86 Mr. E. R. Darnley ou the Transverse Vibrations 



Case III. — One or both ends free. 

 The equation for the periods is obtained from that in 



Case I. bv replacing <j>i by — ^, or cj) n by — ^, or both. 



When there are two bays, one with a free end, otherwise 

 simply supported, the equation is (f) 1 (f) 2 =-.^ 1 or ^ 2 according 

 to which end is free. 



§ 7. The general nature of the functions cj){x) and ^r{x) 

 will be seen from the tables and graph at the end of the 

 paper. 



<j>(x) is zero when x is zero, and has poles (or infinities) 

 when x = tt or any multiple of it. (£>(#) = §# nearly when .1% 

 expressed in radians, is small. <j>(x) exceeds this value by 



7T 



less than one per cent, up to %=- x or 60°. Xear x=tt,. 



1 

 <f>{x) tends to the value - and passes through infinity 



from + to - . When x is greater than tt, $>{x) tends to 

 the value 1 — cot^, exceeding that value by less than *004 

 when x — 7T. </>( i1 ') vanishes when .£ = 3*9266 radians, or 

 224:° 56' 58". At this point it exceeds (1- cot<r) by -0008. 



Similarly, yjr(x) =(p(-\ — (p(d'), and is nearly — ^w when x 



is small. At x = tt, ^r(x) passes through infinity from — to 



3-7T 



+ , and yfr(x') = <j>(x) near x=-~- where both functions are 

 nearly unity. w 



§ 8. The case of two bays simply supported is easily 

 solved approximately from the figure. A more exact 

 solution can then be obtained from the tables. 



The period equation is <£ 1 +<£ 2 = 0, and it is convenient to 

 draw the graph of the function — cf) for values of the argu- 

 ment from 180° to 225°. The problem then reduces itself 

 to finding, by trial and error, a parallel to the axis of x 

 which cuts the graphs of <£ and — <f>m points whose abscissae 

 are in the ratio of the lengths of the bays. 



Thus, if / x =6 feet, / 2 = 4"5 feet, we see from the graph 

 that KZ 2 is about 145°, corresponding to 



KZ 1= 6xlio degrees== 903£°. 



■±'0 



From the tables we find <f> (147°) = 2'5518, 0(196°) = - 2«4853 T 

 and a closer result, say 195|°, may be arrived at by double 

 interpolation. Taking K/ 1 = 195|°. the result in radians 

 is 3-416. 



