Oscillations

JEE Main level, single correct, medium

131 of 194

Question 131 of 194

A simple pendulum of length 1 m1 \mathrm{~m} is oscillating with an angular frequency 10rad/s10 \mathrm{rad} / \mathrm{s}. The support of the pendulum starts oscillating up and down with a small angular frequency of 1rad/s1 \mathrm{rad} / \mathrm{s} and an amplitude of 10−2 m10^{-2} \mathrm{~m}. The relative change in the angular frequency of the pendulum is best given by :

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Answer: (A) 10−3rad/s10^{-3} \mathrm{rad} / \mathrm{s}

Angular frequency of pendulum ω=gℓ\omega=\sqrt{\frac{g}{\ell}}

∴\therefore relative change in angular frequency

Δωω=12Δgg[\frac{\Delta \omega}{\omega}=\frac{1}{2} \frac{\Delta \mathrm{g}}{\mathrm{g}}[ as length remains constant ]]

Δg=2 Aωs2[ωs=\Delta g=2 \mathrm{~A} \omega_{s}^{2}\left[\omega_{s}=\right. angular frequency of support and, A=\mathrm{A}=

amplitude]

Δωω=12×2 AωS2 g\frac{\Delta \omega}{\omega}=\frac{1}{2} \times \frac{2 \mathrm{~A} \omega_{\mathrm{S}}^{2}}{\mathrm{~g}}

Δω=12×2×12×10−210=10−3rad/sec\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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