88 THE ROYAL SOCIETY OF CANADA 



i.e., a cos a+è cos a-^aa sin a, 

 or {a-\-b) cos a-\-aa sin a. 

 The decrease in potential energy when the tilt has decreased to 6 



= mg\ (a + b) (cos a — cos d)-\-a{a sin a — 6 sin 6)]. 

 This must be equal to the gain of kinetic energy of the cylinder 



Yj I\ ( ) =mg{ {a-\-b) (cos a — cos d)-\-a{a sin a — 6 sin d) ). 



Differentiating both sides with respect to the time and substituting 

 we get, by neglecting d'~ and higher powers of 6, and also d ( — j 



\ dt y } 



d'^d 

 {I-\-mb-) — - +mg (a-b) 0-^0, 

 dt'- 



whence as before 



Y ^2t- ^ / + w6- 



mg ia — b) 



Application of the Formula 

 For a solid cylinder 



T = 2^ / T2" ^ T • (2) 



V . gia-b) 



If a rod of rectangular cross-section is used as a vibrator instead 

 of a cylinder and b is its half depth 



/ /" 4 



^^ = 2x / _12 ^__ 



V g{a-b) ^^^ 



If b is small in comparison with / the equation reduces in both cases 

 to 



/ l^ / I' 



y 12 g{a-b) y Q g{A-B) 



where A and B are the diameters of the cylinders (B = thickness in 



