The time period of a simple pendulum in a lift descending with constant acceleration $g$ is

  • A
    $T = 2\pi \sqrt {\frac{l}{g}} $
  • B
    $T = 2\pi \sqrt {\frac{l}{{2g}}} $
  • C
    Zero
  • D
    Infinite

Explore More

Similar Questions

$A$ cylindrical piece of cork of density $\rho$, base area $A$, and height $h$ floats in a liquid of density $\rho_{l}$. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period $T=2 \pi \sqrt{\frac{h \rho}{\rho_{l} g}}$. (Ignore damping due to viscosity of the liquid).

The time period of a simple pendulum inside a stationary lift is $\sqrt{3} \ s$. When the lift moves upwards with an acceleration $g/3$,the new time period will be ($g=$ acceleration due to gravity).

$A$ simple pendulum with a metallic bob has a time period $T$. The bob is now immersed in a non-viscous liquid and oscillated. If the density of the liquid is $1/4$ that of the metal,the time period of the same pendulum will be ...........

The height of a swing changes during its motion from $0.1\, m$ to $2.5\, m$. The minimum velocity of a boy who swings in this swing is ..... $m/s$.

Savitha, a $XI$ standard student, while conducting an experiment to determine the effective length of a simple pendulum $L$, notes down the data of time taken to complete $30$ oscillations as $60 \text{ s}$ and hence calculates the length of the simple pendulum as: (Take $\pi^2 = 9.8$, and $g = 9.8 \text{ m/s}^2$) (in $\text{ m}$)

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