Question

A lift is ascending with an acceleration equal to $g / 3$. What will be the time period of a simple pendulum suspended from ceiling if its time period in stationary lift is $T$ ?

• A. $T/4$
• B. $ \left( \frac{\sqrt{3}}{4} \right)T $
• C. $ \left( \frac{\sqrt{3}}{2} \right)T $
• D. $ \frac{T}{2} $

Answer 1

Answer: (C)

The periodic time of pendulum in SHM is
$T=2 \pi \sqrt{\frac{l}{g}}$ ...(i)
where, $l$ is length of pendulum and $g$ is gravity. Effective gravity when lift accelerates upwards is
$g'=g+\frac{g}{3}=\frac{4}{3} g$
$\therefore T'=2 \pi \sqrt{\frac{l}{(4 / 3) g}}$ ... (ii)
Dividing Eq.(i) by (ii), we get
$\frac{T}{T'} =\sqrt{\frac{\frac{4}{3} g}{g}}=\sqrt{\frac{4}{3}}$
$\therefore T'=\sqrt{\frac{3}{4}}$
$T=\frac{\sqrt{3}}{2} T$