Question

A person standing in a stationary lift measures the periodic time of a simple pendulum inside the lift to be equal to T. Now, if the lift moves along the vertically upward direction with an acceleration of 3g, then the periodic time of the pendulum will now be

• A. T
• B. T/2
• C. 2T
• D. T/3

Answer 1

The correct option is B T/2

When a particle is getting accelerated upwards (against gravity), the net acceleration is g+a = g+3g =4g
The time period of oscillation of a pendulum is $T=2\pi \sqrt{L/g}$, when the lift was stationary.
The time period of oscillation when the lift moves upwards with an acceleration 3g upwards will be $T^{\prime }=2\pi \sqrt{L/4g}=T/2$
The correct answer is (b)