In a stationary lift, time period of a simple pendulum is ' $T$ '.' The lift starts accelerating downwards…
In a stationary lift, time period of a simple pendulum is ' $T$ '.' The lift starts accelerating downwards with acceleration $\left(\frac{\mathrm{g}}{4}\right)$, then the time period of the pendulum will be
$\frac{\sqrt{3}}{2} \mathrm{~T}$
$\frac{2}{\sqrt{3}} \mathrm{~T}$
$\frac{3}{4} \mathrm{~T}$
$\frac{4}{3} \mathrm{~T}$
Solution
Time period of pendulum: $T=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$
$\therefore \quad$ When lift is accelerated downward with acceleration $\frac{\mathrm{g}}{4}$,
$\begin{aligned}
& g=g-\frac{g}{4} \\
& g=\frac{3 g}{4}
\end{aligned}$
$\therefore \quad$ New Time period will be
$\begin{aligned}
& \mathrm{T}_1=2 \pi \sqrt{\frac{4 \mathrm{~L}}{3 \mathrm{~g}}} \\
& \mathrm{~T}_1=2 \pi \frac{2}{\sqrt{3}} \sqrt{\frac{\mathrm{L}}{\mathrm{g}}} \\
& \mathrm{T}_1=\frac{2}{\sqrt{3}} \mathrm{~T}
\end{aligned}$