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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
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle performing uniform circular motion has angular momentum
- Two bulbs when connected in parallel to a source take 60W each, the power consumed,
- By the succesive disintegration of ₉₂U²³⁸, the final product obtained is ₈₂Pb²⁰⁶, then
- If the highest modulating frequency of the wave is 5 kHz, the number
- The value of coefficient of volume expansion of glycerin is 5 x 10⁻⁴ k⁻¹.
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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