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A simple pendulum of length $1 \mathrm{~m}$ is oscillating with an angular frequency $10 \mathrm{rad} / \mathrm{s}$. The support of the pendulum starts oscillating up and down with a small angular frequency of $1 \mathrm{rad} / \mathrm{s}$ and an amplitude of $10^{-2} \mathrm{~m}$. The relative change in the angular frequency of the pendulum is best given by :
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The correct answer is:
$10^{-3} \mathrm{rad} / \mathrm{s}$
Angular frequency of pendulum $\omega=\sqrt{\frac{g}{\ell}}$
$\therefore$ relative change in angular frequency
$\frac{\Delta \omega}{\omega}=\frac{1}{2} \frac{\Delta \mathrm{g}}{\mathrm{g}}[$ as length remains constant $]$
$\Delta g=2 \mathrm{~A} \omega_{s}^{2}\left[\omega_{s}=\right.$ angular frequency of support and, $\mathrm{A}=$
amplitude]
$\frac{\Delta \omega}{\omega}=\frac{1}{2} \times \frac{2 \mathrm{~A} \omega_{\mathrm{S}}^{2}}{\mathrm{~g}}$
$\Delta \omega=\frac{1}{2} \times \frac{2 \times 1^{2} \times 10^{-2}}{10}=10^{-3} \mathrm{rad} / \mathrm{sec}$
$\therefore$ relative change in angular frequency
$\frac{\Delta \omega}{\omega}=\frac{1}{2} \frac{\Delta \mathrm{g}}{\mathrm{g}}[$ as length remains constant $]$
$\Delta g=2 \mathrm{~A} \omega_{s}^{2}\left[\omega_{s}=\right.$ angular frequency of support and, $\mathrm{A}=$
amplitude]
$\frac{\Delta \omega}{\omega}=\frac{1}{2} \times \frac{2 \mathrm{~A} \omega_{\mathrm{S}}^{2}}{\mathrm{~g}}$
$\Delta \omega=\frac{1}{2} \times \frac{2 \times 1^{2} \times 10^{-2}}{10}=10^{-3} \mathrm{rad} / \mathrm{sec}$
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