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 :
Solution
When lift is stationary<br><br>$T = 2\pi \sqrt {{L \over g}}$<br><br>A pseudo force will act downwards when lift is moving upwards.<br><br>$\therefore$ ${g_{eff}} = g + {g \over 2} = {{3g} \over 2}$<br><br>$\therefore$ New time period<br><br>$T' = 2\pi \sqrt {{L \over {{g_{eff}}}}}$<br><br>$T' = 2\pi \sqrt {{{2L} \over {3g}}}$<br><br>$\therefore$ $T' = \sqrt {{2 \over 3}} T$
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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