A simple pendulum of length 1 , m is oscillating with an angular … - Vidyadip

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

✓
$10^{-3} rad/s$
B
$1\,rad/s$
C
$10^{-1} rad/s$
D
$10^{-5} rad/s$

✓

Answer

Correct option: A.

$10^{-3} rad/s$

a
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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