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.
• C.
• D.

Answer 1

The correct option is C

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

In upward moving lift T' = $2\pi \sqrt{\frac{l}{(g+a)}}$

( a = Acceleration of lift )

$\Rightarrow \frac{T^{\prime }}{T}=\sqrt{\frac{g}{g+a}}=\sqrt{\frac{g}{(g+\frac{g}{4})}}=\sqrt{\frac{4}{5}}\Rightarrow T^{\prime }=\frac{2T}{\sqrt{5}}$