In a stationary lift,the time period of a simple pendulum is $T$. If the lift starts accelerating downwards with an acceleration of $\frac{g}{4}$,then the new time period of the pendulum will be:

  • A
    $\frac{\sqrt{3}}{2} T$
  • B
    $\frac{2}{\sqrt{3}} T$
  • C
    $\frac{3}{4} T$
  • D
    $\frac{4}{3} T$

Explore More

Similar Questions

$A$ rubber ball filled with water,having a small hole,is used as the bob of a simple pendulum. The time period of such a pendulum

$A$ simple pendulum is vibrating in an evacuated chamber. It will oscillate with

The acceleration due to gravity on the surface of the moon is $1.7 \; m s^{-2}$. What is the time period of a simple pendulum on the surface of the moon if its time period (in $s$) on the surface of the earth is $3.5 \; s$? ($g$ on the surface of the earth is $9.8 \; m s^{-2}$)

The time period of a simple pendulum is $T$. The angular amplitude is $\beta$. How much time will the bob of the pendulum take to move from the equilibrium position $O$ to position $A$,where the string makes an angle $\alpha$ with the vertical?

Difficult
View Solution

The angular amplitude of a simple pendulum is $\theta_0$. The maximum tension in its string will be

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