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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2019
  • Medium
  • MCQ

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

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Answer & Solution

Correct Answer

(A) \(10^{-3} rad/s\)

Step-by-step Solution

Detailed explanation

Angular frequency of pendulum

\(\omega \propto \sqrt{\frac{g_{e f f}}{\ell}}\)

\(\therefore \quad \frac{\Delta \omega}{\omega}=\frac{1}{2} \frac{\Delta g_{e f f}}{g_{e f f}}\)

\(\Delta \omega=\frac{1}{2} \frac{\Delta g}{g} \times \omega\)

\([ \omega_{s}=\) angular frequency of support] 

\(\frac{\Delta \omega}{\omega}=\frac{1}{2} \times \frac{\Delta g}{g}\)

\(=\frac{1}{2} \times \frac{2\left(\mathrm{A} \omega_{5}^{5}\right)}{10}\)

\(\Rightarrow \frac{\Delta \omega}{\omega}=\frac{1 \times 10^{-2}}{10}=10^{-3}\)

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