JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 5)
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 :
$$\sqrt 3 T$$
$$\sqrt {{2 \over 3}} T$$
$${T \over {\sqrt 3 }}$$
$$\sqrt {{3 \over 2}} T$$
Explanation
When lift is stationary
$$T = 2\pi \sqrt {{L \over g}} $$
A pseudo force will act downwards when lift is moving upwards.
$$ \therefore $$ $${g_{eff}} = g + {g \over 2} = {{3g} \over 2}$$
$$ \therefore $$ New time period
$$T' = 2\pi \sqrt {{L \over {{g_{eff}}}}} $$
$$T' = 2\pi \sqrt {{{2L} \over {3g}}} $$
$$ \therefore $$ $$T' = \sqrt {{2 \over 3}} T$$
$$T = 2\pi \sqrt {{L \over g}} $$
A pseudo force will act downwards when lift is moving upwards.
$$ \therefore $$ $${g_{eff}} = g + {g \over 2} = {{3g} \over 2}$$
$$ \therefore $$ New time period
$$T' = 2\pi \sqrt {{L \over {{g_{eff}}}}} $$
$$T' = 2\pi \sqrt {{{2L} \over {3g}}} $$
$$ \therefore $$ $$T' = \sqrt {{2 \over 3}} T$$
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