Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerates with ${g \over 6}$ vertically upwards then the time period will be :
(Where g = acceleration due to gravity)
Solution
$T^{\prime}=2 \pi \sqrt{\frac{I}{g_{\text {eff }}}}$
<br/><br/>$T^{\prime}=2 \pi \sqrt{\frac{I}{g+\frac{g}{6}}}=2 \pi \sqrt{\frac{6 l}{7 g}}$
<br/><br/>$\Rightarrow T^{\prime}=\sqrt{\frac{6}{7}} T$
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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