 



_ n l- g~* T 



Phil. Mag. S. 6. Vol. 22. No. 129. Sept. 1911. 2 G 



