A pendulum has time period T . If it is taken on to another planet having acceleration due to gravit

English

Gujarati

Hindi

13.Oscillations

easy

A pendulum has time period $T$. If it is taken on to another planet having acceleration due to gravity half and mass $ 9 $ times that of the earth then its time period on the other planet will be

A

$\sqrt T $

B

$T$

C

${T^{1/3}}$

D

$\sqrt 2 T$

Solution

(d) $T = 2\pi \sqrt {\frac{l}{g}}$

$\Rightarrow T \propto \frac{1}{{\sqrt g }}$

$ \Rightarrow \frac{{{T_P}}}{{{T_e}}} = \sqrt {\frac{{{g_e}}}{{{g_P}}}} = \sqrt {\frac{2}{1}}$

$\Rightarrow T' = \sqrt 2 T$

Standard 11

Physics

Share

Similar Questions

In an experiment to determine the period of a simple pendulum of length $1\, m$, it is attached to different spherical bobs of radii $r_1$ and $r_2$ . The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be $5\times10^{-4}\, s$, the difference in radii, $\left| {{r_1} – {r_2}} \right|$ is best given by …. $cm$

hard

(JEE MAIN-2017)

What is linear harmonic oscillator ? And what is non-linear oscillator ?

medium

A simple pendulum is made of a body which is a hollow sphere containing mercury suspended by means of a wire. If a little mercury is drained off, the period of pendulum will

medium

Two simple pendulums of length $1\, m$ and $4\, m$ respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed number of oscillations equal to

hard

(JEE MAIN-2013)

The amplitude of an oscillating simple pendulum is $10\,cm$ and its period is $4\, sec$. Its speed after $1\, sec$ after it passes its equilibrium position, is … $m/s$

easy