170 
Proceedings of the Royal Society of Edinburgh. [Sess. 
Enter Table II for a = 150' with argument 0 = 60", and we 
find that the time up to this point is 0 S 131. Hence the time 
given to free oscillation in this quarter period is 0 S, 500 — 0 S T31 
= 0 S *369. 
IV. What is the phase angle \[s corresponding to time 0 S, 37 from the 
lowest position of the pendulum ? 
Enter any member of Table II with £ = 0 S, 37, e.g., a = 300', 
and we get i//- = 66 0, 6. 
The theory runs thus. If l is the length of the simple pendulum 
swinging true seconds for zero arc we have 
and the equation of its motion when swinging to extreme excursion a is 
l(d6/dt) 2 = 2g (cos 0 - cos a) ; 
hence the time taken to pass from the vertical to the position 0 is 
t = 2 \l( l h) I dof sin 2 f - sin 2 | 
or writing 
6 a • . 
Sill — = Sill - Sill l b 
4) o r 
t — 
1 ft 
7 T 
but 
so that 
dij/(l - sin 2 ^ sin 2 1 fr) ^ 
n 1 ' °) f $ 
xf/ + ^ sin 2 — I sin 2 ifrdif/ + — - | sin 4 if/dif/ + , 
2 .'o 2 4 A 
j ^ sin 2 1 f/dif/ = 2 t~i s i n 2^ 
sin 4 if/dif/ =s4 r ~i s ^ n s i n 
_ 
t = 
l 2 
, , * • oa 1 2 , 3 2 . , a 
1 + — . S 11 V - + _ Sill 4 — + 
7 T 
sin 2 1 J/ 
7 r 
sin 4 1 // 
— . sur — f- „ „ 
')d 2 2 2 > 4 2 
■i • o OL •> • a CL 
i sm 77 + A sm 77 + 
7 r 
A QllT^fi 
2 5 6 blli g ^ * 
+ 
Table I gives for the complete semi-oscillation (\Js = tt) the value of the 
excess of t over 1 sec.; (ta — 1) for each semi-arc a from O' to 300', and also 
the accumulated value of this excess in a day of 86,400 secs. 
Table II gives the values of \Js and t for each value of 0, for the series 
of semi-arcs a = 10', 20', . . a 3002 
