The time period of a simple pendulum inside a stationary lift is $T$. When the lift starts accelerating upwards with an acceleration of $\frac{g}{3}$,the time period of the pendulum will be

  • A
    $\frac{\sqrt{5}}{2} T$
  • B
    $\frac{\sqrt{3}}{2} T$
  • C
    $\frac{2 T}{\sqrt{3}}$
  • D
    $\frac{2 T}{\sqrt{5}}$

Explore More

Similar Questions

In order to find time,the astronaut orbiting in an earth satellite should use

The time period of a simple pendulum is $T$. If its length is increased by $21\%$,what is the percentage increase in its time period?

Difficult
View Solution

$A$ pendulum clock is running slow. In order to correct it, we should:

$A$ simple pendulum of length $L$ is suspended from the roof of a trolley. The trolley moves in a horizontal direction with an acceleration $a$. What would be the period of oscillation of the simple pendulum? [$g$ is the acceleration due to gravity]

The bob of a simple pendulum having length $l$ is displaced from the mean position to an angular position $\theta$ with respect to the vertical. If it is released,then the velocity of the bob at the lowest position is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo