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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
- A force of 10 newton acts on a body of mass 20 kg for 10 seconds. Change in its momentum
- An iron rod of volume 10⁻⁴ m³ and relative permeability 1000 is placed inside a long
- When a current of (2.5±0.5) A flows through a wire, it develops a potential difference
- A particle of mass m is driven by a machine that delivers a constant power k watts
- One mole of an ideal diatomic gas undergoes a transition from A to B along a path
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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