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NEET generalGeneralMedium

Question

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s² at a distance of 5 m from the mean position. The time period of oscillation is

A

2 s

B

\pi \text{ s}

C

2\pi \text{ s}

D

1 s

Step-by-Step Solution

For SHM, acceleration ∣a∣=ω2y|a| = \omega^2 y. Given ∣a∣=20 m/s2|a| = 20 \text{ m/s}^2 and y=5 my = 5 \text{ m}. So 20=ω2(5)  ⟹  ω2=4  ⟹  ω=2 rad/s20 = \omega^2(5) \implies \omega^2 = 4 \implies \omega = 2 \text{ rad/s}. The time period T=2πω=2π2=π sT = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi \text{ s}.

Exam Context & Concepts Covered

This question aligns with the NEET general syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

generalpendulumsufficientlybuildingmovingfreely

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