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  4. A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is T. With what acceleration should the lift be accelerated upwards in order to reduce its period to T/2 ? (g is acceleration due to gravity).

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Q. A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is $T$. With what acceleration should the lift be accelerated upwards in order to reduce its period to $T/2$ ? (g is acceleration due to gravity).

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Solution:

Time period of simple pendulum is given by
$T=2 \pi \sqrt{\frac{l}{g}} \,\,\,\,\,\,\,\,...(i)$
When the lift is moving up with an acceleration $a$, then time period becomes
$T'=2 \pi \sqrt{\frac{l}{g+a}} $
Here,$T'=\frac{T}{2}$
$\Rightarrow \,\,\,\frac{T}{2}=2 \pi \sqrt{\frac{l}{g+a}}\,\,\,\,\,\,\,\,...(ii)$
Dividing Eq. (ii) by Eq. (i), we get
$a=3\, g$