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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
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The correct answer is:
decreases
The time period of a pendulum is given by
$T=\frac{1}{2 \pi} \sqrt{\frac{l}{g}}$
When a lift containing a simple pendulum is accelerated upwards with acceleration $a$, then its time period becomes
$T^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{l}{(g+a)}}$
Clearly, $\quad T^{\prime} \angle T$
Hence, its time period decreases.
$T=\frac{1}{2 \pi} \sqrt{\frac{l}{g}}$
When a lift containing a simple pendulum is accelerated upwards with acceleration $a$, then its time period becomes
$T^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{l}{(g+a)}}$
Clearly, $\quad T^{\prime} \angle T$
Hence, its time period decreases.
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