A simple pendulum has a period T inside a lift when it is stationary. The lift is accelerated upwards with constant acceleration $^{'} a^{'}$ . The period.

Decreases

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Increases

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Remains same

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Becomes infinite

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Solution

The correct option is A Decreases
$T = \frac{1}{2 π} \sqrt{\frac{l}{g}}$
Inside the moving lift
$T = \frac{1}{2 π} \sqrt{\frac{l}{g + a}}$
hence T will decrease.

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A simple pendulum has a period T inside a lift when it is stationary. The lift is accelerated upwards with constant acceleration ′a′. The period.

The correct option is A DecreasesT=12π√lgInside the moving liftT=12π√lg+ahence T will decrease.

A simple pendulum has a period T inside a lift when it is stationary. The lift is accelerated upwards with constant acceleration $^{'} a^{'}$ . The period.

The correct option is A Decreases
$T = \frac{1}{2 π} \sqrt{\frac{l}{g}}$
Inside the moving lift
$T = \frac{1}{2 π} \sqrt{\frac{l}{g + a}}$
hence T will decrease.