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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
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The correct answer is:
$\frac{\sqrt{3}}{2} \mathrm{~T}$
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
$$
\begin{aligned}
\mathrm{T}^{\prime} & =2 \pi \sqrt{\frac{l}{\mathrm{a}}}=2 \pi \sqrt{\frac{3 l}{4 \mathrm{~g}}} \\
\therefore \quad \mathrm{T}^{\prime} & =\frac{\sqrt{3}}{2}\left(2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\right) \\
\therefore \quad \mathrm{T}^{\prime} & =\frac{\sqrt{3}}{2} \mathrm{~T}
\end{aligned}
$$
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
$$
\begin{aligned}
\mathrm{T}^{\prime} & =2 \pi \sqrt{\frac{l}{\mathrm{a}}}=2 \pi \sqrt{\frac{3 l}{4 \mathrm{~g}}} \\
\therefore \quad \mathrm{T}^{\prime} & =\frac{\sqrt{3}}{2}\left(2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\right) \\
\therefore \quad \mathrm{T}^{\prime} & =\frac{\sqrt{3}}{2} \mathrm{~T}
\end{aligned}
$$
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