Physics Circle home Student Login

Home›Chapters›Oscillations PYQs›Simple Pendulum›Question 76

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

Simple Pendulum question from 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:

Show answer and solution

Answer: (b)

  1. Support moves as y=Asin⁡Ωty = A\sin\Omega t, so its maximum acceleration is Ω2A=12×10−2=10−2 m/s2\Omega^2A = 1^2 \times 10^{-2} = 10^{-2}\text{ m/s}^2.
  2. geff=g±Ω2Ag_{eff} = g \pm \Omega^2A and ω∝geff⇒Δωω=12Δgg\omega \propto \sqrt{g_{eff}} \Rightarrow \frac{\Delta\omega}{\omega} = \frac{1}{2}\frac{\Delta g}{g}.
  3. Δω=12×10−210×10=5×10−3 rad/s\Delta\omega = \frac{1}{2} \times \frac{10^{-2}}{10} \times 10 = 5\times10^{-3}\text{ rad/s}, of order 10−310^{-3} rad/s.
  4. So the answer is (b) 10−310^{-3} rad/s.
Report a mistake

More questions on Simple Pendulum