$A$ simple pendulum of length $l$ is made to oscillate with an amplitude of $45^{\circ}$. The acceleration due to gravity is $g$. Let $T_0 = 2 \pi \sqrt{l / g}$. The time period of oscillation of this pendulum will be

  • A
    $T_0$ irrespective of the amplitude
  • B
    slightly less than $T_0$
  • C
    slightly more than $T_0$
  • D
    dependent on whether it swings in a plane aligned with the north-south or east-west directions

Explore More

Similar Questions

The breaking strength of the string of a simple pendulum is twice the weight of the bob. The bob is released from rest when the string is horizontal. At what angle $\theta$ with the vertical will the string break?

Difficult
View Solution

At which position in the string of a simple pendulum is the tension maximum?

In $S.H.M.$,a simple pendulum oscillates with frequency $f$. If the length of the pendulum is increased by three times its original length,then the frequency of oscillation of the pendulum will be

The bob of a simple pendulum of length $200 \ cm$ is released from a horizontal position. If $10 \%$ of its initial energy is lost due to air resistance,then the speed of the bob at the mean position is (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m \ s^{-1}$)

$A$ pendulum suspended from the ceiling of a train oscillates with a time period $2 \ s$,when the train is accelerating at $10 \ m/s^2$. What will be its time period when the train retards at $10 \ m/s^2$?

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