Question

­­A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. If the lift accelerates upwards with an acceleration $g/4$, then the period of the pendulum will be

• A. T
• B. $\frac{T}{4}$
• C. $\frac{2T}{\sqrt{5}}$
• D. $2T\sqrt{5}$

Answer 1

Answer: (C)

$\frac{2T}{\sqrt{5}}$

Answer 2

In stationary lift $T = 2\pi\sqrt{\frac{l}{g}}$

In upward moving lift

$T^{'} = 2\pi\sqrt{\frac{l}{(g + a)}}$

($a =$Acceleration of lift)

$\Rightarrow$ $\frac{T^{'}}{T} = \sqrt{\frac{g}{g + a}} = \sqrt{\frac{g}{\left( g + \frac{g}{4} \right)}} = \sqrt{\frac{4}{5}}$ $\Rightarrow T^{'} = \frac{2T}{\sqrt{5}}$