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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2019 · 11 January · Evening Shift · Physics · Oscillations

A simple pendulum of length 1 ~m is oscillating with an angular...

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 :

Options

  1. A10−3rad/s10^{-3} \mathrm{rad} / \mathrm{s}
  2. B1rad/s1 \mathrm{rad} / \mathrm{s}
  3. C10−1rad/s10^{-1} \mathrm{rad} / \mathrm{s}
  4. D10−5rad/s10^{-5} \mathrm{rad} / \mathrm{s}

Correct answer

A. 10−3rad/s10^{-3} \mathrm{rad} / \mathrm{s}

Step-by-step solution

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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