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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
Related Questions: - The acceleration of an electron in an electric field of magnitude 50 V/cm,
- Two identical flutes produce fundamental notes of frequency 300 Hz at 27° C
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Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The acceleration of an electron in an electric field of magnitude 50 V/cm,
- Two identical flutes produce fundamental notes of frequency 300 Hz at 27° C
- A parallel beam of fast moving electrons is incident normally on a narrow slit.
- A thin circular ring of mass M and radius R rotates about an axis through its centre
- What is the amount of energy released by deuterium and tritium fusion?
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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