﻿Plane 
  Waves 
  of 
  Sound. 
  449 
  

  

  7. 
  We 
  have 
  now 
  to 
  determine 
  when 
  the 
  motion 
  becomes 
  

   discontinuous. 
  For 
  this 
  purpose 
  we 
  must 
  return 
  to 
  the 
  

   exact 
  equations 
  at 
  the 
  end 
  of 
  paragraph 
  5. 
  These 
  equations 
  

   assume 
  that 
  7=3, 
  but 
  it 
  is 
  evident 
  that 
  they 
  give 
  a 
  very 
  fair 
  

   representation 
  of 
  what 
  actually 
  happens 
  in 
  a 
  plane 
  wave 
  of 
  

   sound, 
  — 
  assuming, 
  of 
  course, 
  that 
  the 
  equation 
  (1) 
  takes 
  

   account 
  of 
  all 
  the 
  facts. 
  The 
  form 
  of 
  the 
  second 
  order 
  

   approximation, 
  for 
  example, 
  in 
  the 
  case 
  when 
  7 
  = 
  3 
  is 
  

   exactly 
  the 
  same 
  as 
  when 
  7 
  has 
  its 
  general 
  value. 
  

  

  The 
  motion 
  becomes 
  discontinuous 
  when 
  one 
  of 
  the 
  three 
  

  

  partial 
  derivatives 
  of 
  the 
  second 
  order, 
  ^-^, 
  ^ 
  ' 
  . 
  ^— 
  | 
  

  

  becomes 
  infinite. 
  It 
  is 
  evident, 
  however, 
  that 
  they 
  all 
  

   become 
  infinite 
  together, 
  so 
  that 
  we 
  need 
  consider 
  only 
  

   one 
  of 
  them. 
  

   We 
  have 
  

  

  B^_ 
  (op 
  d*__3p 
  C^\ 
  //df? 
  ^i_Bf 
  ot\ 
  

   O 
  .i' 
  £ 
  ~ 
  \dr 
  o 
  o 
  WAdr 
  o$ 
  30 
  3t> 
  

  

  where 
  p 
  = 
  ^ 
  j 
  so 
  that 
  ^-^ 
  becomes 
  infinite 
  when 
  

   x 
  OX 
  QX~ 
  

  

  o^oO 
  o0 
  o 
  t~ 
  ' 
  

   which 
  after 
  reduction 
  leads 
  to 
  

   (l 
  + 
  Y 
  2 
  ') 
  2 
  [(2+Y 
  1 
  ' 
  + 
  Y 
  2 
  ')-(l 
  + 
  Y 
  1 
  ')(l 
  + 
  Y 
  2 
  ')(Z/-Z 
  2 
  ')] 
  

  

  = 
  (1 
  + 
  Y 
  1 
  ')(^-t 
  + 
  Y 
  1 
  -Y 
  2 
  )[Y 
  2 
  "-Z 
  2 
  "(1 
  + 
  Y 
  2 
  ') 
  2 
  ] 
  ) 
  (13) 
  

   or 
  

  

  (l 
  + 
  Y 
  1 
  '7[(i'+Y/ 
  + 
  Y 
  2 
  ')-(l+Y 
  1 
  ')(l 
  + 
  Y 
  2 
  ')(Z 
  1 
  '-Z 
  2 
  ')] 
  

   + 
  (1 
  + 
  Y 
  2 
  ')^-t+Y 
  1 
  -Y 
  2 
  )[Y 
  1 
  " 
  + 
  Z 
  1 
  "(1 
  + 
  Y 
  1 
  ') 
  2 
  ]=0. 
  (14) 
  

  

  To 
  find 
  the 
  time 
  corresponding 
  to 
  these 
  equations, 
  we 
  

   have 
  

  

  at 
  = 
  (<?_ 
  T 
  + 
  Y 
  1 
  -Y 
  2 
  )/[ 
  I 
  ^ 
  Y 
  -,+ 
  I 
  -i 
  Y 
  - 
  7 
  -Z 
  1 
  ' 
  + 
  Z 
  2 
  'J 
  ; 
  

  

  i.e., 
  

  

  «< 
  = 
  (1 
  + 
  Y 
  2 
  ') 
  3 
  /[Y 
  2 
  "-Z 
  2 
  "(1 
  + 
  Y 
  2 
  ') 
  2 
  ], 
  m 
  . 
  (15) 
  

  

  or 
  

  

  a< 
  = 
  -(H-Y 
  1 
  ')7[Y 
  1 
  " 
  + 
  Z 
  1 
  "(l 
  + 
  Y/) 
  2 
  ]- 
  • 
  (16) 
  

  

  The 
  time 
  required 
  is 
  the 
  smallest 
  positive 
  value 
  of 
  t 
  

  

  