39 
x 7 
N. / 
[^r~ 
Let 0 be the neutral posi- 
tion of the body considered I 
as a point, AOB the path de- 
scribed during oscillation, let ^ 
p be the force on the body 
when at a unit distance from 
O, so that as the force varies papeffw 
uniformly with the distance, 
px will be the force at the 
distance OP=$. 
Take PN perpendicular to 
OP and make PN on some scale equal px, then N will lie 
on a straight line COD, and the area OPN will represent 
the work U done against the force in moving from O to P 
and 
TT pa x op px 2 m 
2 = Ip - 
Let W be the weight and v the velocity of the body, then 
by the conservation of energy 
2g 2g w 
where E is the energy of the system and v 0 the velocity at 
O, or substituting from equation (1) 
Wvl Wv 2 px 2 
ii = ^i + T (3) 
but when P is at A or B v = 0, put 0 A = a, then 
2 /A\ 
p a = ~Y ( 4 ) 
and equation (3) becomes 
W 
— v 2 = pia 2 - x 2 ) (5) 
Describe a circle about 0 as centre with a radius a, and 
let PN produced meet this circle in Q, and let QT the 
tangent at Q meet AB in T, then the triangle QTP will be 
similar to QOP and 
PT PQ 
TQ OQ W 
