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?

Answer 1

T = 2 pi sqrt(L/g)

4.2/(2 pi) = sqrt (L/g)

3.7/(2 pi) = sqrt ( (L-1)/g )

L/g = .447
(L-1)/g = .347 = .447-1/g

1/g = 0.1

g = 10
L = 4.47

Answer 2

Asimple pendulum has a period of 4•2. When the pendulum is shortened by 1m the period is 3•7s from these measurements calculate (1) the acceleration due to gravity (2) the original length of the pendulum

Answer 3

Ans is right

Answer 4

Thanks

Answer 5

How does that the l/g divide and ans come .4.47