The length of a displaced pendulum ball which passes its lowest point twice every seconds is [g = 10ms-2]
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Correct Answer: Option A
Explanation:
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
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