Oscillations Ques 61
A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by $T=2 \pi \sqrt{(l / \sqrt{g^2 + a^2})}$, where $g$ is equal to
[1991]s
$g$
(b) $g-a$
(c) $g+a$
(d) $\sqrt{(g^{2}+a^{2})}$
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Answer:
Correct Answer: 61.(d)
Solution:
- (d) The effective value of acceleration due to gravity is $\sqrt{g^{2}+a^{2}}$
If a simple pendulum is suspended from a trolley that is moving with constant speed $v$ around a circle
$ \text{ circle of radius } r T=2 \pi \sqrt{\frac{l}{g+(\frac{v^{2}}{r})}} $
BETA
