76 Prof. E. H. Barton on Spherical 
we must choose a form of solution corresponding to stationary 
waves. Thus, let rs he everywhere proportional to cos kat. 
Then d 2 (rs)/dt-= —k?a?(rs), and equation (17) transforms into 
^M + F(W)=0 (27) 
ar 
The general solution of this may be written 
rs = (A cos kr + B sin kr) cos kat, . . . (28) 
where A and B are arbitrary constants. These are to be 
determined for each case by the position and nature of the 
ends of the pipe. There are accordingly a number of 
separate cases to consider. 
Open Ends. — First let both ends of the conical pipe be 
open. Then obviously the condition at the end is approxi- 
mately s=0. For, at the ideal open end there can be neither 
condensation nor rarefaction. Let the coordinates of the 
ends of the pipe be r Y and r 2 , measured from the vertex of 
the cone if completed. Then, we have from (28) for the 
terminal conditions, 
A cos h\ + B sin kr x = and Acos£r 2 + B sin kr 2 —0 ; 
whence, by the elimination of A/B, we obtain 
sin k(r 2 — ?'i) = or k(r ? — r 1 ) = nir. 
This may be written 
r 1 -r^nX J 2 or N„= ^J ■ • <W> 
where n is an integer ; for since s is proportional to cos kat, 
&=27rN/a = 27r/\, N being the frequency and \ the wave- 
length of the motion. Thus, for a conical pipe with open 
ends, the pitch of the prime tone and the form of the series 
of other natural tones are like those for an open-ended 
parallel pipe. This might have been anticipated from the 
similarity of the differential equations and the conditions for 
the open ends in each case, There is, however, this slight 
difference that r 2 — ?\ is the slant length of the conical pipe 
and not its axial length. As to the segments into which the 
pipe is divided when emitting its higher natural tones, it 
follows from equation (29) that the antinodes are equidistant. 
When dealing with the next case, it will be seen that this 
simplicity does not extend to the nodes. 
Closed Ends. — The condition at closed ends is obviously 
n = 0; consequently du/dt = there also. But by equation (6) 
du/dt = a 2 ds/d7\ if u denotes the velocity along?'. We may 
thus write as our condition for a closed end ds/d?°=0. 
