The length of a displaced pendulum ball which passes its lowest point twice every seconds is [g = 10ms-2]
-
A.
0.25 m -
B.
0.45 m -
C.
0.58m -
D.
1.00m
Correct Answer: Option A
Explanation
T = 2(pi) (sqrtfrac{L}{g})
1 = 2 x 3.142(sqrtfrac{L}{10})
(since the pendulum pass its lowest point twice in a second, hence the frequency is 1 and period T= (frac{1}{F}))
L= 0.25m