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?

Question

Damon · Accepted Answer

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

Nazifi · Answer

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

Shiv · Answer

Ans is right

Oscar · Answer

Thanks

Anonymous · Answer

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