Medium MCQ +4 / -1 PYQ · JEE Mains 2022

The time period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $\alpha$, is given by :

  1. A $2 \pi \sqrt{\mathrm{L} /(\mathrm{g} \cos \alpha)}$ Correct answer
  2. B $2 \pi \sqrt{\mathrm{L} /(\mathrm{g} \sin \alpha)}$
  3. C $2 \pi \sqrt{\mathrm{L} / \mathrm{g}}$
  4. D $2 \pi \sqrt{\mathrm{L} /(\mathrm{g} \tan \alpha)}$

Solution

<p>$\left| {{g_{eff}}} \right| = \left| {\overline g - \overline a } \right|$</p> <p>$\Rightarrow {g_{eff}} = g\cos \theta$</p> <p>$\Rightarrow T = 2\pi \sqrt {{l \over {{g_{eff}}}}}$</p> <p>$= 2\pi = \sqrt {{L \over {g\cos \theta }}}$</p>

About this question

Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion

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