A simple pendulum is placed at a place where its distance from earth's surface is equal to radiu

English

Gujarati

Hindi

13.Oscillations

hard

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $4 \mathrm{~m}$, then the time period of small oscillations will be ____ $s$. $\left[\right.$ take $\left.\mathrm{g}=\pi^2 \mathrm{~ms}^{-2}\right]$

A

$5$

B

$6$

C

$7$

D

$8$

(JEE MAIN-2024)

Solution

Acceleration due to gravity $\mathrm{g}^{\prime}=\frac{\mathrm{g}}{4}$

$T=2 \pi \sqrt{\frac{4 \ell}{g}}$

$T=2 \pi \sqrt{\frac{4 \times 4}{g}}$

$T=2 \pi \frac{4}{\pi}=8 s$

Standard 11

Physics

Share

Similar Questions

Two simple pendulums of length $0.5\, m$ and $2.0\, m$ respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed …. oscillations.

medium

(AIPMT-1998)

A simple pendulum hangs from the ceiling of a car. If the car accelerates with a uniform acceleration, the frequency of the simple pendulum will

easy

A simple pendulum executing $S.H.M.$ is falling freely along with the support. Then

easy

Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of $2^o$ to the right with the vertical, the other pendulum makes an angle of $1^o$ to the left of the vertical. What is the phase difference between the pendulums ?

hard

How many amplitudes of $SHO$ covers the distance in the half period ?

easy