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A particle executes simple harmonic oscillations with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
Options
(a) T/8
(b) T/12
(c) T/2
(d) T/4
Correct Answer:
T/12
Explanation:
Displacement from the mean position y = a sin(2π / T) t
According to the problem y = a/2
a/2 = a sin (2π / T) t
⇒ π / 6 = (2π / T) t ⇒ t = T / 12
This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.
Related Questions: - If a mass of 20g having charge 3.0 mC moving with velocity 20ms⁻¹ enters a region
- Out of the following which pair of quantities does not have same dimensions?
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Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If a mass of 20g having charge 3.0 mC moving with velocity 20ms⁻¹ enters a region
- Out of the following which pair of quantities does not have same dimensions?
- A astronomical telescope has objective and eye-piece of focal lengths
- A juggler maintains four balls in motion,making each of them to rise
- If the velocity of charged particle has both perpendicular and parallel components
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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