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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
- The Boolean expression P+ P ̅ Q, where P and Q are the inputs of the logic
- During negative β-decay
- A body weighs 50 grams in air and 40 grams in water. How much would it weigh in a liquid
- A coin of mass m and radius r having moment of inertia I about the axis passes
- A cricketer catches a ball of mass 150 g in 0.1 sec moving with speed 20 m/s
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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