Wires of varying Elasticity. 



89 



free bar, of a fixed-free bar, or of a fixed-fixed bar are at once 

 deducible. They are respectively given by the equations 

 sin od = 0, cosa£ = 0, sina£ = 0. 



Suppose, now, that the bar consists of two different iso- 

 tropic elastic solids, the surface of separation being the plane 

 z = l 1} and the whole length l x -f- l 2 . Let the suffix 1 refer to 

 the first material, 2 to the second. Then, from previous 

 remarks on a similar case in the equilibrium problem, it is 

 obvious that for the first material we may take the solution 



where 



u\ — ( Aj cos a x z + B x sin u x z) cos kt, 





and for the second, 



where 



iv 2 = (A 2 cos a 2 z + B 2 sin u 2 z) cos Jet, 

 2 ~ M, 5 



for it is clear that if the bar is not to break up, the two 

 portions must have the same frequency of vibration. 



At the surface of separation, the normal to which is Oz, we 

 must have w 1 =w 2 , and E 1 = E 2 , cf. (10) ; 



.* . A x cos «i/i + B 2 sin u ± l x = A 2 cos a 2 l x + B 2 sin a 2 Z 1? 



(m l + wi)«!( —A! sin a^ + B^osa^) = (m 2 + n 2 )a 2 ( — A 2 sin a 2 ^ 



+ B 2 cos u 2 l{) ; 

 whence we get 



— ! — sm aiti sin ctd-, > 



m 2 + n 2 ct 2 zy $ 



} 



cos a^i cos a 2 /i + 



> "D f • 7 7 % + «!«, 7 • 7 ) 



+ B x -{ sm a^, cos a^ - * cos a^ sm a 2 l\ \ ; 



(_ ??7. 2 + ?l 2 « 2 J 



wi 2 + 7? 2 a 2 ' 



"l 



ai/i COS a 2 Z a > 



(43) 



B 2 = A 1 -J cos« 1 / 1 sina 2 / 1 — 



, t. f . 7 . 7 . r/ii + n x a, ) 



+ B-t -j sm ai^ sm u 2 t\ H : ! cos u x l^ cos a,^ } . 



<- >"2 + n 2 &2 J 



If there are normal forces applied uniformly over the ends, 



forcing the vibration whose frequency is ^— , we can determine 



Ai and B : by equating R to the applied forces at either end. 

 For the various kinds of free vibration we can eliminate A x : Bi 

 and get a resulting equation giving the frequency. 



(44) 



