202 M. F. Lindemann on the Forms of the 



we have to choose ^=0, v= — 1, and to put 



L"L T' L L T* 



Calculating now 8 and 8' from (6)— -that is, from 



2a(L-a) 



5L(L -70(7-70, 



2a(L-a) 



we get for the interval (12): — 



., # 



2*. 



^ for0 = £ = L T 

 _^L(aT-2LQ + T(L-2a> 



■=6 



2a(L-a)T 

 P a 2t ^ x ^ a a 2t 

 for L-f = L = 2_ L-T 



y= — — (L— x) for 2 



#-< 



T = L 



1. 



(13) 



Again only the middle line depends on t; it remains 

 always parallel to (9) ; the point at which it cuts the X axis 

 now lies between the points x=0 and #=L, while it pre- 

 viously lay outside this interval, towards the positive side. The 

 other two segments of the string are parallel to one another, 

 and lie on different sides of the x axis ; the segment adjacent 

 to the point x=0 has its direction unchanged. 



At the time 



*-£ • < 14 > 



the string again consists of only two straight lines, given by 



b(L-2d) x for < A -<2(L-a), 

 ° za(L — a) ~ — 



2/=--(L-tf) „ 2(L~a)<^<L., 



(15) 



The 



The former of these two lines now coincides with (9). 

 string lies entirely on the negative side of the X axis. 



6. The latter holds also for the entire following time-interval 

 in which 



aT T 



JL <t< T 



