A simple pendulum of length 1 m is oscillating with an angular… | JEE Main 2019 (11 Jan, Shift 2)

Q76JEE Main 2019 · 11 JAN · SHIFT 2Difficult

Simple Pendulum

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: (b)

Support moves as $y = A\sin\Omega t$ , so its maximum acceleration is $\Omega^2A = 1^2 \times 10^{-2} = 10^{-2}\text{ m/s}^2$ .
$g_{eff} = g \pm \Omega^2A$ and $\omega \propto \sqrt{g_{eff}} \Rightarrow \frac{\Delta\omega}{\omega} = \frac{1}{2}\frac{\Delta g}{g}$ .
$\Delta\omega = \frac{1}{2} \times \frac{10^{-2}}{10} \times 10 = 5\times10^{-3}\text{ rad/s}$ , of order $10^{-3}$ rad/s.
So the answer is (b) $10^{-3}$ rad/s.

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Simple Pendulum question from JEE Main 2019 (11 Jan, Shift 2)

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