218 Mr. W. Sutherland on a 



Let \ be the mass of unit length, then the gravitational 

 potential energy of the whole rod is 



The gravitational potential energy due to the vertical dis- 

 placement of M is M<7(Z' — X), where X is the x coordinate of 

 the point originally at distance V from the clamped end. 



~ 2 V2B,/ 15 - 

 Hence the gravitational potential energy of the mass M is 



and the total potential energy of the loaded bent rod is 



6 B* ±2 V2B/H 6 + 15 M P 



the two signs corresponding to the two cases of free end up 

 or down. 



For the kinetic energy we have T=-J f par {dy]dt) % da:x neg- 

 lecting, as is usual, the effect of the slight rotatory motion of 

 the bar, 



Hence for n, the number of vibrations per second, we get 



— ©' { HDf +&•} '-*? p± ®"' (? + fi**> 



But this is only an approximation ; in the rigorous theory 

 when M=0 the fraction 33/70 is replaced by 6/1-8751* (Ray- 

 leigh, vol. i. p. 234). Let us so replace it, and we have an 

 equation that goes very near to the truth in all cases, namely 



(A) 47rV| 2M( r j -r jTgTjpj ^ 3 = 6B ±t(^+ J5F> 



When M = we have the case of the bar unloaded but af- 

 fected by its own weight, then 



. „ „ 6ml 3 nTy , 9g ml 2 



47rV F875P = 6B ±TlT 



