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13.Oscillations
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Time period of a simple pendulum is $T$ inside a lift when the lift is stationary. If the lift moves upwards with an acceleration $g / 2,$ the time period of pendulum will be
A
$\sqrt{3} T$
B
$\frac{ T }{\sqrt{3}}$
C
$\sqrt{\frac{3}{2}} T$
D
$\sqrt{\frac{2}{3}} T$
(JEE MAIN-2021)
Solution
When lift is stationary
$T =2 \pi \sqrt{\frac{ L }{ g }}$
When lift is moving upwards $\Rightarrow$ Pseudo force acts downwards
$\Rightarrow g _{ eff }= g +\frac{ g }{2}=\frac{3 g }{2}$
$\Rightarrow$ New time period
$T ^{4}=2 \pi \sqrt{\frac{ L }{ g _{ eff }}}=2 \pi \sqrt{\frac{2 L }{3 g }}$
$T ^{\prime}=\sqrt{\frac{2}{3}} T$
Standard 11
Physics
