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
Verified Correct Answer: A.
Physical Principles & Numerical Solution
T = 2\(\pi\) \(\sqrt\frac{L}{g}\)
1 = 2 x 3.142\(\sqrt\frac{L}{10}\)
(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
Source: Myschool Past Questions Archive (Record #27751)