1. Tardigrade
  2. Question
  3. Physics
  4. A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10-2 m. The relative change in the angular frequency of the pendulum is best given by :

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A simple pendulum of length $1\, m$ is oscillating with an angular frequency $10\, rad/s$. The support of the pendulum starts oscillating up and down with a small angular frequency of $1 \,rad/s$ and an amplitude of $10^{-2} \; m$. The relative change in the angular frequency of the pendulum is best given by :

JEE MainJEE Main 2019Oscillations

Solution:

Angular frequency of pendulum
$\omega = \sqrt{\frac{g_{eff }}{\ell}} $
$ \therefore \frac{\Delta\omega}{\omega} = \frac{1}{2} \frac{\Delta g_{eff }}{g_{eff }} $
$\Delta\omega = \frac{1}{2} \frac{\Delta g}{g} \times\omega $
[$\omega_s$ = angular frequency of support] $ \Delta \omega = \frac{1}{2} \times\frac{2 A \omega^{2}_{s} }{100} \times100$
$ \Delta\omega = 10^{-3} $ rad/sec.