The time period of a simple pendulum inside a stationary lift is ' \(T\) '. When the lift starts accelerating upwards with an acceleration \(\left(\frac{\mathrm{g}}{3}\right)\), the time period of the pendulum will be

Time period of a simple pendulum is \(\mathrm{T}=2 \pi \sqrt{\left(\frac{l}{\mathrm{a}}\right)}\)
In stationary lift, the value of acceleration is \(\mathrm{a}=\mathrm{g}\)
\(
\mathrm{T}=2 \pi \sqrt{\left(\frac{l}{\mathrm{~g}}\right)}
\)
When the lift is accelerating in an upward direction, there is a pseudo force acting in a downward direction.
\(
\therefore \quad \mathrm{ma}=\mathrm{mg}+\frac{\mathrm{mg}}{3}
\)
\(\therefore \quad \mathrm{a}=\mathrm{g}+\frac{\mathrm{g}}{3}\)
\(\therefore \quad a=\frac{4 g}{3}\)
\(\therefore \quad\) Period for a pendulum in accelerating lift is
\(
\begin{aligned}
\mathrm{T}^{\prime} & =2 \pi \sqrt{\frac{l}{\mathrm{a}}}=2 \pi \sqrt{\frac{3 l}{4 \mathrm{~g}}} \\
\therefore \quad \mathrm{T}^{\prime} & =\frac{\sqrt{3}}{2}\left(2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\right) \\
\therefore \quad \mathrm{T}^{\prime} & =\frac{\sqrt{3}}{2} \mathrm{~T}
\end{aligned}
\)