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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 $T = 2 π \sqrt{\frac{l}{g^{'}}}$ , where $g^{'}$ is equal to

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detailed solution

Correct option is D

g'=g2+a2

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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=2πlg', where g' is equal to

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Ready to Test Your Skills?

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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 $T = 2 π \sqrt{\frac{l}{g^{'}}}$ , where $g^{'}$ is equal to