The period of simple pendulum is measured as T in a stationary lift. If the lift moves upwards with an acceleration of 5 g, the period will be

The same

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Increased by 3/5

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Decreased by 2/3 times

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None of the above

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Solution

The correct option is D
None of the above

$\frac{T^{'}}{T} = \sqrt{\frac{g}{g^{'} + a}} = \sqrt{\frac{g}{g + 5 g}} = \sqrt{\frac{1}{6}} \Rightarrow T^{'} = \frac{T}{\sqrt{6}}$

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The period of simple pendulum is measured as T in a stationary lift. If the lift moves upwards with an acceleration of 5 g, the period will be

The correct option is D None of the above T′T=√gg′+a=√gg+5g=√16⇒T′=T√6

The correct option is D
None of the above

$\frac{T^{'}}{T} = \sqrt{\frac{g}{g^{'} + a}} = \sqrt{\frac{g}{g + 5 g}} = \sqrt{\frac{1}{6}} \Rightarrow T^{'} = \frac{T}{\sqrt{6}}$