The period of simple pendulum is measured as $T$ in a stationary lift. If the lift moves upward 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
$T = 2 π \sqrt{\frac{l}{g}} T \propto \sqrt{\frac{1}{g}}$
$\frac{T^{'}}{T} = \sqrt{\frac{g}{g + a}} = \sqrt{\frac{g}{g + 5 g}} = \sqrt{\frac{1}{6}}$

$\Rightarrow T^{'} = \frac{T}{\sqrt{6}}$

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

The correct option is D None of the aboveT=2π√lgT∝√1gT′T=√gg+a=√gg+5g=√16⇒T′=T√6

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

The correct option is D None of the above
$T = 2 π \sqrt{\frac{l}{g}} T \propto \sqrt{\frac{1}{g}}$
$\frac{T^{'}}{T} = \sqrt{\frac{g}{g + a}} = \sqrt{\frac{g}{g + 5 g}} = \sqrt{\frac{1}{6}}$

$\Rightarrow T^{'} = \frac{T}{\sqrt{6}}$