The length of a displaced pendulum ball which passes its ...
The length of a displaced pendulum ball which passes its lowest point twice every seconds is [g = 10ms-2]
0.25 m
0.45 m
0.58 m
1.00 m
Correct answer is A
T = 2π √Lg
1 = 2 x 3.142√L10
(since the pendulum pass its lowest point twice in a second, hence the frequency is 1 and period T= 1F)
L= 0.25m