physics
posted by la roux on .
a simple pendulum has a period of 4.2sec. when the pendulum is shortened by 1 m the period is 3.7s. what is the acceleration of free fall g and the original length of the pendulum?

T = 2 pi sqrt(L/g)
4.2/(2 pi) = sqrt (L/g)
3.7/(2 pi) = sqrt ( (L1)/g )
L/g = .447
(L1)/g = .347 = .4471/g
1/g = 0.1
g = 10
L = 4.47