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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.
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Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A toroid having 200 turns carries a current of 1 A. The average radius of the toroid
- A particle is executing a simple harmonic motion. Its maximum acceleration
- Two cars A and B moving with same speed of 45 km/hr along same direction.
- A radioactive sample S₁ having an activity of 5 μCi has twice the number of nuclei
- Light of two different frequencies whose photons have energies 1eV
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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