Question 105: Oscillations

Question

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 :

Accepted Answer

When lift is stationary
$$T = 2\pi \sqrt {{L \over g}} $$
A pseudo force will act downwards when lift is moving upwards.
$$ 	herefore $$ $${g_{eff}} = g + {g \over 2} = {{3g} \over 2}$$
$$ 	herefore $$ New time period
$$T' = 2\pi \sqrt {{L \over {{g_{eff}}}}} $$
$$T' = 2\pi \sqrt {{{2L} \over {3g}}} $$
$$ 	herefore $$ $$T' = \sqrt {{2 \over 3}} T$$

Answer

$$\sqrt 3 T$$

Answer

$$\sqrt {{2 \over 3}} T$$

Answer

$${T \over {\sqrt 3 }}$$

Answer

$$\sqrt {{3 \over 2}} T$$

Oscillations

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 :

Choose the correct answer: