F. A. P. Barnard on the Pendulum. 19] 
known. This velocity may be represented by k. Then rf(9) 
2 B | 
=j2: For the value of / (¢), we consider the velocity to be that 
hich a body would acquire in falling vertically through the 
height the pendulum has descended from the commencement of 
its motion, which (if « represent the limit of vibration) is 
I(cos ~—cos«). Then v?=2g1(cos y—cos «), and 
2gl 
rf @)=—5 (cos y—cos a). 
The equation then becomes 
d? : 2gl a 
= —7 (sin 9 (cos p—cos a)-+ == 008 *). 
If we suppose the impulse to continue to the distance @ on the 
other side of zero, and integrate between the limits « and —@,. 
We shall have (employing the simpler symbols for the sake of 
convenience) 
dg? 9 : ; 
qn r [ cos ~—cose+ (m—r) (sin«—sin g)-+r cos « (a—g) ] . 
Replacing the trigonometrical functions by their values in 
terms of the ares, and rejecting minute terms of higher orders 
than the second, we shall obtain, after reduction, 
dg? g he f 
438° 7 a2 +2mae—2mgp—H? |. 
Hence, 
rs ud 
di= |! :. 
gV at fima—Img—gr 
L - P+m 
tw fs ae et +¢. 
And, between the limits « and —B, 
_ [lla . —B+m 
=} (j—aresin pat 
™ : int SO 
= like manner, if ¢’ express the time during which the im- 
Pulse weights oppose the motion, 
sash 1 -> : =" 
ie ‘= (; nice eo ‘ 
ting == 24-7, the total time of vibration, 
B—m me et) 
i 
t= | i —arc sin . 
ol (e+ aresin arc sin Lm 
