Reflexion of Sound from a Perforated Wall. 229 



The evanescence of B requires that of both the squares in 

 (24), or that 



£S = -^— - — 2 + k 2 tan l\ = ik 1 cot ik 2 —k% cot fc ly (25) 



or again with elimination of S, 



z'/^tan ik 2 -f- cot z£ 2 ) = k 2 (tan A^ -f cot ki) , 

 whence 



* 1 sin2* 1 + «ife 2 sin2tifc 2 =0, .... (26) 



or in the notation of the hyperbolic sine 



k t sin 2& 1 =& 2 sinh 2k 2 (27) 



If this equation, independent of cr, a ', and cos 6/, can be 

 satisfied, it allows us to find k Y from an assumed k 2 , or con- 

 versely, and thence k by means of (9). 



The next step is to calculate S by means of one of equations 

 (25). If S, so found, > cos 0, we may choose a' /a so that 

 B shall vanish ; but if S < cos #, no ratio a' fa will serve to 

 annul the reflexion. If the incidence be perpendicular, 

 S must exceed unity. If S were negative, the reflexion 

 would be finite, whatever may be the angle of incidence and 

 the ratio <r'/o\ 



It is natural to expect an evanescence of reflexion when 

 the damping is small and the tuning such as to give good 

 resonance. In this case we may suppose k 2 and 7r — 2k * 

 to be small, and then (27) gives approximately 



^=f(l-$ S ), *=V(*.'-V)=f(l-?£). (28) 



By (25) 



kS = k 1 tanh k 2 + k 2 tan & l9 



= &! tanh k 2 + hj tan (2/- 2 2 /7r), 



= J<i+*/+...). 



8=*r 1 {l+(l+ £)*/+...}. . (29) 



Since S is large and positive, the condition for no reflexion 

 can be satisfied by making the perforated area a small 

 enough. 



For a more general discussion we may trace the curves 

 * So that wave-length is 4 times h. 



so that 



