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A simple pendulum, suspended from the ceiling of a lift, has a period of oscillation $T$, when the lift is at rest. If the lift starts moving upwards with an acceleration $a=3 \mathrm{~g}$, then the new period will be
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The correct answer is:
$\frac{T}{2}$
Given that, $T$ be the time period of simple pendulum when lift is at rest. Then,
$T=2 \pi \sqrt{\frac{l}{g}}$ ...(i)
When the lift is moving upwards with an acceleration $a=3 g$, the new time period will be
$T^{\prime}=2 \pi \sqrt{\frac{l}{g+a}}$
$=2 \pi \sqrt{\frac{l}{4 g}}$ ...(ii)
Dividing Eq. (ii) by Eq. (i), we get
$\frac{T^{\prime}}{T}=\frac{2 \pi \sqrt{\frac{l}{g}}}{2 \pi \sqrt{\frac{l}{4 g}}}=\frac{2}{1}$
$\Rightarrow \quad T^{\prime}=\frac{T}{2}$
$T=2 \pi \sqrt{\frac{l}{g}}$ ...(i)
When the lift is moving upwards with an acceleration $a=3 g$, the new time period will be
$T^{\prime}=2 \pi \sqrt{\frac{l}{g+a}}$
$=2 \pi \sqrt{\frac{l}{4 g}}$ ...(ii)
Dividing Eq. (ii) by Eq. (i), we get
$\frac{T^{\prime}}{T}=\frac{2 \pi \sqrt{\frac{l}{g}}}{2 \pi \sqrt{\frac{l}{4 g}}}=\frac{2}{1}$
$\Rightarrow \quad T^{\prime}=\frac{T}{2}$
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