﻿310 
  Mr. 
  H. 
  J. 
  E. 
  Beth 
  on 
  the 
  Oscillations 
  

  

  There 
  are 
  three 
  different 
  cases 
  : 
  — 
  

  

  (1) 
  In 
  one 
  of 
  the 
  extreme 
  rectangles 
  sin^ 
  = 
  0; 
  in 
  the 
  

  

  other 
  cos 
  d>= 
  , 
  The 
  ends 
  of 
  the 
  double 
  curve 
  

  

  y 
  2V?(1-?) 
  

   are 
  double 
  points 
  of 
  the 
  envelope. 
  The 
  envelope 
  consists 
  of 
  

   two 
  segments. 
  

  

  (2) 
  In 
  both 
  extreme 
  rectangles 
  sin<£ 
  = 
  0. 
  The 
  ends 
  of 
  

   the 
  double 
  curves 
  are 
  double 
  points 
  of 
  the 
  envelope. 
  The 
  

   envelope 
  consists 
  of 
  four 
  segments. 
  

  

  (3) 
  In 
  both 
  extreme 
  rectangles 
  cos<f>= 
  , 
  ' 
  

  

  In 
  the 
  system 
  there 
  exist 
  no 
  double 
  curves 
  ; 
  in 
  the 
  envelope 
  

   no 
  double 
  points 
  appear. 
  The 
  envelope 
  consists 
  of 
  two 
  

   closed 
  curves, 
  not 
  intersecting 
  each 
  other. 
  One 
  of 
  these 
  

   curves 
  is 
  the 
  outward 
  envelope, 
  the 
  other 
  the 
  inward 
  

   envelope. 
  

  

  § 
  44. 
  From 
  the 
  form 
  of 
  /(f) 
  for 
  the 
  case 
  7 
  = 
  1, 
  given 
  on 
  

   p. 
  294, 
  we 
  deduce 
  

  

  /W 
  n 
  _ 
  4Kp+l)f 
  2 
  + 
  4<?(p-f-l)f+9 
  2 
  + 
  4r+* 
  2 
  

   7 
  ^ 
  + 
  \- 
  4(^ 
  + 
  <#+r 
  + 
  iZ 
  2 
  ) 
  

  

  Now 
  from 
  the 
  relation 
  (34), 
  

  

  J=/'(?)±V/'XD 
  + 
  i, 
  

  

  we 
  come 
  to 
  the 
  supposition 
  that 
  simple 
  cases 
  will 
  occur 
  when 
  

  

  is 
  a 
  complete 
  square. 
  The 
  condition 
  runs 
  : 
  

  

  Ap(p 
  + 
  l)(q 
  2 
  + 
  4r 
  + 
  1 
  2 
  ) 
  = 
  q\p 
  + 
  1) 
  2 
  , 
  

  

  which 
  condition 
  degenerates 
  into 
  

  

  p 
  + 
  l 
  = 
  

  

  and 
  p{q 
  2 
  + 
  ±r+l 
  2 
  ) 
  = 
  q 
  2 
  (p 
  + 
  l), 
  

  

  which 
  may 
  be 
  written 
  in 
  this 
  form 
  : 
  

  

  p(4.r 
  + 
  l 
  2 
  ) 
  = 
  q 
  2 
  ; 
  

  

  this 
  expresses 
  at 
  the 
  same 
  time 
  the 
  condition 
  that 
  the 
  de- 
  

   nominator 
  of 
  the 
  fraction, 
  written 
  down 
  at 
  the 
  beginning 
  of 
  

   this 
  section, 
  is 
  a 
  complete 
  square. 
  In 
  this 
  latter 
  case 
  /(f) 
  is 
  

   a 
  linear 
  function 
  of 
  f. 
  Therefore 
  we 
  pass 
  to 
  the 
  discussion 
  

   of 
  these 
  two 
  cases 
  : 
  

  

  p=-i. 
  

  

  /(f) 
  is 
  a 
  linear 
  function 
  of 
  f. 
  

  

  