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A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be:
Options
(a) πa / T
(b) 3π²a / T
(c) πa√3 / T
(d) πa√3 / 2T
Correct Answer:
πa√3 / T
Explanation:
Speed v = ? √(a² – x²) , x = a / 2
v = ? √(a² – a² / 4) = ? √(3a² / 4) = 2π / T . a√3 / 2
πa√3 / T
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Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Three identical spheres, each of mass 3 kg are placed touching each other
- A uniform force of (3i+j) Nacts on a particle of mass 2 kg.Hence the particle is displaced
- In a radioactive decay process, the negatively charged emitted β- particles are
- Radius of first orbit of the electron in a hydrogen atom is 0.53 Å. So the radius
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Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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