Question:
Time period of a simple pendulum is $\mathrm{T}$ inside a lift when the lift is stationary. If the lift moves upwards with an acceleration $\mathrm{g} / 2$, the time period of pendulum will be :
Correct Option: , 4
Solution:
When lift is stationary
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$
When lift is moving upwards
$\Rightarrow$ Pseudo force acts downwards
$\Rightarrow \mathrm{g}_{\mathrm{eff}}=\mathrm{g}+\frac{\mathrm{g}}{2}=\frac{3 \mathrm{~g}}{2}$
$\Rightarrow$ New time period
$T^{\prime}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}_{\text {eff }}}}=2 \pi \sqrt{\frac{2 \mathrm{~L}}{3 \mathrm{~g}}}$
