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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
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A student measures the terminal potential difference (V) of a cell
- Light of wavelength λ from a point source falls on a small circular obstacle
- One solid sphere A and another hollow sphere B are of same mass and same outer radii.
- The logic behind NOR gate is that it gives
- In a capillary tube water raises by 1.2 mm. The height of water that will rise
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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