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
$g^{\prime }=g+\frac{g}{3}=\frac{4}{3}g$
$\therefore T^{\prime }=2\pi \sqrt{\frac{l}{(4/3)g}}$ ... (ii)
Dividing Eq.(i) by (ii), we get
$\frac{T}{T^{\prime }}=\sqrt{\frac{\frac{4}{3}g}{g}}=\sqrt{\frac{4}{3}}$
$\therefore T^{\prime }=\sqrt{\frac{3}{4}}$
$T=\frac{\sqrt{3}}{2}T$