556 Dr. W. F. G. Swann on the Pulse 



We may point out that if the energy per unit cross- 

 section of the wave be calculated it will contain a 2 in the 

 denominator. 



Problem (4). Solution o/(19) and (20) when /jl is not zero. 



The solution of (19) subject to ?/ = ?/= z/ =0 when t = is 

 still given by (22) leading to 



. Y e . 



V = 7T— t Sill 27Tft£. 



The solution o£ (20) subject to x = x=x =0 when £ = is 

 _1/Y 6\ 2 f^ 2n 2 (4n 2 -3/* 2 ) . * cos kirnt 



%i sin 47m£ ~1 .„ . 



~7r(4n 2 -^ 2 ) 2 J ' ' * ^ } 



which on differentiation leads after a little reduction to 



dx 1 /Y e\ J 4?2 2 — 3//, 2 9 2 2 sin 2 7r/xi ??£ sin kirnt 

 dt = Ic W"/ \ (4ft 2 -//, 2 ) 2 ~ n t 7r 2 pH 2 '" ir^-fj 2 ) 



27T 2 (4?2 2 — yLt 2 ) 2 J 



When //, is absolutely zero this of course reverts to the 

 expression found on p. 549. When /ju is small compared 

 with n but not necessarily zero, we have 



x =- s -( -^ ) J % sin 2 it ixt — Z nt sin lirnt + - sin 2 27rn« 



.... (39) 



After a few oscillations of the wave the last term becomes 

 unimportant compared with the others. The first term must 

 not be neglected however, since it is multiplied by the large 



2 



factor — In fact this term is all important until a time t 

 has elapsed such that ~nt= -„, or until a number of oscil- 

 lations p of the light-wave have taken place such that 

 p= - -„ (since p is of the order nt). For example, if 

 n='60fi it. would not be until GOO oscillations had taken 



