THEORY OF THE HOWLING TELEPHONE 43 



where 4> is the velocity potential, t the time, a the velocity of sound in 

 the air, x the distance along the tube, p the pressure and p the density 

 of the air. 



For the case in which we are interested, a sinusoidal oscillation is 

 sustained, so that the special solution 



^ = eJo>i (a cos'*''^" +5 sin'^') (37) 



is suitable for our problem. Quantities A and B are arbitrary con- 

 stants which are determined by the end conditions. Substituting 

 this value of 4) in equation (35), there results 



dp^- pjwe^"^' (a cos ^^ +5 sin ^^j (38) 



It remains then to determine the arbitrary constants A and B. 



At the receiver end of the tube, the displacement, ti? of the air 

 diaphragm across the end of the tube is related to the displacement 

 y of the receiver diaphragm. This relationship is established by the 

 following consideration. If q is the cross-section of the tube, the 

 increase in volume in the air chamber is given by 



dVR = {^Rq-yQR). (39) 



Assuming that the air chamber is so small that the pressure change 

 at any instant is the same throughout, and that it takes place adia- 

 batically, we have : 



dpR=-y~^^dVR (40) 



Combining equations (33), (39), and (40), we obtain: 



qSRU = QRZi- (^—SR^QR^ypR (41) 



Similarly, 



qSnr ={-~^^T+ Qr) dpT (42) 



Then the following conditions must be fulfilled at the two ends of a 

 tube of length /. 



At x = 0, dp = dpR and ^ = ^|7 ; 



T 7 7 1 dd) dtr 



atx = l, dp = dpT and j-^ = ^^-^. 



